Physics 21Fall, 2011 Equation Sheet speed of light in vacuo Gravitational constant Avogadro’s Number Boltzmann’s constant charge on electron free space permittivity free space permeability gravitational acceleration F2 on 1 = 3.00 × 108 m/s 6.67 × 10−11 N m2 /kg2 6.02 × 1023 mol−1 1.38 × 10−23 J/K 1.60 × 10−19 C 8.85 × 10−12 C2 /(N m2 ) 4π × 10−7 T m/A 9.807 m/s2 c G NA kB e 0 µ0 g 1 q1 q2 (r1 − r2 ) 4π0 |r1 − r2 |3 λ σ ; Eplane = r 20 σ for plate 0 capacitor A A Q = CV ; C = 0 K = d d 2 Q 2 Ucap = 12 CV = 12 C Uind = 12 LI 2 ρL V = IR R= A P = IV P = I 2R 1 2π0 Q E= = 0 A 1 dQ (r − r ) 4π0 |r − r |3 E = −∇V ∂V ∂V ˆ ∂V = − ˆi + ˆj +k ∂x ∂y ∂z V f − Vi = − f E · dl i 1 Q dQ 1 ; dV = 4π0 r 4π0 |r − r | 1 uelec = 12 0 E 2 , umag = B2 2µ0 V= Work = R = mv⊥ /(qB) A × B = F · dl E · dA = Q 0 d dt E · dl = − u du = a2 + u 2 a2 + u 2 du 1 = tan−1 a2 + u 2 a v= T /ρ u a ln a2 + u 1 2 (T =tension) v = (347.4 m/s) T /300 v = λf = ω/k ω = 2πf k = 2π/λ T = 1/f (T =period) P = 12 ρA2 ω 2 v 2 1 ωC RC time constant = RC LR time constant = L/R Q(t) for RLC circuit Q0 exp(−Rt/2L) cos ωt ω2 = = sin θ cos π 2 1 R2 − LC 4L2 ± cos θ sin E · dA u du = a2 + u 2 XR = R, XL = ωL, XC = solenoid B = µ0 nI solenoid L = µ0 N 2 A/l C = 2πr C = πd A = πr2 A = 4πr2 V = 43 πr3 π 2 circumference of circle circumference of circle area of circle surface area of sphere volume of sphere cos(a ± b) = cos a cos b ∓ sin a sin b du √ = ln u + a2 + u2 2 2 a +u π ) 2 long wire: B = ξeffective = ξ1 + ξ2 = ± cos θ B · dA d B · dl = µ0 I + µ0 0 dt √ sin(θ ± µ0 I 2πR center loop: B = µ0 I/2R dQ I= I = −neAvd dt Vs Ns Is Np = , = Vp Np Ip Ns µ χm = −1 µ0 τ =µ×B µ = IA ξi =Ci (parallel) or Ri (series): sin(a ± b) = sin a cos b ± cos a sin b B · dA = 0 F = qv × B; dF = Idl × B µ0 Idl × (r − r ) dB = 4π |r − r |3 1 1 1 = + ξeffective ξ1 ξ2 circ. orbit ˆj ˆ k Ay Az By Bz ˆi Ax Bx 6.626 × 10−34 J s 1.055 × 10−34 J s 9.11 × 10−31 kg 1.6726 × 10−27 kg 1.6749 × 10−27 kg 1.6605 × 10−27 kg 8.99 × 109 N m2 /C2 h h ¯ = h/2π me mp mn u k ξi =Ci (series) or Ri (parallel): Eline = F = qE dE (at r) = Planck’s constant Planck’s constant/(2π) electron rest mass proton rest mass neutron rest mass atomic mass unit 1/(4π0 ) a−b a+b sin a + sin b = 2 cos sin 2 2 du u = √ (a2 + u2 )3/2 a2 a2 + u 2 e du = ln u u 2π cos2 θ dθ = 0 ∂ 2D 1 ∂ 2D = 2 2 ∂x v ∂t2 1 S= (E × B) µ0 √ c = 1/ 0 µ0 c 2 S= = B0 µ0 E0 B0 Erms Brms = = 2µ0 µ0 λ = h/p 1 2 1 un+1 n+1 du 1 = ln(a + bu) a + bu b 1 du = eau a ln u du = u ln u − u 1 cE02 2 0 un du = u du 1 = −√ (a2 + u2 )3/2 a2 + u 2 au ax2 + bx + c = 0 ⇒ √ −b ± b2 − 4ac x= 2a E×B∝v ∆x∆p > ¯ h ∼ − 2π sin2 θ dθ = π 0 KE = p2 /(2M ) (plane wave) p=¯ hk (¯ h = h/2π) eiθ = cos θ + i sin θ E=¯ hω = hf (de Broglie) ¯ 2 ∂ 2ψ h ∂ψ = i¯ h 2M ∂x2 ∂t August 16, 2011 Physics 21 Fall, 2011 Solution to HW-2 21-13 Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = −3.00 nC and is at x2 = 4.50 cm. Charge q1 is at x1 = 1.00 cm. What is q1 (magnitude and sign) if the net force on q3 is zero? q3 q1 q2 x3=0 x1 x2 We can work this problem without using vectors by thinking it through. Since q2 and q3 have opposite sign, the force on q3 exerted by q2 is attractive (towards the right). If the total force on q3 is to be zero, the force exerted by q1 must be repulsive (toward the left). Thus q1 and q3 must have the same sign, and q1 must be positive. We can find the magnitude of q1 by equating the magnitude of the forces on q3 exerted by q1 and q2 : 1 |q1 q3 | 1 |q2 q3 | = 2 4π0 x1 4π0 x22 Cancelling like terms on both sides of the equation, we find 2 2 x1 1.0 cm |q2 | = (3 nC) = 0.148 nC. |q1 | = x2 4.5 cm We already concluded that q1 was positive. A more general way to solve this problem is to use the vector expressions for the Coulomb force. We want 0 = F1 on 3 + F2 on 3 , where 0= 1 q2 q3 (r3 − r1 ) 1 q1 q3 (r3 − r1 ) + . 4π0 |r3 − r1 |3 4π0 |r3 − r2 |3 and we can evaluate the forces using the locations of the charges. Because q1 is at the origin, r3 = 0. Also, r1 = x1ˆi and r2 = x2ˆi. Substituting for the vectors gives 1 q1 q3 (0 − x1 )ˆi q2 q3 (0 − x2 )ˆi 0= + 4π0 |0 − x1 |3 |x2 |3 q2 x2 ˆ −q3 q1 x1 + = i 4π0 |x1 |2 |x2 |2 21-15 Three point charges are located on the positive x axis of a coordinate system. Charge q1 = 1.0 nC is 2.0 cm from the origin, charge q2 = −4.0 nC is 4.0 cm from the origin and charge q3 = 6.0 nC is located at the origin. What is the net force (magnitude and direction) on charge q1 = 1.0 nC exerted by the other two charges? This problem is very similar to 21-13, and the same diagram applies. Here we need the sum F of F2 on 1 and F3 on 1 , which is q1 q2 (r1 − r2 ) q1 q3 (r1 − r3 ) 1 + F= , 4π0 |r1 − r2 |3 |r1 − r3 |3 where, as before, r3 = 0, r1 = x1ˆi, and r2 = x2ˆi. Then 1 q2 (x1 − x2 )ˆi q3 (x1 − x3 )ˆi F= q1 + 4π0 |x1 − x2 |3 |x1 − x3 |3 q2 (−0.02 m) q3 (0.02 m) ˆ 1 q1 + = i 4π0 (0.02 m)3 (0.02 m)3 Substituting the other numbers leads to −4 nC(−.02 m) 6 nC(.02 m) ˆ 9 F = (9 × 10 )(1 nC) + i (.02 m)3 (.02 m)3 = 2.25 × 10−4 N ˆi 21-11 In an experiment in space, one proton is held fixed and another proton is released from rest a distance d away. What is the initial acceleration of the proton after it is released? From Physics 11 you know that F = ma, or a = F/m. So just find the electrostatic force on one proton due to the other proton, and then divide by the mass. We’ll drop the vector notation and just find the magnitude: a= e2 1 4π0 mp d2 2 1.602 × 10−19 C = 9 × 10 N m /C (1.67 × 10−26 kg) d2 9 2 2 When you substitute for d, don’t forget to convert to meters. For d = 3 mm = 0.003 m, the result is 2 a = 1.54 × 104 m/s . Note that the terms in the denominators are lengths and must be positive. The quantity in brackets must be zero, so we obtain 2 3 x2 x1 4.5 1.0 q1 = − q2 = − (−3.0 nC) = 0.148 nC x1 x2 1.0 4.5 This formula agrees with the previous result. Because we were careful with the signs, the formula gives the correct answer for any combination of signs of the three charges and for the two vector components x1 and x2 . (We took x3 = 0.) August 31, 2011 21-46 Two particles having charges q1 = 0.600 nC and q2 = 5.00 nC are separated by a distance of d = 1.60 m. At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero? q2 q1 0 x d Since both charges are positive, it’s easy to keep track of the direction of the electric field. The field at x from q1 points to the right, and the one from q2 points to the left. These two fields must be equal in magnitude for their vector sum to be zero. Therefore q2 1 1 q1 = 2 4π0 x 4π0 (x − d)2 Cancelling the common factor of 1/(4π0 ), we can rewrite the above equation as (d − x)2 q2 q2 d−x = = ⇒ 2 x q1 x q1 YF 21-50 mod A point charge q1 = −4.00 nC is at the point x = 0.60 m, y = 0.80 m, and a second point charge q2 = +6.00 nC is at the point x = 0.60 m, y = 0. (a,b) Calculate the x and y components of the net electric field at the origin due to these two point charges. (c,d) Calculate the x and y components of the net electric field at the point x = 0.90 m, y = 0.40 m due to these two point charges. Use the vector expression given in class for the field E at r due to a charge Q at point r . Apply this formula to get the field at r due to Q1 ; apply it again to get the field at r due to Q2 , and then add the results (superposition). E (at r) = Remember, r = field point; r = charge point. y Q1 = -4.0 nC (0.6, 0.8) (0.9, 0.4) Q2 = +6.0 nC (0.6, 0.0) x Solving for x, we find x= d 1 + q2 /q1 Substituting the specific numbers given above leads to (a,b) Find E at origin, r = 0ˆi+0ˆj. For Q1 , r = 0.6ˆi+0.8ˆj, so r − r = −0.6ˆi − 0.8ˆj and |r − r | = 1.0 m. For Q2 , r = 0.6ˆi, so r − r = −0.6ˆi and |r − r | = 0.6 m. x = 0.412 m. Note that instead of taking the square root and solving a linear equation for x, one could also set up a quadratic equation. One must identify the correct root of the quadratic equation, but the result is the same. 1 Q (r − r ) 4π0 |r − r |3 −4 nC(−.6 ˆi − .8 ˆj)m 6 nC(−.6 ˆi)m + (1.0 m)3 (.6 m)3 3.2 nC 1 ˆ 2.4 3.6 = − i + ˆj 4π0 13 (.6)3 13 m2 4 ˆj. r = 0.5 m)3 (0.8) + ˆj(1. r = 0.8ˆj. For Q2 .6ˆi. and |r − r | = 0.2 ˆj = 4π0 m2 = −128. so r − r = 0.5 m.6 + 2. so r−r = 0.3 ˆi + 3.4 ˆj and |r − r | = 0.3 ˆi + 0.2 + 1. For Q1 . −4 nC(.4 ˆj)m 6 nC(.4 ˆj.9 ˆi+0. nC 1 −14.5 m)3 ˆi(−1.3 ˆi − .3 ˆi−0.6ˆi+0.77 ˆj N/C 1 E= 4π0 (c.4) nC 1 = 4π0 0.4 ˆj)m + (0.125 m2 nC 1 .3 ˆi + 28.d) Find E at point r = 0.3 ˆi + .5 m. 8 i + 32 ˆj = 4π m2 0 = 43.7 ˆj N/C 1 E= 4π0 .2 ˆi + 287. ˆ 4. y dQ = λds = λadθ From this result we see the E has a magnitude of Q 2π 2 ǫ0 a2 and points in the −y direction or downward. To determine q2 we calculate the the total force F = F1 on 3 + F2 on 3 on q3 . We already know that there is no net y-component. r3 = 3. The result is r1 = 0. 2 2 0 4π ǫ0 a We will look at each component of the integral over θ separately. b) Determine the magnitude of the net force F on q3 . (a) We first determine the sign of the charge q1 . the original equation becomes 1 ( πa adθ)(−a cos θˆi − a sin θˆj) dE = 4πǫ0 a3 −Q (cos θˆi + sin θˆj)dθ. which is the sum of the x components of F1 on 3 and F2 on 3 . Then the position vectors r1 and r2 of q1 and q2 are trivial. Since the force is directed in the negative x-direction we can infer that q1 must be negative and q2 must be positive. F2 on 3 . We see that Z π Z π sin θ dθ = 2. as expected by the reasoning above. we must relate dQ to dθ so that we can integrate over the angle θ spanned by the semicircle. F1 on 3 = 3 4πǫ0 |r3 − r1 | 4πǫ0 |r3 − r2 |3 We need the position vectors of each charge. cos θ dθ = 0 and 0 0 so we get E=− Q 4π 2 ǫ0 a2 (0ˆi + 2ˆj) = − Q 2π 2 ǫ0 a2 ˆj.2 cm ˆi + 2. |F| = 56. so r = 0. a) Calculate the magnitude of q2 . September 10. The arc length ds swept out by an angle dθ is adθ. and the charge point (where dQ is located) is r′ = a cos θˆi + a sin θˆj. (b) Using q2 . From the sum of the y-components we find that the charge q2 = +0. so the net force has only an xcomponent. The x component of the field is zero. dθ x We start with the equation from the equation sheet that gives the field at the field point r in terms of a charge dQ at the charge point r′ : dE(r) = 1 dQ (r − r′ ) 4πǫ0 |r − r′ |3 The field point r (where we want to know E) is at the origin.Physics 21 Fall.4 cm ˆj Knowing these position vectors we can write F1 on 3 and F2 on 3 . Along with this and Pythagorean’s theorem we can find both x and y-components of r3 . where θ is the angle above the x-axis. so the charge dQ on ds is dQ = λds = λa dθ. as one would expect from symmetry. Let q1 be at the origin. Since we want to add up all the contribution from all dQ. 2011 . where λ is the linear charge density (charge per unit length): λ= Q . For r3 .00 µC.844 µC. 2011 Solution to HW-3 21-96 Positive charge Q is uniformly distributed around a semicircle of radius a. The magnitude of the total force is the absolute value of its x component. we notice that cos θ = 4/5 = x/4 where x is the x-component of the position vector of q3 . so r − r′ = −a cos θˆi − a sin θˆj. and the net force F on q3 is entirely in the negative x-direction. and |r − r′ | = a. Since there is no y-component of the force on q3 we know that q1 and q2 must have opposite signs. 21-105 Three charges are placed as shown in the figure. Charge q3 is +4. the sum of the y-components must be 0. The charge is positive. πa With all these substitutions.00 µC. Expressions for F1 on 3 and F2 on 3 follow from the general expression for F2 on 1 on the equation sheet: 1 q3 q2 (r3 − r2 ) 1 q3 q1 (r3 − r1 ) . r2 = 5 cm ˆi. = 4π 2 ǫ0 a2 Q We can find the total field at the origin by integrating: Z Z π −Q E = dE = (cos θˆi + sin θˆj) dθ. in terms of the unknown charge q2 : F1 on 3 = −36 N ˆi − 27 N ˆj F2 on 3 = (−24 × 106 N/C) q2 ˆi + (32 × 106 N/C) q2 ˆj Since the net force on q3 is in the negative x-direction.26 N. but its sign and the value of the charge q2 are not known. We can do this by thinking about which direction the force will be in for the different combinations of signs for charges q1 and q2 . The magnitude of q1 is 2. we can now determine the net force on q3 by adding the components together. Find the electric field (magnitude and direction) at the center of curvature P . use our expression for vy0 . we must resolve the intial velocity vector into x and y components: v0 = vx0 ˆi + vy0 ˆj. you have studied how a constant force influences the motion of an object. m Since the electric field is uniform. (a) Find the maximum distance hmax that the proton descends vertically below its initial elevation.0◦ . Using vy = vy0 + at. as shown in the figure. One segment carries 2. In this exercise.20 m nonconducting wires meet at a right angle. and a is half the length of the line. 21-99 Two 1. yields hmax = 8. (a) When the proton reaches its maximum “height”. By the choice of coordinate system.21-87 A proton with the mass m is projected into a uniform electric field that points vertically upward and has magnitude E.72 × 10−2 m. which is 60. The acceleration of the proton will be in the same direction as the electric field. t2 . This leads to the same type of problem as that studied in projectile motion near the surface of the earth. recall that Newton’s Second Law tells us how force and acceleration are related through the equation F = ma. the two wires have the same length. You can ignore gravitational forces. the distance travelled in the x direction. and α = 35. the acceleration will be constant. (c) and (d) Substituting in the given values. and the proton will follow a parabolic trajectory. where y1 is the position at maximum “height”. along with the fundamental constants e = 1.0 cm from each wire. the y component of the velocity reaches 0 (see v1 in the figure). this will be a negative position. eE This expression is also the answer for d. where vx0 = v0 cos α and vy0 = −v0 sin α. Doubling this result gives us t2 = 2v0 sin α . the position when the proton comes back up to its original height.00 × 104 m/s. and α = 35. we need to find the time it takes to get there. it experiences a constant force according to the relationship F = eE. Combining these two equations allows us to calculate the acceleration: a= eE . when it enters a region of uniform electric field.00 × 104 m/s. Find (a) the magnitude and (b) the direction of the electric field these wires produce at point P . what is (c) the magnitude and (d) the direction of the net force that these wires exert on it? y direction y = y0 + vy 0 t + 12 at2 vy = vy 0 + at 2 + 2a∆y vy2 = vy0 Before we go any further. the same magnitude of charge.673 × 10−27 kg (see the equation sheet). . and the other carries −2. Thus. (a) In class we worked out the vector electric field due to a finite line of charge at a field point located directly outward from the midpoint of the line. 2πǫ0 d d2 + a2 where d is the distance of the point from the line of charge. Because of the symmetry of the motion. The magnitude of the field is E= 1 λ a √ . a Substitute this expression into x = x0 + vx0 t. there is zero acceleration along the x direction. We get an answer of hmax = mv02 sin2 α . The only difference is that the acceleration is upward instead of downward. as shown in the figure. and set x0 = 0 to get x2 = 2mv02 sin α cos α . since d = x2 − x0 = x2 − 0. “height” is a scalar quantity. and solve for t1 . In particular. use our expression for a.0◦ . t2 is just twice the time t1 needed to reach the “peak”.26 × 10−3 m and d = 4. (b) After what horizontal distance d does the proton return to its original elevation? (c) Find the numerical value of hmax if E = 520 N/C. 2eE (b) In order to calculate x2 . Because the proton is a charged particle (charge e).00 µC distributed uniformly along it. If an electron is released at P .602 × 10−19 C and m = 1. we substitute in vy = 0 at t1 . In previous physics classes. so we can solve for y1 and take the absolute value. The initial velocity of the proton has a magnitude v0 and is directed at an angle α below the horizontal.00 µC of charge distributed uniformly along its length. (d) Find the numerical value of d if E = 520 N/C. 2 Substituting into vy2 = vy0 + 2a∆y yields 0 = v02 sin2 α + 2ay1 . v0 = 5. v0 = 5. The equations we need to use are: x direction x = x0 + vx0 t vx = vx 0 = constant However. θ = 315◦ . It will have a contribution to the electric field 1 dQ x dE = ˆi 4πǫ0 (x2 + r′2 )3/2 Before we can integrate this. so simply enter ¢ ¡ σπ R22 − R12 (b) This problem is similar to Example 21. dQ = σdA = σ2πr′ dr′ . The disk has a uniform positive surface charge density σ on its surface. (a) Given a uniform positive surface charge density σ.00 × 10−15 N. A = πR22 − πR12 . We can represent the charge distribution as a collection of concentric rings of charge dQ. so enter σ 1 1 q −q 2ǫ0 2 2 1 + (R1 /x) 1 + (R2 /x) (c) Here we select the direction of the electric field that was found in part (b). surface area. (c) The force acting on a charge q is F = qE.12 in the textbook. of radius r′ . we find E= 1 1 Q p 4πǫ0 d d2 + (L/2)2 = (9 × 109 Nm2 /C2 ) = 35355N/C. where A is the .60 × 10−19 C)(5. We will now assume an infinitesimally thin ring. while the field E+ due to the positively charged wire is straight down. called an annulus. ¸R2 · Z σx R2 r′ dr′ ˆi σx − √ 1 E = ˆi = 2ǫ0 R1 (x2 + r′2 )3/2 2ǫ0 x2 + r′2 R1 1 σ 1 q −q = ˆi 2ǫ0 2 2 1 + (R1 /x) 1 + (R2 /x) Note that the integral used here is on the table of integrals on the equation sheet. Thus. the force will be directed opposite (180◦ ) from the electric field. (c) Find the direction of the electric field E. Recalling that λ = Q/L and a = L/2. x is the distance along the axis from the ring to the point at which we are finding the field. the positive x -direction. with its center at the origin. and 135◦ counterclockwise from the +y axis. 2 × 10−6 C 1 p 0. 21-104 A thin disk with a circular hole at its center. Counterclockwise from the +y axis.00 × 104 N/C = E− + (b) Since the field vectors are perpendicular and of equal magnitude. Consider points above the annulus in the figure. find the magnitude of the electric field E. the net electric field will be 45◦ from either vector. so at P the magnitude of the force on the electron is Fnet = eEnet = (1. Mastering Physics is expecting a symbolic expression. we integrate this expression from r′ = R1 to r′ = R2 . From lecture. For a single ring. and a is the radius of the ring.and are the same distance from the point P . Our target is the electric field along a symmetry axis of a continuous charge distribution. (b) The annulus lies in the yz -plane. we can adapt the formula above to the charge and spatial orientation of each wire by putting in the correct direction for each field.00 × 104 N/C) = 8. so all we have to do is add the contributions of the rings. Ering = ˆi Qx 1 4πǫ0 (x2 + a2 )3/2 where Q is the total charge on the ring. We will approximate the area of the ring as circumference × thickness (this approximation works because the ring is infinitesimally thin). Consider points above the annulus in the figure. We can use superposition to find the net electric field Enet Enet = E− + E+ = −35355 N/C ˆi − 35355 N/C ˆj q 2 + E 2 = 5. For an arbitrary point on the x -axis (the axis of the annulus). We now have r′ dr′ σx dE = ˆi 2ǫ0 (x2 + r′2 )3/2 To get the net electric field. (a) Determine the total electric charge on the annulus.6 m)2 The field E− due to the negatively charged wire is directed to the left.6 m)2 + (0. For an annulus. we need to come up with an expression that tells us how much charge is on our infinitesimally thin ring. and thickness dr′ . we know the field of a single ring on its axis of symmetry. (d) Since the electron has a negative charge. with charge dQ on it. has inner radius R1 and outer radius R2 . Mastering Physics is only asking for the magnitude of the electric field here.6m (0. So. the total electric charge on any surface is σA. 00 × 10−9 C = −678 NC−1 m2 . so Q = q1 + q2 : q1 + q2 4. Φ= λl Qenclosed = = 614. 0 (a) Because the electric field lines are perpendicular to the line of charge.465 m that has an infinite line of positive charge running along its axis.00 nC is on the y-axis at y = 1. (b) Surface Sb encloses q2 but not q1 .500 m.00 m. Because increasing the radius does not change how much charge is enclosed.00 nC is located on the x-axis at x = 2. 700 Nm2 /C 0 0 (b) We already found in (a) that the flux is independent of r. Φ= where the left hand side is the total electric flux ΦE through a surface S. they will also be perpendicular to the surface of the curved or “barrel” part of the cylinder.50 m? q2 Sa q1 Sb Sc The problem asks for total electric flux through a surface. The total electric flux is zero (in any units).Physics 21 Fall. so Q = q2 : ΦE = q2 −6. Then the electric flux can just be calculated by: Φ = EAbarrel = λ 2πrl = 375. the new flux must be calculated because now more charge on the line is enclosed.85 × 10−12 = −226 NC−1 m2 . = 0 8. We can also use Gauss’ law to do the calculation and obtain the same expression. the field strength will also be constant along the barrel part of the cylinder. Gauss’s Law helps here.00 × 10−9 = 0 8. (b) 1. 000 Nm2 /C 0 0 September 12. (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to r = 0. Because the line of charge is concentric with the cylinder. 2011 . whose radius is 0. and parallel to the ends of the cylinder. The Equation Sheet gives Gauss’s Law in the form: Q E · dA = . (c) Surface Sc encloses both charges.00 m.50 m. Φ = 375. (c) 2. encloses no charge. Gauss’ law also tells us that the flux through the cylinder only depends on the charge enclosed. 2011 Solution to HW-4 22-7 The electric field due to an infinite line of charge is perpendicular to the line and has magnitude E = λ/2π0 r.760 m? 22-10 A point charge q1 = 4. 700 Nm2 /C (c) If the length of the cylinder is increased.200 m and length l = 0.515 m? c) What is the flux through the cylinder if its length is increased to l = 0.15µC/ m. and Q is the charge enclosed by S. The charge per unit length on the line is λ = 7. the answer for part (b) should be the same as the answer for part (a).00 × 10−9 − 6. Consider an imaginary cylinder with a radius of r = 0. and a second point charge q2 = −6. (a) Surface Sa . the flux is independent of the radius of the cylinder. ΦE = λl Qenc = = 375.500 m.85 × 10−12 C2 Nm− 2 Note the units: NC−1 is electric field and m2 is area. What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) 0. 700 Nm2 /C 2π0 r Note that the r’s cancel. Hence Qcond = 4.3 × 10−5 N/C.0948 m)3 − (. like that shown below.00-cm section of the line? The equation for the electric field of an infinite wire is E= 1 λ .11×10−6 C+(. which we can calculate as ρV .000736 C/m3 )(. By symmetry.50 cm inside an insulating spherical charged solid.4 × 840 = 1. the minus sign we obtained for E above tells us that E and dA must be antiparallel.36 × 10−4 C/m3 . which consists of the point charge qpoint and the charge Qinsul on the insulator. E points into the surface. insulated from the conductor.48 × 10−2 −3. 4 4 3 4 3 3 πr − πrcav = π(r3 − rcav ) 3 3 3 4 = π (. carries a total charge of Qcond = +4.20 nC. so we can evaluate the surface integral in terms of the unknown E. Qencl E · dA = .74 × 10−10 C 22-26 A conductor with an inner cavity. that is.8541 × 10−12 × 0.8 nC − 6. 3 V = Thus the total charge enclosed is given by Qencl = qpoint + Qinsul = qpoint + ρVencl = − 2. At every point on S the vector dA and E are perpendicular to S and are therefore parallel (or antiparallel – we’ll provisionally assume the former).0650 m)3 = . 2π0 r 22-24 A point charge of −2.3 × 10−7 C m)2 (8.86 × 10−8 C m If we muliply λ (in C/m) by the length (in m) we get charge: λ × 0.02 = 3. Qencl E · dA = E dA = 4πr2 E = . How much charge is contatined in a 2.8 nC − qinner = 4.20 nC = −1. and thus the total electric charge Qencl enclosed by G must be zero. 0 S S All that remains is to determine the charge enclosed by the Gaussian surface. (a) How much charge is on the inner surface of the conductor? (b) How much charge is on the outer surface of the conductor? (a) The Gaussian surface G shown includes the inner surface of the conductor. the electric field has the same magnitude E everywhere on the surface. The electric field must be zero inside the conductor. 0 S where we choose the Gaussian surface S to be a sphere of radius r = 9. Calculate (a) the magnitude and (b) the direction of the electric field inside the solid at a distance r = 9. making G E · dA = 0.8 nC = qinner + qouter qouter = 4. so Qencl = 0 = qinner + q ⇒ qinner = −q = 6. The charges contained within the surface are the charge in the cavity q and the charge on the conductor’s inner surface qinner . . The charge density in the solid is ρ = 7. is q = −6.3 × 10−7 C. (b) Since the magnitude E must be positive.) The volume V of the insulator inside S is the volume of S less the volume of the cavity. An additional charge within the cavity.48 cm centered at the point charge.00242 m3 .00242 m3 ) = − 3. (b) The total charge on the conductor Qcond is the sum of the charge on the inner surface qinner and the charge on the outer surface qouter .20 nC. (Note that the charge on an insulator can be evenly distributed throughout its volume with density ρ.80 nC.22-20 The electric field 0.400m from a very long uniform line of charge is 840 N/C.11 µC is located in the center of a spherical cavity of radius rcav = 6. and λ is charge per unit length.48 cm from the center of the cavity. where r is the perpendicular distance to the wire. Solving for λ gives λ = 2π0 rE = 2π × 8.854 × 10−12 C2 /(Nm2 )) = − 3.4 nC (a) To find the electric field we can use Gauss’ Law. Now we can use the relation between E and Qencl : E= = Qencl 4πr2 0 4π(9. insulated from the conductor.41x dy dz = −4. (a) How much charge is on the inner surface of the conductor? (b) How much charge is on the outer surface of the conductor? qouter qinner qtot = +5.29z dx dy = −3. S3 .118 N m2 /C Surface S3 is similar to S1 . qtot = qinner + qouter =⇒ qouter = qtot − qinner qouter = 5.41L3 = 0.80 nC − 7. S4 . Because dA is always perpendicular to the surface. For surface S4 . qinner = +7. The so we need to take the negative of the k flux is that component integrated over the surface.Physics 21 Fall. 2011 Solution to HW-5 22-4 A cube has sides of length L = 0. We can prove this by placing a Gaussian surface within the conductor that encompasses the entire cavity and point charge. so the electric flux Φ1 = 0. The induced charge will have the same magnitude as the center charge. (b) To find the total electric charge inside the cube. For surface S1 . It is placed with one corner at the origin as shown in the figure.54 · 10−13 C.330m. Surface S6 is similar.10 nC = −1.) For each of the six surfaces.29zL2 0 But evaluating the electric field at z = 0 results in Φ4 = 0. so take only the k component of E.158 N m2 /C. so the flux Φ3 = 0. which tells us that the total flux through this cube is equal to the charge enclosed. the normal direction is in the −z direction. 2011 . ˆ component.80 nC Gaussian Surface qcenter -7. Φ5 = − Z 0 L Z L 4.118 + 0 + 0 − 0. The integral over the surface is Z L 0 Z L 3. there is no electric field component in the normal direction ˆj. like that shown below. the normal direction is in the +x direction. we only need the component of the field that is normal to the surface of interest. divided by ǫ0 .80 nC.10 nC (a) The insulated point in the center of the cavity will induce an opposite charge on the inner surface of the conductor. S5 . For surface S2 .10 nC. The electric field is not uniform but is given by For surface S5 . it will work out that the component normal to the surface is constant. Φ5 = −4. so we integrate the ˆi component over the surface.10 nC.41 N/(C m)] x ˆi + [3. 0 To evaluate the electric field at the surface we set z = L. (a) Find the electric flux through each of the six cube faces S1 .29L3 = 0. The charge within the cavity. gives the flux: ˆ E = [−4. and therefor Qencl = qcenter + qinner = 0. S2 .29 N/(C m] z k. carries a total charge of qtot = +5. there is no component of the electric field normal to the surface (in the y direction ˆj).30 nC L 3. but the surface is located at x = 0 so the flux evaluates to Φ6 = 0. September 15. and then the flux is Φ2 = 3.158 + 0)ǫ0 = −3.29zL2 . the component of the electric field normal ˆ to the surface is in the +z direction. Inside the conductor E = 0. is qcenter = −7.29z dx dy = 3. (The dot product of dA with the parallel component of the field will be zero. so the surface integral will just be the constant value of the normal component times the surface area L2 .41xL2 0 Evaluating this at x = L. S6 . (b) Find the total electric charge inside the cube. R (a) The electric flux is defined as Φ = E · dA. we can apply Gauss’ Law. Φ1 + Φ2 + Φ3 + Φ4 + Φ5 + Φ6 = Qencl ǫ0 Solving for Qencl and substituting the numbers gives Qencl = (0 + 0. Φ4 = − Z 0 L Z 22-26 A conductor with an inner cavity. but we must find how it is divided between the inner and the outer surface. (b) The total charge qtot on the conductor is a constant and does not change. 2. and then we integrate over the length of the charged rod (from x′ = −a to x′ = 0). If we label the three charges 1. When the two spheres are ri = 0. with total charge Q. Etot has the same value at any other value of rf : Etot = 12 mvf2 + 1 q1 q2 . For part (b). The key steps are x+a a a ln = ln(1 + ) ≈ x px px ¶ µ a a2 + y 2 + a y2 + a a ≈ . r = y ˆj and r − r′ = −x′ ˆi + y ˆj. what does your result reduce to as y becomes much larger than a? We will express the potential dV at points P and R from charge dQ at the charge point r′ = x′ ˆi.689841 J.80 µC and mass 1. carrying a net charge of q1 = −2. we know that when q2 is closest to q1 . we see 2 1 2 mvf + 1 q1 q2 = 0. and as it approaches particle 1. The result is vf = 15. We can evaluate the total energy Etot by substituting the given initial values ri and vi : Etot = 21 mvi2 + 1 Q dx′ 4πǫ0 a (x − x′ ) dV = (1) Another expression for the argument of the log follows from p y a2 + y 2 + a p = y a2 + y 2 − a (c. the KE reaches its minimum value of zero. r = x i and r − r = (x − x′ ) ˆi. Take the potential to be zero at infinity. Particle 2 must therefore start with positive kinetic energy KE.90 µC. (b) Find the potential at the point R. Hence For this problem we will use the conservation of energy. what does your result reduce to as x becomes much larger than a? (d) In part b. (c) In part a. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart. we can consider the whole charge Q on the line to be a point charge at the origin.d) For x or y large compared to a. 4πǫ0 |r1 − r2 | 1 q1 q2 = 0.80 g.2 × 10−6 µC)2 1 3q 2 = (9 × 109 ) = 0. then U= 3(1.500m 23-5 A small metal sphere.800 m apart. with a net charge of q2 = −7. so we must set vf = 0 in Eq.689841 J.078 J 4πǫ0 r 0. We know KE = 21 mv 2 and U= 1 q1 q2 .500 m long. The general formula is dQ 1 . 23-79 Electric charge is distributed uniformly along a thin rod of length a. (a) Find the potential at the point P . ≈ ln ≈ ln 1 + ln y y y y .20µC point charges are placed at the corners of an equilateral triangle whose sides are 0.) The total potential energy is obtained by summing up the potential energy of interaction between each pair of charges (superposition). is projected toward q1 . You can ignore the force of gravity.23-8 Three equal 1.0 m/s. Assume that the two spheres can be treated as point charges. A second small metal sphere. Mathematically the result follows if we use the approximation ln(1 + δ) ≈ δ for δ ≪ 1. (a) What is the speed of q2 when the spheres are rf = 0. is held in a stationary position by insulating supports. dV = 4πǫ0 |r − r′ | The charge dQ can be related to the length dx′ using the linear charge density: Q dQ = λdx′ = dx′ . The potential is then 1/(4πǫ) times Q/x for c or times Q/y for d.5 m/s. and we need to solve Eq. a distance y above the right-hand end of the rod. a distance x to the right of the rod. so the Coulomb force is repulsive. (1) for vf . 4πǫ0 ri Since energy is conserved.295 m. q2 is moving toward q1 with speed vi = 22.430 m apart? (b) How close does q2 get to q1 ? For part (a) we are given rf . 4πǫ0 rf V = Z 0 −a dV = 1 Q 4πǫ0 a Z 0 −a dx′ 1 Q x+a = ln x − x′ 4πǫ0 a x (b) For point R. a ′ ˆ (a) For point P . Hence dx′ 1 Q p 4πǫ0 a (−x′ )2 + y 2 Z Q dx′ y 1 Q 0 p = ln p V= 2 4πǫ0 a −a (x′ )2 + y 2 4πǫ0 a a + y2 − a dV = where |r1 − r2 | is the distance between the charges. We first notice that the two charges are both negative. 4πǫ0 rf By equating the two expressions for Etot . The result is rf = 0. and 3 then U = U12 + U13 + U23 q1 q3 q2 q3 1 q1 q2 ( + + ) = 4πǫ0 r12 r13 r23 Since all three charges are identical and the separation of any pair is the same. KE will diminish and the potential energy U will increase by the same amount. (1) and solve for rf . To find the total potential energy of several charges. 2q 2 Using the velocities given and remembering that q = −e for an electron gives us the answer V2 − V1 = 164 V. Note that on Mastering Physics. (a) K1 + U1 2 1 2 mv1 = K2 + U2 + qV1 = 21 mv22 + qV2 q(V2 − V1 ) = − 21 m(v22 − v12 ) The final equation shows that the change in potential energy is the negative of the change in kinetic energy. Since the first solution is not between the two other point charges.11 m| = −3. A positively charged particle would slow down when the potential goes up (like a mass going uphill).50 × 106 m/s to a velocity of 8.21 m| |0 − . the particle slows down. the question asks for the value of V1 − V2 .1 nC is to be placed on the x-axis between q1 and q2 . So the total potential energy here is Utotal = U12 + U13 + U23 ¶ µ q1 q3 q2 q3 q1 q2 1 + + = 4πǫ0 |0 − . 4πǫ0 |r1 − r2 | where |r1 − r2 | is the distance between the two charges.0 cm? b) Where should q3 be placed to make the potential energy of the system equal to zero? (a) The general formula for the potential energy between any two point charges labeled 1 and 2 is U12 = 1 q1 q2 .9) (4.59x + 180. The only forms of energy involved are electrical potential energy (U ) and kinetic energy (K).) a) What is the potential energy of the system of the three charges if q3 is placed at x = +11.0 cm.1)(2. In this case. we must find the potential energy due to each pair of point charges. (b) With the same equation we found for part (a). 2011 Solution to HW-6 23-9 A point charge q1 = 4. and a second point charge q2 = −2. not V2 − V1 . we can recalculate for the new velocities that V2 − V1 = −182 V.25.21 m x (.9 nC is placed on the x-axis at x = +21. we can ignore it. so a positively charged particle would go through a positive change in potential. and that the relationship U = qV is not on your equation sheet.81 = 0 Using the quadratic formula we find x = 38. Since all the terms have exactly the same units.Physics 21 Fall.5662x2 − 26. So q3 must be placed in between the two charges at x = 8.53 × 10−7 J (b) We can say that q3 is placed at some point x between the other two charges and determine which value of x gives the system a total potential energy of zero.1) + + =0 21 x (21 − x) Multiply through by x(21 − x) to get a quadratic equation: 23-31 (a) An electron is to be accelerated from a velocity of 2.00 × 106 m/s. but again the electron does the opposite. (Take as zero the potential energy of the three charges when they are infinitely far apart. Remember that these values are in centimeters.1)(−2. Incorporating this into the expression for the total potential energy gives: ¶ µ q1 q3 q2 q3 q1 q2 1 =0 + + Utotal = 4πǫ0 .25 cm to give the system a total potential energy of zero.21 m − . Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from 8. So. but an electron behaves in the opposite way.21 m − x) We can save ourselves some trouble by eliminating several overall factors that won’t affect the solution x. 0. September 16. Think carefully about the sign.00 × 106 m/s to a halt? This problem is a conservation of energy problem.9)(2.72 or 8. we can drop the conversion factors to SI units and write distances in cm and charges in nC. the numerical answers you should give are −164 V and 182 V. as expected.11 m| |.1) (−2. The result is (4.10 nC is placed at the origin. 2011 . Note that “potential difference” means the “change in voltage” (V2 −V1 ) from the initial value V1 to the final value V2 . A third point charge q3 = 2. Dividing by the charge gives V2 − V1 = − m 2 (v − v12 ). First. Now. the potential decreases as one goes further away from the positively charged cylinder.00 cm carries a uniform linear density of 15. we can calculate the potential difference (the quantity measured by the voltmeter) by evaluating the line integral of the field between the probes of the voltmeter. Find Ex . where dA is the product of the length (circumference) 2πr and width dr of the ring of radius r. we must determine dQ in terms of known quantities. It is equal to the surface charge density times the surface area dA of the thin ring. and Ez . . B. That makes sense. 2 2 a +u Evaluating it at the limits results in i σ hp 2 x + R2 − x . µ ¶ −2πǫ0 ∆V d = ln 1 + λ r0 ¶ µ d −2πǫ0 ∆V =1+ exp λ r0 · µ ¶ ¸ −2πǫ0 ∆V − 1 = 2. This will make E and dl parallel. We will evaluate Z f E · dl Vf − Vi = − From the second equation. Ey . we used ∆V = −175 V.) This electric 1 λ where λ field is listed on the equation sheet as Eline = 2πǫ 0 r is the linear charge density and r is the distance away from the line. = 1− √ 2 2ǫ0 x + R2 Notice that as R → ∞. outside of the cylinder. Where is the electric field zero (multiple choice)? ∂V = −Ay + 2Bx ∂x ∂V Ey = − = −Ax − C ∂y ∂V Ez = − =0 ∂z Ex = − y = −2BC/A2 . Ey = 0 when x = −C/A. beginning at the surface of the cylinder. so the logarithm is positive. (a) By regarding the disk as a series of thin concentric rings. The electric field lines are directed radially outward from the surface of the cylinder. this result reduces to the field of an infinite sheet. Let’s call the radius of the cylinder r0 and call the distance away from the cylinder’s surface that we are looking for d. Thus ∆V is negative. Then (from the first equation) Ex = 0 when −Ay + 2B(−C/A) = 0 Be careful with the signs. Now it is possible to do the integration Z Z R Z R 1 2πσr dr r dr σ √ √ V = dV = = 2 2 4πǫ0 0 2ǫ x +r x2 + r2 0 0 The integral is on the equation sheet: Z p udu √ = a2 + u2 . dQ = σdA = σ2πrdr. (We can see this by applying Gauss’s law to a Gaussian surface like that used for the uniform line of charge. the electric field is the same as though all the charges were concentrated on a line along its axis. Now that we know what the electric field looks like. 23-47 In a certain region.0 nC/m. Z r=r0 +d Z f 1 λ dr Edl = − ∆V = − 2πǫ 0 r r=r0 i Z r=r0 +d λ dr λ = − =− [ln (r0 + d) − ln r0 ] 2πǫ0 r=r0 r 2πǫ0 ¶ µ d λ ln 1 + = − 2πǫ0 r0 ⇒ dV = 1 dQ √ . y. V = 2ǫ0 (b) The x component of the electric field is equal to −∂V /∂x: · ¸ ¢ 1 ∂V σ 1¡ 2 2 −2 − =− 2x − 1 x +R ∂x 2ǫ0 2 ¸ · σ x . 23-66 A disk with radius R has a uniform charge density σ. which simplifies the dot product in the integral. 4πǫ0 x2 + r2 Because electric potential is a scalar quantity. That analysis gives x and y.) (b) Find Ex = −∂V /∂x. any value of z is OK. Assume that the potential is zero at infinity. If you put one probe of a voltmeter at the surface. In fact. we solve for d and evaluate. we can modify the result for the finite ring to determine the potential dV due to a thin ring of radius r and charge dQ. how far from the surface must the other probe be placed so that the voltmeter reads 175 V? The symmetry of such a uniformly charged cylinder is the same as the symmetry of a uniform line of charge. (a) Using the hint. the electric potential is V (x. The argument of the logarithm is greater than one. and C are positive constants.74 cm d = r0 exp λ Again. calculate the electric potential V at a point on the disk’s axis a distance x from the center of the disk.23-36 A very long insulating cylinder of charge of radius 3. (Hint: Use the result that the potential at a point on the ring axis at a distance x from the center of the ring is V = Q 1 √ 4πǫ0 x2 + a2 where Q is the charge of the ring. where A. we can integrate dV without keeping track of vectors. z) = Axy −Bx2 +Cy. We have i along a path radially outward. 4 = 200 µC.50 mm.4 µF.7 × 2. so this is the potential difference across both capacitors. (a) what is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.70 is inserted between the plates of the capacitor. the potential difference across the equivalent capacitor is the same as the potential difference across each individual capacitor.63 µF.1 × 10−4 C = 610 µC. we find C3.2 µF.4 . we get µ ¶2 µ ¶2 ¶µ 2 V 300 V −12 C 1 1 = 2 8. The separation between the plates is 1. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3. so we know Q1 = Q5 = 610 µC. We can find the potential difference by applying V = Q/C to find V1 = V5 = 73 V.4 = C2 + C3.3 µF and C2 = C3 = C4 = 4.60 µF. Next we combine this with C2 .3. and d are µ ¶ ¡ ¢ 4 V −12 Q = C0 Ed = 5.85 × 10 u = 2 ǫ0 d N · m2 . C3. Finally. The charge is K times the previous value: ¡ ¢ Q = KC0 Ed = 2. Energy density is calculated using the formula u = 21 ǫ0 E 2 .4 = Q3 = Q4 = 200 µC. The equivalent capacitance for capacitors in parallel is the sum of each capacitor. Ceq C1 C2. The applied potential is Vab = 230 V. C2.00 × 104 V/m? (a) Without the dielectric.25 × 10−10 C = 225 pC (b) With the dielectric. calculating the charge and potential at each point.4 = 7. By applying V = Q/C we find V3 = V4 = 42 V. Q = C0 V and V = Ed.3.4 = 84 V. not the electric field strength between the plates. But the voltage and electric field strength are related. 2011 .4 is made up of C2 and C3. Also. and C5 . and C5 which are in series. (a) What is the equivalent capacitance of the network between points a and b? (b) Calculate the charge on each capacitor and the potential difference across each capacitor. we know the voltage on the capacitor. Again. C2.Physics 21 Fall.08 × 10−10 C = 608 pC 24-59 In the figure.00 × 104 V/m? (b) A dielectric with K = 2. we combine C1 .4 the same way to find V2. the charge for the given values of C0 . parallel-plate. In this problem.4 C3 C4 C3 + C4 Substituting.25 × 10−10 C = 6.3. the charge on the equivalent capacitor is the same as the charge on each individual capacitor.3. C2.3. C3. Capacitor 1 2 3 4 5 Charge 610 µC 400 µC 200 µC 200 µC 610 µC Potential Difference 73 V 84 V 42 V 42 V 73 V September 21.4 . Using these rules we can break down the circuit by combining each capacitor until we are left with a single equivalent capacitance.8 µF. 3 J/m . C1 = C5 = 8.3.00 mm and is charged to a potential difference of 300 V.0015 m) m = 2. E.4 is made up of C3 and C4 in series so we know Q3.3. so this charge is the charge on each of these three capacitors. and applying Q = CV gives Q2 = 400 µC and Q3. So we know V2 = V3. which are in parallel. for capacitors in parallel. (b) We know that for capacitors in series. Keeping these rules in mind. We can determine the total charge on Ceq since we know the potential across a and b. and V = Ed still holds.00 pF when there is air between the plates. we will determine the charge and potential difference across each capacitor by rebuilding circuit in the opposite way that we broke it down in part (a).00×10 (. completely filling the volume between the plates. We can determine the potential across C2. air capacitor has a plate separation of 3.4 C5 We find Ceq = 2. 2011 Solution to HW-7 24-25 A 6. Ceq is made up of C1 .4 . C3 and C4 are in series so we can find the equivalent capacitance: 1 1 C3 C4 1 = + =⇒ C3. which are in series. The capacitors are in parallel so we find: C2. 1 1 1 1 = + + . Because the electric field between the plates is uniform and directed straight across the gap between the plates. Calculate the energy density in the region between the plates.4 = . Therefore. C increases to KC0 . Qtot = Ceq Vab = 6. it is easy to show that V = Ed where d is the separation between the plates.43 × 10 −2 (a) The reciprocal of the equivalent capacitance for capacitors in series is the sum of the reciprocals of each capacitor. Doing the substitutions and evaluating for the given values.003 m = 4.4 = 2.4 = 84 V. 24-38 A parallel-plate capacitor has capacitance C0 = 5.00×10 F 3. (d) As in part (b).0 µF. with the positively charged plates connected together.04 × 10−6 F Note one gets the same answer using the form Ucap = 21 CV 2 . This time use the form of the equation that involves the voltage and the capacitance: Ucap = 21 C1 V 2 + 12 C2 V 2 + 21 C3 V 2 = 1 2 (C1 +C2 +C3 ) V 2 = 21 Ceff V 2 = 21 (2. 8.2 × 10−5 C. Q = C0 V0 . (c) When the capacitors are disconnected from the serial circuit.1 V = 1 K +2 + 3K . The effective capacitance is found using 1 1 1 1 1 1 1 = + + = + + Ceff C1 C2 C3 8. without any of the charge leaving the plates. When a dielectric is inserted between the plates.1 µF capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. but a new V is measured. Equating both expressions for Q.3 µF 4.0 µF 8. Use the form of the equation that involves the charge and the capacitance: µ ¶ 1 Q2 Q2 Q2 1 1 Ucap = 12 + 12 + 21 = 21 Q2 + + C1 C2 C3 C1 C2 C3 ¡ ¢2 −5 8. where V0 has been measured. and 4. What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one third of the space between the plates? (a) With no dielectric. ⇒ (2/3) C0 (1/3) KC0 Ceff = 23 C0 + 31 KC0 Therefore the measured voltage will be V = Q = Ceff 2 3 C0 C0 V0 = + 31 KC0 2 3 V0 3V0 = 22. With the dielectric.0 V when placed across the capacitor. They are then reconnected in parallel with each other. so Q = KC0 V .3 + 4.5 mJ. C0 V0 = KC0 V Q = Ceff V = (2. the charge does not change. The voltage V = Va − Vb is then given by ¡ ¢ 3 8. When they are reconnected in parallel.57 V V 11. After the charges equilibrate. the potential difference Va − Vb must be the same across every one of the capacitors.3 µF.04 × 10−5 F)(12 V)2 = 1. the total energy stored in all the capacitors can be found by adding up the energies on each capacitor. the charge on each one remains the same (+Q). the voltmeter reads 11.04 µF) × (40 V) = 8. using Qi = Ci V .5 V. (a) A voltmeter reads 41. (a) What is the charge on the 4. completely filling the space. We can determine the effective capacitance of the three capacitors and use this to determine the charge Q. with Ceff and V = 40 V.1 µF are connected in series across a 40 V potential difference.1 µF Ceff = 2. K= V0 41 = = 3. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors? The parallel combination is equivalent to a single effective capacitance Ceff = C1 + C2 + C2 with the same total charge 3Q. the total charge (3Q) is free to move around on the three positive plates shown in the diagram.0 + 8.04 µF Then 24-65 A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery.24-61 Three capacitors having capacitances of 8.5 (b) This situation is essentialy two capacitors in parallel: (b) The total energy stored in all the capacitors can be found by adding up the energies on each capacitor. (a) In a series connection the charges on the plates is as shown.6 mJ Ceff 2. The capacitor is then disconnected from the battery.2 × 10−5 C 3Q Qtot = = = 12 V V = Ceff C1 + C2 + C3 (8.1) µF One could determine the charges on each capacitor with this result.2 × 10 C Q2 = 12 = 12 = 1. 1 V × 0. 2011 Solution to HW-8 25-44 If a “75 W” bulb (75 W are dissipated when connected across 120V) is connected across a 220 V potential difference (as is used in Europe). 0.167 A) 10 kΩ = 13.82 W The total energy used over a time of two hours is: Etot = P t = 1. From the equation sheet we know that P = IV and V = IR.31 × 104 J 25-70 A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 14 kV power supply. The total resistance of resistors in series is the sum Rtot = Rint + Rperson . We already found the current going through the person in part (a).99 MΩ. which we don’t know. and get a relation between power.2 A = 1.167 A.001 A (a) We can calculate the energy use over a period of time if we know the rate that the device uses electrical energy (power). how much power does it dissipate? (a) The power dissipation P and potential difference V across the bulb are variables.1 V draws a current of 0. but the resistance R across the light bulb is a physical constant whether you are in America or Europe. we have safe Rint = 14 kV V − Rperson = − 10 kΩ = 13. 2011 . which we can find using V = IRperson : 2 P = IV = I 2 Rperson = (1.82 J/s = 1. kids). Isafe = safe Rint + Rperson safe Solving this equation for Rint .0 h? (a) We can model the power supply as a battery with V = 14 kV in series with an internal resistor of Rint . R R P Using the US values P = 75 W and V = 120 V.001 A 0. a) How much electrical energy does it consume during a time of 2.Physics 21 Fall. and the current going through them will be the same as the current going through the equivalent Rtot . R 192 Ω 25-46 A battery-powered global positioning system (GPS) receiver operating on a voltage of 9. voltage. We can combine these formulas to eliminate the current I.82 J/s × 7200 s = 1. When the person holds the terminals of the power supply (don’t do this at home.20 A. what is the current through the person’s body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance. It is the potential across just the person’s body. The power can be determined from: P = IV = 9. (c) To make the battery safe we need to increase the internal safe resistance to Rint . From the second equation in part (a) we can see that V . (a) If the internal resistance of the power supply is 2000 Ω. This current is I= V 14 kV V = = = 1. he completes the circuit shown in the diagram. 75 W We can then use this resistance to solve for the power dissipated when the bulb is connected to the new potential difference: P = V2 (220 V)2 = = 252 W. what should the internal resistance be for the maximum current in the above situation to be 1. we find R= (120 V)2 = 192 Ω.6 kW. However the potential V here is not the same potential as the battery. Rtot Rint + Rperson 2 kΩ + 10 kΩ (b) The power dissipated in a device with current running through it is given by: P = IV.00 mA or less? Rperson Rint V Power supply September 23. and resistance: P = IV = V2 V2 V V = ⇒R= . 26-8 Three resistors having resistances of R1 = 1. I1 I3 9V 8V I2 1Ω 4Ω 2Ω 1 2Ω 2 1V 3I1 + 4I2 = 9 −2I1 + 6I2 = 7 (1) (2) Multiply (1) by 2 and (2) by 3. I2 . what is the value of the current I1 ? Remember that I1 may be positive or negative.0 V (i.c.j. OE 41-1 For this problem. R1 V I2 = = 8.5 A.5 A . P3 = 147 W.9 A. node: loop 1: loop 2: I2 = I1 + I3 9 − I1 − 2I1 − 4I2 = 0 8 − 2I3 − 1 − 4I2 = 0 (b) Determine the currents by explicit solution of the equations. (a) Find the equivalent resistance of the combination.60 Ω are connected in parallel to a 26.h) Find the voltage across each resistor. Req = 0. and R3 = 4. you must write and then solve the loop and node equations needed to find the currents I1 . P1 = 423 W.g. Results are I1 = 1. R2 = 2. therefore V = IR ⇒ I = V /R. I2 . R3 (e) The current from the battery is the sum of the currents through each resistor: Itotal = I1 + I2 + I3 = 30. (c) What is the value of the current I3 ? Remember that I3 may be positive or negative.h) The voltage is the same across resistors in parallel.d) The voltage is the same across parallel resistors. (f.k) Find the power dissipated through each resistor. Thus. (b. You must show your work. Since the voltage is the same across all the resistors. P2 = 233 W.842 Ω) = 30.0 V battery that has negligible internal resistance. Substitute back for I1 then I3 . R2 V I3 = = 5.90 Ω. add and solve for I2 . (i.0 A.c. V I1 = = 16.842 Ω. (f. (b) What is the value of the current I2 ? Remember that I2 may be positive or negative.65 A.g. 1 1 1 1 = + + .k) Any formulation of P = V I = I 2 R = V 2 /R will work. (l) Which resistor dissipates the most power: the one with the greatest resistance or the least resistance? (a) The equivalent resistance for resistors in parallel is given by the sum of their reciprocals. Indicate clearly the loop used to determine each loop equation.60 Ω. I3 = 0.9 A One could also use the equivalent resistance. Itotal = V /Req = 26 V/(0. Rearrange loop 1 and loop 2 equations: loop 1: loop 2: 3I1 + 4I2 = 9 4I2 + 2I3 = 7 Use the node equation to eliminate I3 = I2 − I1 : Thus. Req R1 R2 R3 (a) Write the loop and node equations needed to determine the currents I1 .j. and I3 in the circuit shown. V1 = V2 = V3 = 26.d) Find the current in through each resistor. the power is greatest in the smallest resistor.97 A. I2 = 1. (a) In the circuit shown in the figure. (l) P = V 2 /R. (b. and I3 shown in the figure.3 A. (e) Find the total current through the battery. 2011 Solution to HW-9 Set up the loop equation for this new loop: 25 V − (50 Ω)I = 0 =⇒ I = 26-27 In the circuit shown in the figure the batteries have negligible internal resistance and the meters are both idealized. it would imply that the assumed polarity in the drawing was incorrect.4375 A Substituting the currents found above into the right loop equations gives 0 = − 15. (a) Find the emf E of the battery. With the switch S open. The solution is I1 = 0.0 Ω(I1 ) + E − 48. but this time instead of solving for three currents.6 A)− 75.Physics 21 Fall. 2011 . there are three unknowns in this system. and I2 .32 A. let’s write out the loop and node equations: Node : Left loop : I1 = I2 + I3 E − (20 Ω)I1 − (75 Ω)I2 = 0 Right loop : E − (20 Ω)I1 − (30 Ω)I3 − (50 Ω)I3 = 0 Since we already know I3 .0 Ω(1. we only need to consider the one loop shown.4375 A) + E − 48. Next we simplify the equations and solve for E. V = IR =⇒ I3 = V 15 V = = 0. September 30.0 Ω(−. E. I2 = Left loop : E − 20I1 − 75I2 = 0 Right loop : E − 20I1 − 24 = 0 E = 36.0 Ω(I2 ) − 55.0 Ω Now substitute this I2 into the node equation to solve for I1 : Now we have three equations for three unknowns. since the switch is open and there will be no current through it. 48.60 A = −.60 A in the direction shown.0 Ω(I2 ) + 12 Ω(1.8 V = 1. Using Ohm’s Law. Conveniently.1625 A) = 49. We assign currents I1 and I2 to the remaining branches and loops as shown in the diagram: We can find I3 since we were given the measured voltage across the 50 Ω resistor.3 The left loop equation is 0 = 48. we will solve for two currents and the emf. (b) The circuit is different for this part and is shown below. Here are the equations after substituting for I3 and combining some terms: Node : I1 = I2 + 0. R 50 Ω Now. showing also the currents in each branch and the loops we will use to write Kirchhoff’s equations: 25 V = 0. (b) The polarity of the battery as shown is correct.24 V.30 A. we will use Kirchhoff’s rules. (a) Find the emf E of the battery. If the calculated E had been negative. I1 .0 Ω(1.62 A.8 V. The node equation is I2 = 0.60 A.1625 A. So let’s draw the circuit that we are concerned with. (b) Is the polarity shown correct? (a) To solve this problem.1625 A) E = 15.0 Ω(I2 ) We can solve the left loop equation directly for I2 : 55.4375 A) + 48. I1 = I2 − 1.0 Ω(−. because the E we calculated was positive.50 A 50 Ω 26-28 In the circuit shown in the figure both batteries have insignificant internal resistance and the idealized ammeter reads 1. and the right loop equation is 0 = −15.0 V = 48.4 V. the voltmeter reads 15 V. (b) What will the ammeter read when the switch is closed? (a) For this part we can ignore the battery with the switch. I2 = I1 + 1. to find the new current in the ammeter. We found q(t) by solving the differential equation obtained from Kirchoff’s loop equation.700 A. What is the value of the resistance R? (c) How long after the switch is moved to position 2 will the charge on the capacitor be equal to 99.7 A)(87 Ω) = 230 µC. and 100.0% of the final value found in part (a)? where the final charge Qf = CV = 7. the charge on the capacitor is measured to be 110 µC. E = 28.35 × 10−5 2.90 µF and E = 28. so the capacitor begins to charge. The time constant τ = RC = 11. so we can determine the current as a function of time by differentiating the expression for q(t) above: i(t) = dq Qf −t/RC = e = I0 e−t/RC dt RC where we substituted Qf = CV and noted that the initial current I0 that flows is the battery voltage V divided by the resistance R.9 × 10−6 F) ln(1 − 0. As the capacitor charges. C Now we can solve for the charge on the capacitor as a function of the current: h i q(t) = C[E − i(t)R] = 3. C = 5. In other words we want to solve for t when q = 0. t q(t) = CE(1 − e− RC ) Since we know q(t = 3 s) we can rearrange this equation to solve for R.80 × 10−5 1. (a) We know that after a long time the circuit will approach a steady state where the charge on the capacitor will be simply given by Q = CE. Substituting C = 5.00 ms. (b) We know that the resistance R is part of the time constant in the function q(t). the formula for the charge on a charging capacitor as a function of time is: ´ ³ q(t) = Qf 1 − e−t/RC . (a) Compute the charge on the capacitor at the following times after the connections are made: 0 s.392 s.26-40 A 12.0 s.8µF capacitor is connected through a 0.0 V we find the charge on the capacitor after a long time will be Q = 165. I0 is 6.99 × CE. (a) As derived. and the emf has negligible resistance.68 × 10−4 C.16 × 10−5 1. 1 t R=− q C ln(1 − CE ) We insert our value for q(t = 3 s) = 110 µC and find the resistance R = 464 Ω.90 µC 120 V − (0. (b) The relationship between charge and current is i = dq/dt. 10. 26-48 In the circuit shown below.74 × 10−5 4. 20. 5. t (s) 0 5 10 20 100 q(t) (C) 0 2.90 µF. The table below gives q(t) at the times specified. Then E − VR − VC = 0 q(t) E − i(t)R − = 0. Let q(t) and i(t) be the instantaneous charge on the capacitor and current in the circuit. and q we find: t = −(464 Ω)(5.0 V.0 s.99) = 0. and at that point the capacitor is essentially fully charged. and the current from the battery is essentially zero.0 s. Initially the capacitor is uncharged and the switch S is in position 1.90 µF are connected in series.0 V.0 Ω.0 s. The switch is then moved to position 2. a resistor with a resistance of R = 87.49 × 10−4 6. We rearrange our equation for q(t) to get: t = −RC ln(1 − q ).74 × 10−5 A. (b) Compute the charging currents at the same instants.68 × 10−4 i(t) (A) 6. CE Substituting our values of R. C. (c) We want to find at what time t will the charge on the capacitor be 99% of its final value which we found in part (a).73 × 10−4 4. Note that 100 s is about nine times the time constant. Here is a table of the charge and current at various times.890 MΩ resistor to a constant potential difference of 60. (a) What will be the charge on the capacitor a long time after the switch is moved to position 2? (b) After the switch has been moved to position 2 for 3.2 µC. what is the magnitude of the charge on each plate of the capacitor? (a) The simplest way is to apply the loop equation.0126 s .35 × 10−4 7. when the current in the resistor is 0. and a capacitor with a capacitance of C = 3.00 × 10−8 26-43 An emf source with a magnitude of E = 120 V. in this case. the thumb of your right hand is pointing to the right (eastward). and west is to the left. We can evaluate the cross product v × B by using the cross product of each pair of unit vectors: ˆj × ˆi = −k ˆ ˆj × ˆj = 0 Then ´i ³ h ˆ F = qv × B = q vyˆj × Bxˆi + Byˆj = −qvy Bx k Substituting the given quantities we get: F = −(1.22 × 10−8 C)(3.81×10−3 kg and a charge of q = 1. (a) We need to apply the right hand rule to see if the direction of the force is consistent with a positive charge or a negative charge.63 T) ˆ = −5. The force on a charged particle moving in a magnetic field is given by: F = q (v × B) For the cross product.63 T) ˆi + (0. so curl the fingers of your right hand upward. Imagine you are seated so that north is in front of you. The magnetic force on a charged particle moving in a magnetic field is given by the equation F = qv × B Since. we notice that many components of v and B are zero: v = vyˆj and B = Bxˆi + Byˆj. (a) What is the magnitude of the particle’s acceleration produced by a uniform magnetic field B = (1.00 × 104 m/s) ˆj. In order to do this. Then. the magnitude of F is simply given by F = qvB. your palm must be facing upward. the particle must have a positive charge.90 µC)(4.980 T) ˆj? (b) What is the direction of the particle’s acceleration? To determine the acceleration of the particle we need to know what force is acting on it. with the direction determined by the right hand rule. Here is a diagram: N v B F W E 27-4 A particle with mass m = 1. v. F.22 × 10−8 C has. The magnetic field is upward. at a given instant.72 km/s is deflected toward the east.0567 N September 30.00 × 104 m/s)(1. 2011 . and B are mutually perpendicular.72 km/s)(1. The other directions are then determined: east is to the right. We can assume the only force is due to the magnetic field. 2011 Solution to HW-10 27-3 In a 1. Point the fingers of your right hand straight forward. we can find the acceleration of the particle: ˆ a = F/m = −0.35 T magnetic field directed vertically upward. S (b) F = qvB = (8. a velocity v = (3.97 × 10−4 N k Using Newton’s second law.330 m/s k.35 T) = 0. a particle having a charge of magnitude 8. So northward velocity means the particle is moving forward.90 µC and initially moving northward at 4.Physics 21 Fall. Thus. south is behind you. Eastward is the direction the particle is deflected. (a) What is the sign of the charge of this particle? (b) Find the magnetic force on the particle. the components of v parallel and perpendicular to B (vk and v⊥ ) both have the √ √ value v0 / 2 (since sin θ = cos θ = 1/ 2 for θ = 45◦ ). where it moves in a circular arc with a radius of 0. we have πR = v⊥ t =⇒ t = πR/v⊥ Then ¡ −3 ¢ π 6. but we have to find v for an electron accelerated through a potential difference of ∆V = 1950 volts. the speed is written as v⊥ to remind us that only the component of the velocity vector that is perpendicular to the magnetic field direction contributes to the circular motion.62 × 107 m/s. We have an equation from the equation sheet that describes this motion. (b) Find the time required for it to make 21 of a revolution.90 × 10 m πR = = 2. The particle’s gain in kinetic energy is equal to its loss in potential energy. where v0 is the given speed 2.8 µm r= eB (1.52 × 10−8 s = 25.2 ns t= v⊥ 8. By definition.90 mm in a magnetic field with a magnitude of 2. and the speed v⊥ of the particle in its circular orbit is constant. v⊥ eBv⊥ eB The pitch is the distance travelled parallel to B (at speed vk ) in time ∆t (one loop): p = vk ∆t = 2πme vk = 1. We are given numbers for all of the other quantities in this equation.3 T at a 45◦ angle to B.60 × 10−19 C (2. assuming its speed is 2 × 106 m/s.594 × 105 m/s (c) This type of problem can be solved by conservation of energy.6 × 10−19 )(0.602 × 10−19 J/eV Solving for v gives v = 2.60 × 10−19 C. The deuteron travels in a circular path with a radius of 6. Since d = πR. we know that B = mv/(eR).30) The time to make one loop of circular motion is ∆t = 2πr 2πme v⊥ 2πme = = . In this problem. the particle starts from rest.71 kV 27-25 An electron in the beam of a TV picture tube is accelerated by a potential difference of 1. (a) Find the speed of the deuteron. ¡ ¢¡ ¢ 6.11 × 10−31 )(2.90 × 10−3 m 1. eR OH11-05 An electron enters a uniform magnetic field with magnitude 0.69 × 10−4 m = 169 µm eB .7 in the textbook for a reminder.34 × 10−27 kg 8.95 kV. K0 + U0 = K1 + U1 K1 − K0 = U0 − U1 2 1 2 mv⊥ = q (V0 − V1 ) = −q∆V ¡ ¢¡ ¢ 2 3.34 × 10−27 kg and a charge of 1. Then it passes through a region of transverse magnetic field. we are told it is circular motion.0 × 106 / 2) me v ⊥ = = 26. R= mv⊥ qB Here. Then B= mv = 8.0 × 106 m/s.594 × 105 m/s (b) For this part.60 T.60 T) RqB = v⊥ = m 3. What is the magnitude of the field? From problem 27-24.34 × 10−27 kg = 8. We get the radius of the circular motion using v⊥ : √ (9. Assuming the electron starts from rest.27-21 A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3. Determine the radius r and pitch p (distance between loops) of the electron’s helical path. so we don’t have to worry about getting the signs right. the force on a moving charged particle due to a magnetic field causes it to travel in a circular path. so we know that the velocity is purely perpendicular to the magnetic field direction. we must recall concepts from Physics 11. Since v makes an angle of 45◦ with B. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed? Note that Mastering Physics is only asking us to provide |V |. so simply solve it for v⊥ . or to travel a distance d of half the circumference of the circular orbit. the electron gains an energy 1950 eV.60 × 10−19 C) |V | = 7709 V = 7. In this case. we have 2 1 2 mv (a) As is seen in the figure.594 × 105 m/s mv⊥ ∆V = − =− 2q 2 (1. = e∆V = (1950 eV) × 1. See example 23. We are looking for the time t to complete half a revolution.179 m.32 × 10−4 m. 00 cm that is in this uniform magnetic field. vertical wire carries a current of I = 1.010 m)(0. that points in the direction of the current. and its direction will be 90◦ clockwise from B.470 T. Solving for V . R October 10. A battery and a resistor of resistance 26. Fmag is the same as in part (a).03 mN and the direction is south. (c) 34◦ south of west? 27-39 A thin.19 A)(0. LB (b) When the resistance R drops to a lower value R . (c) For B pointing east. the magnetic force will also be horizontal. (a) For B pointing east. 51. two metallic supports in a uniform magnetic field with a magnitude of 0. (b) south. the current in the wire increases.0 cm long metal bar with mass 770 g rests on. R Substituting the expression for V derived in part (a) into the expression above leads to mgR LB R − mg = ma ⇒ g − g = a LB R R R − R 2 a=g = 84. If the resistor suddenly gets partially short-circuited. R where I is the current through the wire. One can see that for any horizontal B.19 A downward in a region between the poles of a large superconducting electromagnet. 2011 . if the magnetic field direction is (a) east. for an arbitrary orientation of B in the horizontal plane: N B W I S E Fmag We use the right hand rule to evaluate the magnetic force Fmag = Il × B. and the direction of Fmag is 34◦ west of north. but is not attached to. we find V = mgR = 819 V. 2011 Solution to HW-12 27-36 A straight. Fmag is the same as in part (a).70 Ω. (b) For B pointing south. (a) What is the largest voltage the battery can have without breaking the circuit at the supports? (b) The battery voltage has the maximum value calculated in part a. What are the magnitude and direction of the magnetic force on a section of the wire with a length of l = 1. and the direction of Fmag is west. The diagram shows the view looking down in the direction of the current. as shown in the figure.591 T) = 7. decreasing its resistance to 2. and L is a vector that points along the portion of the wire that is in the field.Physics 21 Fall. (a) The largest force that won’t break the circuit occurs when the upward magnetic force just balances the downward gravitational force: V Fmag = IL × B = ILB = LB = mg. Then the magnetic field exerts a force on the currentcarrying wire that is in the upward direction.0 Ω are connected in series to the supports. find the initial acceleration of the bar.591 T and is horizontal. the magnitude of Fmag is Fmag = IlB = (1. the length of the wire section.6 m/s . and the upward force on the wire will exceed the downward gravitational force. leading to a net upward force ma: V LB − mg = ma. where l is a vector of length l. The voltage V causes a current I = V /R to flow in the wire. where the magnetic field B has a magnitude of B = 0. 0 cm is parallel to a magnetic field of magnitude 0. and the magnitude of that force. π The magnitude of the maximum torque is τmax = πr2 IB = 5.602 × 10−19 C = 1.03 × 10−3 N m (c) The maximum torque would be achieved for the loop with the largest area for the available length of wire. then the radius r of the circle satisfies 2πr = 2(a + b) ⇒ r= a+b = 0.3 × 10−11 m)2 (. ˆ F = IL × B = ILk ˆ × ˆi = ˆj. The magnitude is µ = 2.13 m.52 × 10−2 A m2 (a) The torque on the loop is τ = µ × B.3 A. We get F = IL −Byˆi + Bxˆj ⇒ Fx = 2.5 × 10−16 s (c) The magnitude of the magnetic moment of a current loop is the current I times the area of the loop: µ = πr2 I = π(5.38 T . We can work out the cross products using k ˆ ˆ ˆ ˆ ˆ k × j = −i.482 T F = Fx2 + Fy2 + 02 = 2.0 cm long lies along the z axis and carries a current of 8. Find the x. The shape of the optimum loop is a circle. and Bz = −0. (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop.327 T.3 × 10−11 m. Because µ is perpendicular to B. I = e/T = 1.42 × 10−3 N m. in the lowest energy state the electron orbits the proton at a speed of v = 2.16 T.2 × 106 m/s (b) Current is the amount of charge ∆Q passing a point in time ∆t.0 cm and a height of 8.27-42 The plane of a rectangular loop of wire with a width of 5.200 T. The force on the wire is given by (b) The magnetic moment of the loop µ has magnitude IA and direction perpendicular to the loop.00106 A) = 9. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field? 27-74 A wire 28.60 A in the +z direction. The magnetic field is uniform and has components Bx = −0.33 T. By = −0. what is the current I? (c) What is the magnetic moment of the atom due to the motion of the electron? (a) The time T to travel distance 2πr at speed v is T = 2πr/v = 2π(5. and z components of the magnetic force on the wire. y. Since the electron with charge e passes any point on its orbit once in time T . Fy = −0.5 × 10−16 s 2.3 × 10−11 m) = 1.3 × 10−24 A m2 ˆ × (Bxˆi + Byˆj + Bz k).968 T.1 mA 1. 27-43 In the Bohr model of the hydrogen atom.2 × 106 m/s in a circular orbit of radius r = 5. If the sides are a and b. the magnitude of τ is τ = µB = 4. and k × k = 0. The loop carries a current of 6. 0)(0.0) = 10−7 0. B=0 (b) At point (b). the sines of the angles in each case is 2. a distance a below the bottom wire. Each wire carries current I in the +x direction. The distances will be multiples of a.00 cm apart and carry currents in opposite directions. in each case. and (c) at a distance a below the lower wire.76 × 10 T. The vector sum is zero.64 × 10−7 T. 3πa (c) At point (c). For the present problem. (b) I dl y 2a (a) I r-r' dl r-r' z x (c) The magnitude of the magnetic field near a long wire is B= Use the Biot-Savart Law: dB = µ0 Idl × (r − r′ ) . 2011 . and r − r′ points from the current point to the field point (P ). and (c). B=− 2µ0 I ˆ k. one above the other. both dBtop and dBbottom point into the page. Hence (12. the field from the bottom wire is µ0 I/(2πa) into the page. straight wires. The simplest way to evaluate the cross product is to use the right hand rule for the direction.0. By the right hand rule.08 m. respectively. Let the +y axis be in the plane of the wires in the direction from the lower wire to the upper wire.5/8.0015 m and 0. Since both contributions to the field point into the page. midway between the wires.083 −7 = 1.0015)(0. Find B.5/8.080)(2. 2πR where R is the perpendicular distance to the wire.5/8. and the field from the top wire is µ0 I/(2π3a) into the page. (b) at a distance a above the upper wire. a distance a above the top wire. and the field from the bottom wire is µ0 I/(2πa) out of the page. and the field from the bottom wire is µ0 I/(2π3a) out of the page. 3πa October 12. B= 2π a 3a 3πa In terms of vector components.080)(2. we must add the vector contributions to the field from each wire at each point (a). so does the sum. B= 2µ0 I ˆ k. respectively) separately and then add them. 3 4π |r − r′ | where dl points in the direction of the current I. µ0 I .0)(0. Since the wires are 5.79 × 10−8 T (24. The diagram shows the vectors dl and r−r′ for each of these contributions. and dBtotoal = dBtop + dBbottom = 2. We can evaluate dBtop and dBbottom (the contributions to the total field from the top and bottom wires. The sum is µ ¶ 2µ0 I 1 µ0 I 1 = + (out of the page). the field from the top wire is µ0 I/(2πa) out of the page.0015)(0.0 cm apart. are separated by a distance 2a and are parallel to the x axis. as shown in the figure. the field from the top wire is µ0 I/(2πa) into the page.0) 10−7 0. 2011 Solution to HW-13 28-12 Two parallel wires are 5.083 dBtop = dBbottom = 8.50-mm segments of wire that are opposite each other and each 8. (a) At point (a). In terms of vector components. the net magnetic field of the two wires at the following points in the plane of the wires: (a) midway between the wires. and to get the magnitude from the products of the magnitudes |dl| and |r − r′ | and the sin of the angle between the vectors. 28-18 Two long.00 cm from P . Find the magnitude and direction of the magnetic field at point P due to two 1. The magnitudes of dl and r − r′ are 0. The direction of the field is given by the right hand rule.Physics 21 Fall. (b). B= 2µ0 I 3πa (into the page). Then ¶ µ I1 I2 I2 L = =⇒ x = x L+x I1 − I2 This solution won’t work in our case. By looking at the directions of the two fields in various locations.0 A.0 cm apart.39 m. Let out of the page be plus. 2011 . The currents running through the wires are 8. The magnitude of the field B a distance R from a long wire with current I is B = µ0 I/(2πR). (b) (a) L x I1 L x I2 I1 I2 (a) Panel (a) of the diagram shows the two wires end on when the currents are in the same direction. so x < 0.0-A and I2 = 78. In this case the vector sum of the fields can only be zero on the line connecting the wires. October 14. (b) Panel (b) of the diagram is similar to panel (a). Therefore there are no other solutions.0-A. which contradicts our initial assumption that x is a positive distance. The magnetic field lines due to each wire separately are shown by the concentric circles (dashed for I1 . which is L − x = 29. x= I> − I< where I< (I> ) is the lesser (greater) of I1 and I2 .0 A. What about a point to the right of I2 in panel (b)? We can set up the equation. I2 > I1 . L = 38. Then µ0 I2 I1 I2 µ0 I1 = =⇒ = 2πx 2π(L − x) x L−x Solving this equation and substituting values leads to µ ¶ ¶ µ I1 23 x= 38 cm = 8. We consider first a distance x to the left of I1 . The point where the field is zero is outside the two currents. which is L + x = 53. The vector field will be zero at a point on the line between the wires a distance x from the left wire and L − x from the right wire. but outside the wires.0 cm. and let I > 0 correspond to up: B out of page = µ0 (−10 + I − 8 + 20) 2πd Solving. We just have to keep track of the direction of each field using the right hand rule. carry currents I1 = 23. Find all locations where the net magnetic field of the two wires is zero if these currents are (a) in the same direction or (b) in opposite directions. Let this distance be d = 21 × 0. 2011 Solution to HW-14 Cancelling Magnetic Field Four very long.Physics 21 Fall.3 cm in our notation. parallel transmission lines. 10. 28-22 Two long. where the magnitudes of the fields are equal. The direction of the field follows from the right hand rule and is shown at selected points by an arrow next to each circle.0 A.9 cm in our notation. a point a distance x to the right of I2 would be x + L from I1 . and I. we get I = −2 A. Find the magnitude and direction of the current I that will make the magnetic field at the center of the square equal to zero. in this case the condition is µ0 I2 I1 I2 µ0 I1 = =⇒ = 2πx 2π(L + x) x L+x The field point at the center of the square is equidistant from all four wires. L is the distance between the wires. the Solving this equation and substituting values leads to µ ¶ ¶ µ I1 23 x= 38 cm = 15. currentcarrying wires in the same plane intersect to form a square with side lengths of 39. a distance x from the wire with the smaller current. vector sum of the fields can only be zero on the line between the wires. and the minus sign means I is directed downward. it’s easy to see that for case (a).7 cm L= I1 + I2 23 + 78 MasteringPhysics asks for the distance from the 78 A wire in the direction of the 23 A wire. Comparing the two solutions we have obtained for part (b). solid for I2 ). 20. it shows the two wires and the fields when the currents are in opposite directions. as shown in the figure. one can see that a general way of writing the solution is ¶ µ I< L.9 cm L= I2 − I1 78 − 23 MasteringPhysics asks for the distance from the 78 A wire in the direction of the 23 A wire. These forces are obtained from F = Il × B. say 0. two sides of the square loop are parallel to the wire and two are perpendicular as shown. The magnitude of the magnetic field B1 of wire 1 at wire 2 is µ0 I1 .5×107 4. B1 = 2πd and from the rh rule. Hence the net force is the vector sum of the forces to the left and to the right. (a) If the cord carries current to a 100 watt light bulb connected across a 120 V potential difference. what force per meter does each wire of the cord exert on the other? (Model the lamp cord as a very long straight wire. (a) What is the current in the second wire? (b) Are the two currents in the same direction or in opposite directions? I1 B1 I2 Let’s assume that the currents flow in the directions shown.7 In general.Wire and Square Loop A square loop of wire with side length a carries a current I1 .833 A.027 2π F d = 0.70 cm. The infinite wire and loop are in the same plane. much larger than the magnetic force. .1 kg)9. so the forces make the wires repel each other. B1 points into the page. The force per unit length is F µ0 I1 I2 = L 2π d The question asks for I2 .10× 10−5 = 7. and the wires repel each other. where R is the distance to the wire. The current in one wire is I1 = 0. (a) The force per unit length is given by the formula derived in problem 28-26. If we guess that a meter of wire weighs a few ounces. (c) No. then 2 mg = (0.) (b) Is the force attractive or repulsive? (c) Is this force large enough so it should be considered in the design of lamp cord? Since P = IV .1 kg. The diagram shows the forces on the left and right hand side of the loop from this field. The magnitude of the force F that B1 exerts on a length L of wire 2 is µ0 I1 F = I2 LB1 = I2 L . (a) The B field is into the page everywhere on the right of the wire in the plane of the square loop. 28-27 The wires in a household lamp cord are typically d = 2. and we’ll show that the force between the wires is repulsive. Find the magnitude F of the force on the loop from Part (a) in terms of the magnitude of its magnetic moment.700 A. One can go through a similar argument to find that the force on the upper wire has the same magnitude and is upward.81 m/s ∼ 1 N. The net force is to the left and has magnitude µ ¶ 1 1 µ0 I1 I2 a − Fleft − Fright = 2π d − a/2 d + a/2 µ0 I1 I2 a2 = 2 2π d − a2 /4 (b) The magnitude of the magnetic moment µ of the loop is the current times the area. 2πd and by the rh rule one can see that the direction is downward. The force is small compared to the gravitational force. the currents must flow in the opposite direction.5 mm apart center to center and carry equal currents in opposite directions. parallel wires are separated by a distance of d = 2.56 × 10−5 N/m L 2π d 0.91 A I2 = µ0 L I1 0. The center of the loop is located a distance d from an infinite wire carrying a current I2 . or 2 µ = I1 a .833)2 = = 2 × 10−7 = 5.0025 (b) The force will be repulsive.10 × 10−5 N/m. The force per unit length that each wire exerts on the other is 4. (a) What is the magnitude F of the net force on the loop? (b) The magnetic moment µ of a current loop is defined as the vector whose magnitude equals the area of the loop times the magnitude of the current flowing in it (µ = IA). with both currents equal: F µ0 I 2 (0. Its magnitude is given by B = µ0 I2 /2πR. but they are equal in magnitude and oppositely directed. We can write the net force in terms of µ as Fleft − Fright = µI2 µ0 2π d2 − a2 /4 28-26 Two long. so µ ¶ ¡ ¢ 0. and whose direction is perpendicular to the plane in which the current flows. the current in each wire is I = P/V = 100 W/120 V = 0. and the magnitudes are µ0 I2 2π(d − a/2) µ0 I2 = I1 aB = I1 a 2π(d + a/2) Fleft = I1 aB = I1 a Fright There is also an upward force (Ftop ) and (Ftop ). 2011 . The central conductor and tube carry equal currents I in opposite directions. the total magnetic field will be N times this magnitude. The only current enclosed is the current I in the center conductor.059)2 N= (a) This problem is best answered with Ampere’s Law I B · dl = µ0 Iencl Chose a circular line integral path with radius r between the center and outer conductor.0592 = 57 = (4π × 10−7 )(3. choose a current loop with radius r > c. the magnetic field is 6.30)(. (b) Derive an expression for the magnitude of the magnetic field at points outside the tube (r > c). but inside the tube (a < r < b). So we can determine the number of turns by solving for N to get: 2B(x2 + a2 )3/2 µ0 Ia2 ¢3/2 ¡ 2(. solid conductor. How many turns must it have if. Boutside = 0 October 18.30 A.58 × 10−4 T? The magnitude of the magnetic field at the point x on the axis of a single circular loop with a radius a and current I is B= µ0 Ia2 . 2(x2 + a2 )3/2 For a loop with N turns. The currents are distributed uniformly over the cross sections of each conductor. Binside = ¡ ¢I µ0 I = 2 × 10−7 2πr r (b) Taking a similar approach for this problem. 2πrBinside = µ0 I Solving for B.90 cm and carries a current I = 3. at a point on the coil axis 6. 28-34 A closely wound coil has a radius a = 5.Physics 21 Fall.0622 + . then by symmetry we expect the value of B to be constant around the path. The total current enclosed is now zero because the two currents I in each conductor are going in opposite directions. (a) Derive an expression for the magnitude of the magnetic field at points outside the central.20 cm from the center of the coil. 2011 Solution to HW-15 28-32 A solid conductor with radius a is supported by insulating disks on the axis of a conducting tube with inner radius b and outer radius c.000658) . I B · dl = µ0 Iencl . which is the circumference. where the positive direction of current flow is out of the page since the integral is done counterclockwise. (a) I Iencl = 0 ⇒ 28-45 A wooden ring whose mean diameter is 16. B · dl = 0 (b) Iencl = −I1 ⇒ I B · dl = −5.016 m) = 39. To calculate n. The total number of loops is simply the number of turns per length times the total length of the solenoid N = nl.16 585 0. L = N 2πr. (d) Iencl = I2 + I3 − I1 ⇒ I B · dl = 5. several conductors that carry currents through the plane of the figure.1 m Use Ampere’s Law.655 A. and d. (a) In the center of the windings.00 × 10−2 T B = 1180 = . and I3 = 2. b.0 cm is wound with a closely spaced toroidal winding of 585 turns. we use the number of turns divided by the circumference of the toroid: n= B = µ0 585 π 0.0 A. Wb −7 µ0 Imax m (4π × 10 A·m )(13. and the directions shown. The currents have the magnitudes I1 = 4.655 = 9. and the wire can carry a maximum current of 13. Solving this equation for n = B/µ0 I. are shown.5 A. I2 = 6.60 cm and a length of 33.16 .00 × 10−2 T at its center. (c) Iencl = I2 − I1 ⇒ I B · dl = 3. where n is the number of turns per unit length.78 × 10−6 T m.33 m)(2π)(. a toroid is basically a rolled up solenoid.1 A. n is the number of turns per unit length along the toroid.14 × 10−6 T m. (a) Compute the magnitude of the magnetic field at the center of the cross section of the windings when the current in the windings is 0.58 × 10−4 T π 0. we can see that the minimum n will occur when the current is at its maximum.5 A) (b) The total length of wire L will be the total number of loops times the length of one loop.28-36 The figure shows. It has a radius of 1.0 cm.03 × 10−6 T m. c. in cross section.H labeled a. What is the line integral B · dl for each of the four paths? The integral involves going around the path in the counterclockwise direction. (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required? (a) The magnitude of the magnetic field inside a solenoid is given by B = µ0 nI. Four paths. the strength of the magnetic field can be found using the formula derived in the class lecture: B = µ0 nI This is the same formula used for a solenoid. Thus the total length of the wire is L = nl2πr = (1180 m−1 )(. n= turns 2.5 A. 28-41 A solenoid is designed to produce a magnetic field of 2. à ! µ0 Ia2 µ0 Ia2 F = qv 2 = qv 3/2 3/2 2 (x2 + a2 ) (x2 + a2 ) ¡ ¢ = 1. There are N = 500 turns and the mean radius is r = 25. F = qv × B.110 m) + (. 28-52 A long. Line ab is normal to the plane of the loops and passes through their centers.175 m) = 4. and because the loops carry the same current in the same direction. and the direction of B is along the loop axis.0 cm apart. An electron is traveling in the vicinity of the wire.175 m) × ´3/2 ³ 2 2 (. Note that the negative sign of the electron makes F in the opposite direction from v × B.5) N 2π(.4 (b) The magnetic susceptibility is χm = K m − 1 thus the answer is 2020.940 T. v. we get Km = Using SI units. and B are mutually perpendicular. We only need to find the magnitude of the force. At the instant when the electron is 4.50 A.00 × 104 m/s directly toward the wire. straight wire carries a current of 2. Find the magnitude of the magnetic force these loops exert on the proton just after it is fired. (b) Calculate the magnetic susceptibility of the material that fills the toroid. so the magnitude of F is F = evB = evµ0 I 2πr October 21. we know that line ab is on the axis of both loops. Solving the last equation for Km . From the description of the loops. These loops are parallel to each other and are 22.94 = = 2021. F = qvB sin θ = qvB sin 90◦ = qvB. what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron? B F v I The magnetic field due to the wire has magnitude B= µ0 I 2πr and direction (from the rh rule) out of the page at the location of the electron as shown. 28-55 Two identical circular.15: B= µ0 Ia2 2 (x2 + a2 ) 3/2 The direction of the magnetic field is along the axis as determined by the right hand rule.50 cm from the wire and traveling with a speed of 6. for a loop of radius a. 2011 .60 × 10−19 C (2750 m/s) ¡ ¢ 2 4π × 10−7 T · m/A (2.30 A) (. 2πrB 2π · 0.400 A. In particular. each loop contributes the same magnitude and direction of magnetic field. The total magnetic field is thus twice the magnetic field due to a single loop. The vectors F.07 × 10−19 N The magnetic field exerts a force in the same direction as the current.30 A in the same direction. Notice that the direction of the force is the same regardless of which side of the loop is particle is located at. A proton is fired at 2750 m/s perpendicular to line ab from a point midway between the centers of the loops. Since the particle is located equidistant from the two loops. the magnetic field along the axis a distance x from the loop is given by equation 28. The particle is located on line ab. The magnetic field inside the windings is found to be 1. 2πr where n is the number of turns per unit length and N is the total number of turns. 2011 Solution to HW-16 F = 28-48 The current in the windings of a toroidal solenoid is 2. carrying a current I.42 × 10−21 N F is in the direction shown in the diagram. µ0 N I (4π × 10−7 ) 500 · 2.25 · 1. (1.045) = 1.00 cm. The force on the electron is given by F = qv × B = −ev × B. The toroidal solenoid is filled with a magnetic material. (a) Calculate the relative permeability.0 cm in diameter each carry a current of 2. (a) The magnetic field inside a tightly wound toroidal solenoid is µ0 N I B = Km µ0 nI = Km .Physics 21 Fall.602 × 10−19 )(60000)(4π × 10−7 )(2. wire loops 35. This problem involves the magnetic force on a moving charged particle. since the direction of v is perpendicular to the loop axis. Section 28-5 in the textbook shows how to calculate the magnetic field along the axis of a loop. The trick is to find the magnetic field at the position of the particle due to the two loops. 10 × 10−2 kg/m. When I is increased. tension. is the induced current in the loop clockwise or counterclockwise? (a) The induced emf is given by Faraday’s Law: ¯ ¯ ¯ ∆Φ ¯ dΦ ¯.0 cm and oriented in the horizontal xy plane is located in a region of uniform magnetic field. B = µ0 I/(2πr). The sum of the forces can then be written as X Fy = T cos θ − mg = 0 =⇒ T = mg/ cos θ X Fx = F − T sin θ = 0 =⇒ F = T sin θ = mg tan θ. parallel wires hang by 4. one can find the force per unit length on one wire due to the current in the other: µ0 II ′ F = .362 × 10−3 m.8 A.00-cm-long cords from a common axis (see the figure ). =⇒ |E| ∼ ¯¯ E =− dt ∆t ¯ The change in flux ∆Φ is the difference between the final flux and the initial flux. Now we substitute this r and µ0 = 4π × 10−7 T m/A.0◦ into the expression for I and find I = 21.00◦ ) = 8. Therefore the current is counterclockwise. the angle θ from the vertical increases. (a) What is the current in each wire if the cords hang at an angle of 6.0 ms. λ = m/L.14 m) × 1. A large current is required for even the small displacement seen here.00-cm cord hangs at 6◦ with respect to the vertical.8 T = = 55 V. L 2πr Viewing a wire end-on allows us to construct a free body diagram involving the gravitational force.0 µs. find the average of the emf that will be induced in the wire loop during the extraction process. (b) If the coil is viewed looking down on it from above. 29-5 A circular loop of wire with a radius of r = 14. the induced current will be in the direction that will restore an upward pointing field. Originally. and the force F on a length L of wire from a magnetic field B perpendicular to the wire. (a) If the loop is removed from the field region in a time interval of 2. µ0 I 2 L = mg tan θ = λLg tan θ 2πr r µ0 I 2 L 1 = λLg tan θ =⇒ I = 2πrλg tan θ 2πr µ0 F = Note the Ls cancel. and θ = 6. where F denotes the force on a length of wire L.00◦ with the vertical? (a) Two parallel wires with currents running in opposite directions exert a repulsive force on one another. g = 9.8 T is directed along the positive z direction. . λ = 1. and ∆t is 2. A field of magnitude B = 1. F = ILB. r = 2 l sin θ = 2(0. Since the field B is initially perpendicular to the loop. We get the absolute value of the emf from ¯ ¯ ¯ ¯ 2 ¯ Φf − Φi ¯ ¯ 0 − Φi ¯ ¯=¯ ¯ = Φi = πr B |E| = ¯¯ ¯ ¯ ¯ ∆t ∆t ∆t ∆t 2 π(0. which is upward. Now we use the expression for the magnetic force F and the linear mass density. the flux was determined by the field B pointing up (as one looks down on the loop).040 m) sin(6. the initial flux Φi through the loop is just the area of the circular loop times the magnetic field: Z Φi = B · dA = πr2 B. and the magnetic force.8 m/s2 . The wires have a mass per unit length of 1. the final flux Φf is zero.28-63 Two long. When the loop is removed from the field.10 × 10−2 kg/m and carry the same current in opposite directions.002 s (b) The direction of the induced current I is obtained using Lenz’s Law. After the loop is removed from the magnetic field. The direction of I must oppose the change in flux. Using the formula for the magnetic field at a distance r from a long wire. The distance r between the two wires is twice the base leg of the right triangle that is formed when the 4. 0. the induced emf will cause a current to flow in a direction that will cause a magnetic field upward.057t =⇒ t = ln 20/(0. what are the magnitude and direction of the field B at a distance r to the right of the wire? (c) What is the flux dΦB through the narrow shaded strip? (d) What is the total flux through the loop? (e) What is the induced emf in the loop? (f) Evaluate the numerical value of the induced emf if a = 12. circular. the induced emf will cause current to flow in the direction that opposes the change in magnetic flux. 2011 Solution to HW-17 29-7 The current in the long.60 A/s. dt dt (a.24 m)(9. b = 36.0 cm. steel loop of radius 75 cm is at rest in a uniform magnetic field. 1 = e−0. straight wire AB shown in the figure is upward and is increasing steadily at a rate di/dt. 2011 . and di/dt = 9. (a.4)(−0. Because the area of the loop doesn’t change with time. October 27.057t We usually take E to be the magnitude of the emf. Because the magnitude of the magnetic field directed upwards through the loop is decreasing with time. and don’t worry about the sign.122e−0. dΦ/dt = πr2 dB/dt.0 cm.6 s 20 (c) According to Lenz’s Law. (b) When is the induced emf equal to 1/20 of its initial value? (c) Find the direction of the current induced in the loop. For that to happen.057)e−0.4)(−0. according to −1 B(t) = (1.0 cm. as viewed from above the loop.4 T)e−(0. we can solve for t at a particular emf. the current must flow in a counterclockwise direction.057)e−0. (b) Now that we have an expression for the induced emf as a function of time.06 × 10−7 V What will be the direction of the current? Counterclockwise. we can write Z B · dA = Φ = B(t)A sin(60) where A = πr2 is the area of the loop.6) ln 12 ¶ = 5. Because the magnetic field makes a 60 degree angle with the loop.b) B= µ0 i(t) 2πr into page (by right hand rule) (c) µ0 i(t) L dr 2πr dΦ = B dA = (d) Integrate Φ= Z b a dΦ = µ0 i(t)L 2π Z a b dr µ0 i(t)L b = ln r 2π a (e) dΦ µ0 L di b |E| = = ln dt 2π dt a (f) (2 × 10 −7 µ 36 )(0.057 s−1 ) = 52. The negative sign is related to the polarity of the voltage. as shown in an edge-on view in the figure.Physics 21 Fall. (a) We can find the emf induced in the loop using I Z d dΦ E · dl = − B · dA = − . L = 24. 29-8 A flat.057t dt Putting these pieces together results in: E = −πr2 sin(60)(1.057s )t . The field is changing with time. where d B(t) = (1. where the integral of E · dl is the induced emf.057t = 0.b) At an instant when the current is i. (a) Find the emf induced in the loop as a function of time (assume t is in seconds). Φ = A (B0 + bx) Using the following equation. a magnetic field points in the +x direction (toward the right). dt dx dt Now the emf can be calculated using dΦ = Abv. the current will be going in a clockwise direction (CW). E =− dt (b) The direction of the current can be determined with Lenz’s Law. Because the flux through the loop is decreasing.60 × 102 V rad Emax = = 7.80× 102 T. and so the induced current will induce a magnetic field pointing towards from the origin. We can solve for the angular speed that would create the stated emf: Emax = ωN BA 2.0195 m)2 s . we can get the time derivative of the flux at the location x of the loop: dΦ dΦ dx = = (Ab)(−v) = −Abv. Its magnitude varies with position according to the formula Bx = B0 + bx. the opposite direction from part (b). the current will be in a counter-clockwise direction (CCW). E = −N dΦB = ωN BA sin ωt dt The maximum induced emf occurs when sin ωt = 1. So. B(x) = B 0 + bx +X v (a) This problem should be solved with Faraday’s Law: E =− d Φ. (c) If the coil is moving from left to right instead. what would be the answer to part (b)? 29-13 The armature of a small generator consists of a flat. ¿From Faraday’s law. where B0 and b are positive constants. This means.29-11 In a region of space. but the magnetic flux will change with time because the loop moves towards a weaker field. ΦB = B · A = BA cos φ = BA cos ωt. what is the direction (clockwise or counterclockwise) of the current induced in the coil? (c) If instead the coil moved from left to right. the problem is similar to part (a). thus.95 cm. square coil with 120 turns and sides with a length of 1. dt The area of the loop remains constant. for x ≥ 0. what would be the answer to part (a)? (d) If instead the coil moved from left to right. as viewed from the origin (on the left). we must remember we have a coil with N = 120 turns. The direction of the induced current will be in a direction to oppose the change in flux. (a) What is the emf induced in this coil while it is to the right of the origin? (b) As viewed from the origin (which we take to be to the left). Note that our coil has 120 turns.80 × 102 T)(0. Since no sign convention is defined for the emf enter Abv for Mastering Physics. A flat coil of area A moves with uniform speed v from right to left with the plane of its area always perpendicular to this field. (d) Because the coil is moving in the opposite direction compared to parts (a) and (b). the induced current will induce a magnetic field inside the loop pointing to the right. Thus dφ/dt = ω.60 × 102 V? (a) The figure below shows the general configuration of the square coil and the magnetic field. viewed from the origin (on the left).31 ω= N BA 120(7. The magnetic field is constant and to the right. while the coil (and consequently. the vector A perpendicular to the loop) changes with time. except the velocity is positive. and so the answer is −Abv. the flux through the loop is increasing. The coil rotates in a magnetic field of 7. (a) What is the angular speed of the coil if the maximum emf produced is 2. But here. The angle φ between the magnetic field B and the vector A is given by φ = ωt. and the flux through the coil at any given time can be written as. the induced emf is given by the change of the flux with respect to time. (c) Calculate the current through the resistor. The flux at any time is the area of the loop times the magnitude of the magnetic field B. The E induced in coil B will oppose the change in magnetic flux by causing current that will create additional magnetic field to the left. since the battery voltage V does not change.0 Ω.0 Ω = 0.21 A . You can ignore the resistance of the bar and the rails. Lenz’s Law says that the E induced in the coil B will oppose the change in magnetic flux by causing current that will create an additional magnetic field to the right. the current in coil B will flow from b to a. 29-20 A 1. the current will stop flowing and the magnetic field it produces will decrease. where ∆x and ∆t are related to the speed of the bar by v = ∆x/∆t. (a) Calculate the magnitude of the emf induced in the circuit. (b) When S is closed. When S is then opened. (b) Using Lenz’s law. the relation I = V /R tells us that the current will increase. (a) The current will flow from the positive terminal of the battery when S is closed. By the right hand rule.44 V (b) Pulling the rod increases the flux by adding magnetic field directed into the page. [This argument is exactly the same as the one used in part (b)].60 m × 4. (c) Using Lenz’s law. In a time ∆t the area of the circuit will increase by L∆x.755 T magnetic field. determine the direction of the current in resistor ab of the figure when the resistance of R is decreased while the switch remains closed. (a) Using Lenz’s law. the current in coil B will flow from b to a. The E induced in coil B will oppose the change in magnetic flux by causing current that will create additional magnetic field to the left.755 T × 1. Thus. (c) If the resistance in R is decreased. The current must therefore be counterclockwise. so the flux becomes larger. The bar rides on parallel metal rails connected through R = 26. determine the direction of the current in resistor ab of the figure when coil B is brought closer to coil A with the switch closed. By Lenz’ Law.29-17 Consider the system shown below.44 V/26. Thus. by the right hand rule the current in coil B will flow from a to b. coil A produces a magnetic field inside the coils pointing to the right.60 m long metal bar is pulled to the right at a steady 4. the current flowing in coil A will produce a magnetic field inside coil A pointing to the right (and also to the right inside coil B). determine the direction of the current in resistor ab of the figure when switch S is opened after having been closed for several minutes. E = IR ⇒ I = 5. When the current through coil A increases. By the right hand rule. As coil B is brought closer to A. the induced current must then produce a magnetic field inside the loop directed out of the page.5 m/s = 5. (c) By Ohm’s Law. (a) The emf is given by E = ∆Φ/∆t = BL∆x/∆t = BLv = 0. the magnetic field inside coil B becomes stronger. (b) Find the direction of the current induced in the circuit. 0.5 m/s perpendicular to a uniform. so the apparatus makes a complete circuit. and we let L be the length of the bar. as shown in the figure. the magnetic field directed to the right will also increase. 70 × 10−2 H. = (2. ω=√ 1 2π = .00 V and negligible internal resistance. (d) In the LC oscillator.90 × 10−3 A = 6.90 mA. The period of oscillation of the circuit is then measured to be 9.5A. R 8Ω 30-29 A 7.0 V. T LC (The frequency formula is given on the equation sheet. At that time. dt dt L E 6V = = 0. (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 405 turns.60 × 10−9 F)(13 V) = 9.5 A × 8 Ω) di = = = 0. dt L 2. and the period of the oscillations T by N iA. Then i= di E − iR di = 0 =⇒ = .256 H) (0.500 A.88 × 10−8 C.715 A) Li = N 405 −4 = 4. L=N ΦB . (b) the rate of increase of current at the instant when the current is 0. (30. and we defined the self-inductance L by E = −L di dt Here we aren’t concerned with the direction of the emf. (d) Calculate the maximum current in the circuit. U = 21 LI 2 . and the total energy of the oscillator is the energy initally stored in the capacitor: U = 21 CV 2 = 21 (7.65 × 10−2 V E = di/dt 6.250 s after the circuit is closed. At one point.60 nF capacitor is charged up to 13. only the magnitude. so at that time E 6V di = = = 2. The self-induced emf is E = −N dΦ dt . C.00×10−5 s. 2011 . (a) Calculate the inductance of the coil.65 × 10−2 V.42 × 10−7 J) = 6.413 A.00 × 10−5 s)2 T2 = = 2.6)].5 H)0.75 A.45 × 10−2 A/s.42 × 10−7 J. i is in fact the definition used in the book for L [Eq. we have ´ 6V ³ ´ E ³ i= 1 − e−(R/L)t = 1 − e−(8Ω/2.52 × 10 Wb ΦB = 30-19 An inductor with an inductance of 2. dt L 2. (a) Applying Kirchhoff’s Law. 4π 2 C 4π 2 (7. and we can use the expression for energy in the inductor. (d) The general solution shows that as t → ∞. Therefore we can substitute the given values to obtain 1.60 × 10−9 F)(13 V)2 = 6. Find (a) the initial rate of increase of current in the circuit.50 H and a resistance of 8. l we can rewrite the expression for the inductance L given on the equation sheet in terms of the flux. (d) the final steady-state current.00 Ω is connected to the terminals of a battery with an emf of 6.60 × 10−9 F) (b) The capacitor has its maximum charge when it is initially charged to V = 13 V: Q = CV = (7. (b) Calculate the maximum charge on the capacitor. the current through the inductor is a maximum. the total energy U of the circuit is all in the inductor. We can use it to answer this problem by solving for ΦB . the magnitude of the self-induced emf is 1. and solve for I: r 2U I= L s 2(6. (a) The resonant frequency of an LC circuit is related to L. we find the loop equation is E − iR − L E − iR 6 V − (0. 2011 Solution to HW-18 30-7 At the instant when the current in an inductor is increasing at a rate of 6. (c) Energy is conserved. the exponential term vanishes. (0. the energy swings back and forth between the capacitor and the inductor. l l i Note that the relation above.45 × 10−2 A/s = 0.70 × 10−2 H) The initial condition is that i = 0 at t = 0. L = µ0 (b) Using the loop equation again for i = 0.715 A? (a) In the class lecture we applied Faraday’s Law to a B solenoid with N turns.4 A/s. for this case R = 0. (c) Calculate the total energy of the circuit.Physics 21 Fall.256 H L= (b) For this part we must notice that since the flux through one loop of a solenoid of length l is given by ΦB = BA = µ0 N ΦB N2 A = N µ0 A = N .) Solving for L leads to L= (9.8 A/s.5 H (c) Using the general solution for the loop equation. what is the average magnetic flux through each turn when the current is 0.5 H October 28.25 s R 8Ω = 0. then disconnected from the power supply and connected in series through a coil. (c) the current 0. (a) Calculate the maximum charge on the capacitor.760 A) (0. a) When the current has reached its final value. the second equation above gives a relationship between Qmax and Imax : √ Qmax = Imax /ω = Imax LC p = (0. S1 is opened and S2 is closed.60 ms is given by i(t) = I0 exp(−Rt/L). = 21 (0. How much time does it take for the energy stored in the inductor to decrease to half of its original value? 30-33 An LC circuit containing an 86. respectively.) The current is given by (c) The energy stored in the inductor at t′ = 2.760 A.60 ms of oscillation. the energy stored in the inductor is 0.115 H) q(t) = Qmax cos(ωt) dq = −ωQmax sin(ωt) = −Imax sin(ωt) i(t) = dt √ where ω = 1/ LC. The inductance is L = 0.260 J)(120 Ω)2 2 = 255 V.54 × 10−4 Hz (b) When the switches are changed.115 H. and the resistance is R = 120 Ω. the energy stored in the inductor is given by U0 = 21 LI02 = 12 L µ E R ¶2 .086 H)(1. Since q(t) and i(t) are related by a time derivative.115 H ln 2 = ln 2 = 0. (0. Solving this equation for t we find t= L 0. (a) The charge on the capacitor oscillates as a function of time.30-22 In the figure below. We want to find the time t when the inductor has half of its initial energy.0026 s) where I0 is the steady state current from part (a). and this will lead to a current that decays exponentially with time constant L/R (see page 1044 of the text.25 × 10−9 F) = 1.086 H)(0.0760 A)2 £ ¤ × sin2 2π(1. (b) Calculate the oscillation frequency of the circuit. switch S1 is closed while switch S2 is kept open.332 ms. What is the emf E of the battery? b) After the current has reached its final value.260 J. 2R 2 (120 Ω) U (t′ ) = 21 L[i(t′ )]2 2 = 12 LImax sin2 (2πf t′ ) = 7. exp(−2Rt/L) = 21 . calculate the energy stored in the inductor after 2.88 × 10−6 C (b) The oscillation frequency of the circuit is related to ω by: f= 1 ω 1 p √ = = 2π 2π LC 2π (0. then we know what the charge and current will look like as functions of time: (a) When the current has reached its final value I0 . We can use this result to determine the energy stored in the inductor as a function of time: U (t) = 21 Li(t)2 = 12 LI02 exp(−2Rt/L) = U0 exp(−2Rt/L). (c) Assuming the capacitor had its maximum charge at time t = 0. Assuming the capacitor starts with the maximum charge at t = 0 (as stated in part c).54 × 10−4 hz)(0. We can rearrange this to solve for the emf and substitute our known values to find E= µ 2 UR L ¶1 2 2 = µ ¶1 2(0. where we used Ohm’s Law to express the current in terms of the emf and resistance. The maximum quantities correspond to the charge and current at times when the cosine and sine are equal to one.25 nF capacitor oscillates with a maximum current of 0. At that time. the inductor will release the energy stored in the magnetic field.25 × 10−9 F) = 7.0 mH inductor and a 1.03 × 10−3 J .086 H)(1. 2011 .1042 Mastering physics wants an answer in revolutions per minute.1 cm. then the flux through the loop will change because number of field lines going through the loop will be changing. To have an acceptable coil resistance. Besides plenty of wire.Physics 21 Fall. If the loop is rotated within the magnetic field.7 rad/sec = N BA 450 × 1.7 rad 60 sec 1 rev · · = 130. you are assigned the project of designing a generator of sinusoidal ac voltage with a maximum voltage of 120 V. the flux will change with time: ΦB (t) = N BA cos(ωt) The time derivative of this gives the induced EMF: E =− dΦB = N BAω sin(ωt) dt The amplitude is then N BAω. Solving for ω ω= 120 V E = 13. where A is the area. At what time does the displacement current in the dielectric equal 23 µA? 29-51 As a new electrical engineer for the local power company. The basic design should consist of a square coil turning in the uniform magnetic field. you have two strong magnets that can produce a constant uniform magnetic field of 1.8 T over a square area with a length of 10. For t > 0 the electric flux through the dielectric is (7800 V m/s3 )t3 . dt dt The flux through a square loop perpendicular to the magnetic field with the maximum possible dimensions and N loops of wire will be ΦB = N BA. we multiply ǫ0 by the dielectric constant K to get the permittivity of the material ǫ = Kǫ0 .5 s = 3.8 × . Thus we E have iD = ǫ dΦ dt . The dielectric is ideal and nonmagnetic. the coil can have at most 450 loops. Rotating at an angular frequency of ω. The function for the electric flux can then be substituted: ¡ 3¢ ¢ d ¡ 3 3 3 d t (7800 V m/s )t = ǫ(7800 V m/s ) iD = ǫ dt dt = ǫ(7800 V m/s3 )(3)t2 Solving this for time: s iD t= ǫ(7800 V m/s3 )(3) s 23 × 10−6 = 5.4 cm on a side when the magnets are separated by a distance of 12. When a dielectric is between the capacitor plates. 2011 Solution to HW-19 29-34 A dielectric of permittivity 3.3 × 10−11 F/m completely fills the volume between two capacitor plates. so the conversion is: 13. For a charging capacitor we have the equation for displaceE ment current iD = ǫ0 dΦ dt . I Z d dΦ E · dl = − B · dA = − . the conduction current in the dielectric is zero.8 rpm sec 1 min 2π rad November 4.3 × 10−11 (7800 V m/s3 )(3) We can use Faraday’s law to calculate the induced EMF. 4 V (e) Using the values of VL and VR just calculated. we can easily find the phase angle from the diagram above: φ = arctan 14. VR 24.70µF. (a) What is the impedance of the circuit? (b) What is the current amplitude? (c) What is the voltage amplitude across the resistor? (d) What is the voltage amplitudes across the inductor? (e) What is the phase angle φ of the source voltage with respect to the current? (f) Does the source voltage lag or lead the current? (g) Construct a phasor diagram. (a) At what angular frequency will they have the same reactance? (b) If L = 4.0 Ω.0 V and an angular frequency of 250 rad/s.430 H inductor. (a) What frequency is required? (a) The frequency f is related to angular frequency ω by the equation ω = 2πf .430 H)2 = 210 Ω.8◦ .00 mH (part of the circuitry for a radio receiver) to be 2. 1 = ωL = 36. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 28. is p p V = (ωLI)2 + (RI)2 = (ωL)2 + R2 I = ZI.4V VL = arctan = 30.0V VL = f= −3 2πIL 2π(2.80 × 10−3 )(3. (b) We’ve already written the relation between V and I.10 × 10 A)(5.134 A)(200 Ω) = 24.70 × 10−6 ) 31-14 You have a 180 Ω resistor and a 0.00 × 10−3 H) = 182 kHz φ VR = RI The diagram is simpler since there is no capacitor. Therefore the impedance Z is given by p Z = (180 Ω)2 + (250 rad/s × 0. The voltage V of the power supply.134 A) = 14. We can find it by drawing the phasor diagram: VL = ωLI = 7.0 V V = = 0. ωC 31-10 You want the current amplitude through an inductor with an inductance of 5. (g) The diagram is shown above.0 V is applied across the inductor. so I= 28. (Remember the current is in phase with VR ).0V (f) From the phasor diagram above it is clear that the power supply voltage V leads the current. Thus the frequency is 12.0 V (d) The voltage amplitude across the inductor is the inductive reactance XL times the current: VL = ωLI = (250 rad/s)(0. shown by the dashed line. The amplitude of the voltage across the inductor is VL = ILω = IL2πf .134 A = 134 mA.50 × 103 rad/s. . V = ZI (c) Again substituting numbers. (a) The impedance relates the peak values of the voltage and current.80 mH and C = 3. Z 210 Ω (c) To find the voltage amplitude across the resistor we use Ohm’s law and the current from part (b): VR = IR = (0. LC (b) Substituting numbers.10 mA when a sinusoidal voltage with an amplitude of 12. what is the numerical value of the angular frequency in part (a)? (c) What is the reactance of each element? (a) Equate the reactances and solve for ω: 1 = ωL ωC 1 = ω 2 LC ⇒ ⇒ 1 ω=√ .31-6 A capacitance C and an inductance L are operated at the same angular frequency. ω=p 1 (4.430 H)(. (a) We need to look at the relationship V = IZ.1 V. (c) Now. and a variable-frequency ac source with an amplitude of 2. a 0. This means that voltage lags current (or.408 H inductor.7 V V is the “vector sum” of the rms voltages across all the circuit elements. November 4.408 H) − 1 (410 rad/s) (4.9 V. current leads voltage) at the power supply. Z depends on ω via XL and XC : q 2 Z = R2 + (XL − XC ) Clearly.99 µF) (b) When XL − XC = 0. Z = R. will the source voltage lead or lag the current? 31-23 In an LRC series circuit. Since V is fixed.Physics 21 Fall. 2011 . However we cannot just add the rms voltages together since they are all not in phase with each other.93 V 2 (207 Ω) + (−322 Ω) 2 = 7.9 V)2 + (50. in order to maximize I we need to minimize Z. What is the rms voltage of the source? Since this is a series circuit (there is only one loop).2 V.408 H) (4. across the capacitor is 89. We can use a phasor diagram to add the voltages together vectorially. (a) At what frequency will the current in the circuit be greatest? (b) What will be the current amplitude at this frequency? (c) What will be the current amplitude at an angular frequency of 410 rad/s? (d) At this frequency.99 µF) = − 322 Ω I= V =q Z 2. a 4.99 µF capacitor. so the rms source voltage is V0 = 49. the voltage supplied by the source will be equal to the sum of the instantaneous voltages across all the other circuit elements.93 V. p V = (VR )2 + (VL − VC )2 p = (30.66 × 10−3 A (d) Since our calculation of XL − XC in part (c) came out negative. You connect all four elements together to form a series circuit. the circuit is mostly capacitive. the rms voltage across the resistor is 30. so I= 2.93 V V = = 14. 1 = ωL ωC 1 ω2 = LC 1 1 = 701 rad/s ω=√ =p LC (0. 2011 Solution to HW-20 31-21 You have a 207 Ω resistor. Z is minimized when XL − XC = 0.2 V − 89. equivalently.2 × 10−3 A Z 207 Ω From this we see we can use the Pythagorean theorem to determine the total voltage across all three circuit elements. XL − XC = (410 rad/s) (0. Using the definitions of XL and XC . and across the inductor is 50.7 V.1 V)2 = 49. 7 W 31-59 In an LRC series circuit the magnitude of the phase angle is 50.3◦ ) (120 V)2 V2 = = 440 Ω P 32. The average power delivered by the source is 135 W.73 A R 4.3◦ ) = ωLI − 348I ωL − 348 = .861 A (c) Now use Pav = 135 W = Irms Vrms cos φ. (b) Find the rms current. The reactance of the capacitor is 348 Ω.0 V)(0. Irms = 12 V Vrms = = 2.861 A) cos(−50.40 Ω.7 W (d) We want the power on both ends of the transformer to be the same. So.3◦ with the source voltage lagging the current. and the resistor resistance is 182 Ω.73 A)(12 V) = 32.0 V Irms = = = 0. (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the source line (which has a voltage of 120 V) would draw the same power as the transformer? (a) The ratio of the number of primary to secondary turns is equal to the ratio of the voltages of the primary to the secondary part of the transformer. ωLI 31-37 A transformer connected to a 120 V ac line is to supply 12. (a) Find the reactance of the inductor. (b) The rms current is related to the average power by p p Irms = Pav /R = 135 W/182 Ω = 0.742 A Z 120 Ω and the power factor is given by cos φ = 74.0 Ω R = = 0.7 W . We know everything in this equation except Vrms .742 A)(0. so we solve to get 135 W = 245 V Vrms = (0. we can solve the average power as Pav = Vrms Irms cos φ = (89. The load resistance in the secondary is 4. Np Vp 120 V = = = 10.617 Z 120 Ω Thus.0 Vrms to a portable electronic device.4 Ω (c) We calculate the average power delivered to the load by using: Pav = Irms Vrms = (2. (c) Find the rms voltage of the source. 182I 182 We can easily solve for the reactance of the inductor: ωL = 129 Ω.0 Ω and an impedance at this frequency of 120 Ω. Ns Vs 12 V (b) We can determine the rms current supplied since we know the load resistance and the rms voltage supplied by the transformer.617) = 40.0 V. The circuit has a resistance of 74.3o V R= I = 348 I ωC (a) From the diagram we see that tan(−50. the root-mean-square current is given by Vrms 89. so we can calculate the resistance that would draw this power since we know: P = IV = RI = 182 I V2 R Solving for R and substituting the power found in part (c) and the source voltage we find: φ = -50. What average power is delivered to the circuit by the source? The average power in an ac circuit is given by Pav = Vrms Irms cos φ In an L-R-C circuit.31-28 An L-R-C series circuit is connected to a 120 Hz ac source that has Vrms = 89. 5. From the graph. Because the earth is solid. where ρ is the mass per unit length. 4 November 11. t) = A sin(kx + ωt + φ). Since the maximum amplitude on the graph is 1.183 × 1 = . it can support both longitudinal and transverse seismic waves.0 Hz is given. travel at a slower 4500 m/s. and f = 5.49 m v = f λ ⇒ f = v/λ = 180/4.0458 W = 4. Let d be the distance from the epicenter of the earthquake to the detector. 2 .40 rad/m.90 g and length 75. where T is the tension (the equation sheet uses µ for the mass per unit length): 30.0 min = 120 s) gives d= 4500 × 8000 × 120 = 1230 km. v = T /ρ. and A is the wave’s amplitude.75 (b) If the wave amplitude is halved. 8000 − 4500 15-20 A piano wire with mass 2.0 v = T /ρ = = 88. Then tS − tP = ∆t is the delay time for the S wave. These travel at different speeds.9 × 10−3 ) /0.0 Hz wave traveling to the left. (b) What happens to the average power if the wave amplitude is halved? D(x. then the average power 2 will change by a factor of 12 = 14 because power depends on the square of the amplitude. v increases by √ 2. What are the speed and phase constant of the wave? K20-52 Earthquakes are essentially sound waves traveling through the earth. We already know the graph is at t = 0. and d d − = ∆t vS vP 1 1 d − = ∆t vS vP vP − v S = ∆t d vS vP vS v P ∆t d= vP − v S Substituting numbers (2. (2. (3) in the waves handout. called S waves.0029 (. The speed of longitudinal waves. We need to calculate the velocity of the wave with v = T /ρ.0 m. (a) Calculate the average power carried by the wave. What is its wavelength? What is its frequency? k = 2π/λ ⇒ λ = 2π/k = 2π/1.0015)2 (2π × 110)2 (88. so if the tension T is doubled. They are called seismic waves. although actual seismic waves follow more complex routes. 2011 .0 sin(k0 + ω0 + φ) = sin φ ⇒ φ = 30 .5 = 1. A seismograph records the two waves from a distant earthquake. What is the speed if the tension is doubled? From Eq.75 Using ρ = m/L we can calculate the average power P = 1 . A wave with frequency 110 Hz and amplitude 1. we have ◦ 0.1 m/s. Its wave number is 1. 2011 Solution to HW-22 K20-7 The wave speed on a string under tension is 170 m/s. K20-11 A wave travels with speed 180 m/s. If the S wave arrives 2.50 mm travels along the wire.0 N.1 Hz K20-42 This is a snapshot graph at t = 0 of a 5. Hence v = λf = 10 m/s.58 × 10−2 W. (a) The average power carried by the wave is given by the formula P = 12 ρA2 ω 2 v. Then the travel time for the P wave is tP = d/vP and the travel time for the S wave is tS = d/vS .49 = 40. called P waves.0 min after the P wave.Physics 21 Fall. the amplitude is abut 0. is 8000 m/s. λ = 2.40 = 4.1) = 0. At x = 0. Transverse waves.183 W.0 cm is stretched with a tension of 30. how far away was the earthquake? You can assume that the waves travel in straight lines. 0. 33 mm.504 × 10−3 kg/m)(106.15-24 A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is D(x.35 m so T = ρv 2 = (2. length 1. (c) the wavelength. k 6. (e) the direction the wave is traveling. (b) the frequency is given by f= 742 rad/s ω = = 118 Hz 2π 2π (c) The wavelength is given by λ= 2π 2π = = 0. Being more practical-minded. (b) the frequency.35 m and a mass of 3.504 × 10−3 kg/m.39 W P = . Determine (a) the amplitude.38 × 10−3 kg mass = = 2. (f) We get the tension T from v = T /ρ ⇒ T = ρv 2 The linear mass density ρ is ρ= 3.38 g.98 rad/m)x + (742 rad/s)t]. and (g) the average power transmitted by the wave. t) = (2.33 mm) cos[(6.3 m/s)2 = 28.3 N (g) The average power transmitted by the wave is given by: 1 2 2 1 1 ρA ω v = ρA2 ω 2 T /ρ = ρT A2 ω 2 2 2 2 1 = (2. (f) the tension in the rope. k 6.98 rad/m (d) The wave speed is v= ω 742 rad/s = = 106 m/s.98 rad/m (e) The wave travels in the −x direction since the argument has the form of kx + ωt.50 × 10−3 )(28. (a) Comparing the above equation with the general form given in class. t) = A sin(kx − ωt + φ). (d) the wave speed.3) (2. one sees that the amplitude is A = 2. you measure the rope to have a length of 1.90 m.33 × 10−3 )2 (742)2 2 = 0. D(x. then she draws her bow across it. At a depth of 1000 m. We can relate the changes in velocity and temperature: ∆v ∼ 4.11 × 10−2 ◦ C.50 cm at the same time.51 cm = 8. K 21-40 A violinist places her finger so that the vibrating section of a 1. What is the wavelength of the waves emitted by the source? We know the velocity of a wave is given by: v = λf We know the frequency of the speed of sound as it propagates through water so we can determine the wavelength emitted.0 cm. We also need to know that the speed of a wave is equal to the product of wavelength and frequency. dv v v where we dropped the minus sign.0 kHz. λ= 1482 ms v = = 0. A listener nearby hears a note with a wavelength of 60. ⇒ ∆T ∼ 2 v D(4.0 m/s)∆T. v = 1480 m/s. the sound waves travel 7700 km.0 cm.Physics 21 Fall.0 s. 1 1 1 1 − fbeat = v − = 344 m/s λa λb 6.3 m × 572 Hz)2 = 130 N.0◦ C increase in temperature.10 g/m string has a length of 30. fbeat = fa − fb .0 m/s for every 1.0 m/s) Substituting ∆t = 1 s. Since in general λf = v. The speed of sound in water (assumed to be uniform at 20◦ C) is 1482 m/s.0 m/s.6 m Substituting numbers. Then their travel time is t = D/v. v = λf . Take the speed of sound in air to be 343 m/s. 2011 Solution to HW-23 K 20-53 One way to monitor global warming is to measure the average temperature of the ocean. where sounds generated near California are detected in the South Pacific. what is the smallest change in average temperature that can be measured? Let D be the distance the waves travel and v be their velocity. we have f= 343 m/s vair = 572 Hz = λair 0. (4. ∆T so ∆v ∼ (4. In one experiment.0 m/s)∆T. T = (1.51 cm instead.0Hz D v2 ∆t. If the smallest time change that can be reliably detected is 1. 2011 . we obtain ∆t ∼ YF 16-75 The sound source for a ship’s sonar system operates at a frequency f = 23. Substituting the expression for ∆v into the expression above for ∆t. Researchers are doing this by measuring the time it takes sound pulses to travel underwater over large distances. D dt D = − 2 ⇒ ∆t ∼ 2 ∆v. where ocean temperatures hold steady near 4◦ C.10 × 10−3 kg/m)(2 × 0. the average sound speed is 1480 m/s. It’s known from laboratory measurements that the sound speed increases 4. and D = 7700 km gives ∆T ∼ 7. What is the tension in the string? The key point is that the sound in air will have the same frequency as the vibrating violin string.0644 m f 23000 Hz 16-38 Two guitarists attempt to play the same note of wavelength 6.50 cm 6. What is the frequency of the beat these musicians hear when they play together? Beats are heard when two tones with slightly different frequencies fa and fb are sounded together. but one of the instruments is slightly out of tune and plays a note of wavelength 6. November 18. Thus. 000215 m) π 21400 kg/m3 = 74. If a vibrating tuning fork of just the right frequency is held next to the wire. K 20-50 One cue your hearing system uses to localize a sound (i. multiply the number of seconds by 106 . where v = T /µ. To convert to µs. The wire is under tension T = mg because of the gravitational force acting on the weight. Your ears are spaced approximately 20 cm apart.5 m)(0.440 kg)(9. (a) The problem here is to find the vibrational frequencies of a stretched wire held fixed at both ends.80 m/s2 ) 1 = (0.500 m. What tuning-fork frequencies will cause this to happen? You may assume that the bottom end of the wire (to which the mass is attached) is essentially stationary. The result is 412 µs. We find µ by dividing the total mass of the wire by its length: µ= volume × density (πd2 /4)Lρ πd2 ρ mass = = = . while a mass m = 440 g is attached to the other end so that the wire hangs vertically under tension. we have T /µ 4mg/(πd2 ρ) v 1 mg f= = = = λ 2L 2L Ld πρ (0.3 YF 16-67 A platinum wire (density ρ = 21. and µ is the mass per unit length of the wire. and a heavy weight is at the other end..e. L L L 4 Putting all these pieces together. One end of the wire is attached to the ceiling. and that the tension in the wire is essentially constant along its length.5 Hz. What is the difference in arrival times? Give your answer in microseconds. to tell where a sound is coming from) is the slight difference in the arrival times of the sound at your ears. the wire begins to vibrate as well. Any positive integer multiple of this frequency is a vibrational frequency of the wire. Both ends of the wire are fixed: one end is attached to the ceiling. Consider a sound source 5.0 m from the center of your head along a line 45◦ to your right. .4 g/cm ) has diameter d = 215 µm and length L = 0. The fundamental mode will have a wavelength λ = 2L and a frequency f that satisfies λf = v. 02 − x. 1. We solve for this frequency using x= fmin = 344 m/s v = = 86 Hz.86 m. and lowest frequency is from n = 0. you will find that the maximum value of LA − LB is 2.8 m.26 m. λ 4m November 18. (a) The speakers are in phase.0 m.0 m B Speaker A is 3. To calculate the wavelength λ. 3. For any frequency such that 12 λ is greater than 2. x2 =1. so λ = 2. Consider point Q along the extension of the line connecting the speakers. LA −LB must be an integer number of wavelengths. the wavelength increases. Both speakers emit sound waves that travel directly from the speaker to point Q.0 m. At what distances from B will there be (a) destructive interference and (b) constructive interference? (c) If the frequency is made low enough. Now solve the equation above for x: x2 + 2. so the path difference is 2. there will be no positions where LA −LB is 2. By calculating several values of LA − √ LB + x2 + 2.00 m to the right of speaker A. and that LA − LB gets smaller and smaller as x increases.3 m. use λ= 344 m/s v = = 0. x3 =2. (b) For constructive interference. The frequency is 784 Hz and the speed of sound in air is 344 m/s.] Hence take λ = 4. nλ. 2011 Solution to HW-24 16-33 Two loudspeakers. or LA − LB = n + 12 λ.0 m (b) For destructive interference the path difference must be 0. we get x0 =0.0 m. 4. How low must the frequency be for this to be the case? A LA 2.0 m from Q. Hence λf = v ⇒ f = 343 m/s v = = 172 Hz λ 2. x1 =0.02 − x = (n + 12 )λ.71 m. and x3 =4. 2nλ Evaluating x for several n gives x0 =0. A similar analysis then gives 4 − n2 λ2 . there will be no positions along line BC at which destructive interference occurs. x2 =1. 2.00 m to the right of speaker B. are driven by the same amplifier and emit sinusoidal waves in phase. We choose n = 1 to get the lowest frequency. and x4 =9. [(n+ 12 )λ gives destructive interference. A and B. Speaker B is 2.27 m.00 m and f= v 343 m/s = = 86 Hz λ 4. x1 =0. and speaker B is 1.0 m from Q.439 m. A small microphone is moved out from point B along a line perpendicular to the line connecting A and B (line BC in the figure).00 m apart. (a) What is the lowest frequency for which constructive interference occurs at point Q? (b) What is the lowest frequency for which destructive interference occurs at point Q? 16-70 Two identical loudspeakers driven by the same amplifier are located at points A and B.Physics 21 Fall. 2011 . so for constructive interference. 1.026 m.5 wavelengths. 12 λ = 2 m or λ = 4 m. f 784 s−1 The condition for destructive interference is LA − LB = x2 + 2.02 = x + (n + 12 )λ x2 + 4 = x2 + (2n + 1)λx + (n + 12 )2 λ2 x= 4 − (n + 12 )2 λ2 2(n + 12 )λ Substituting n=0. 2. the path difference must be an integral number of wavelengths (nλ). (c) As the frequency decreases. Hence at the cutoff frequency fmin . and the mimimum value LA − LB = 12 λ needed for destructive interference increases.0 m. where n is an integer.53 m.0 m at x = 0.0 m LB = x (a) There will be destructive interference when the difference of the path lengths from each speaker tothe microphone is a half integer wavelength.0 m. 32-11 Radio station WCCO in Minneapolis broadcasts at a frequency of 830 kHz. At a point some distance from the transmitter, the magnetic field amplitude of the electromagnetic wave from WCCO is 4.92 × 10−11 T. (a) Find the wavelength. (b) Find the wave number. (c) Calculate the angular frequency. (d) Calculate the electric field amplitude. (a) For any wave, v = f λ, and for a radio wave v = c. Hence λ= 3.00 × 108 m/s c = = 361 m. f 8.30 × 105 Hz (b) Since we now have the wavelength, λ, we can use k= 2π = 1.74 × 10−2 m−1 . λ (c) We were given the frequency of the wave, f , so to find the angular frequency, ω, we use ω = 2πf = 5.22 × 106 rad/s. (d) For an electromagnetic wave, the amplitude of the electric field is c times the amplitude of the magnetic field: E0 = cB0 = (3.00 × 108 m/s)(4.92 × 10−11 T) = 1.48 × 10−2 V/m. 32-16 Consider each of the electric- and magnetic-field orientations. (a) What is the direction of propagation of the wave if E = E ˆi, B = −B ˆj. (b) What is the direction of propagation of the wave if E = E ˆj, B = B ˆi. (c) What ˆ is the direction of propagation of the wave if E = −E k, B = −B ˆi. (d) What is the direction of propagation of the ˆ In each case, express the diwave if E = E ˆi, B = −B k. rection of the propagation vector as a unit vector. Its three components should be entered as x, y, z, separated by commas. For example, if the wave propagates only in the −x direction, enter −1, 0, 0. The direction of propagation is the direction of E × B. By remembering cyclic order, it’s easy to evaluate the product ˆ ˆj × k ˆ = ˆi, and k ˆ × ˆi = ˆj. of any two unit vectors: ˆi × ˆj = k, ˆ ˆ ˆ ˆ ˆ (a) The direction is i × (−j) = −(i × j) = −k. Enter 0, 0, −1. ˆ Enter 0, 0, −1. (b) The direction is ˆj × ˆi = −k. ˆ × (−ˆi) = k ˆ × ˆi = ˆj. Enter 0, 1, 0. (c) The direction is (−k) ˆ ˆ ˆ ˆ = ˆj. Enter 0, 1, 0. (d) The direction is i × (−k) = −(i × k) Physics 21 Fall, 2011 Solution to HW-25 32-15 We can reasonably model a 60 W incandescent lightbulb as a sphere 5.2 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. a) What is the visible light intensity at the surface of the bulb? b) What is the amplitude of the electric field at this surface, for a sinusoidal wave with this intensity? c) What is the amplitude of the magnetic field at this surface, for a sinusoidal wave with this intensity? (a) The intensity of light is given by the power per unit area, so the visible light intensity Ivis at the surface of the light bulb is the power due to visible light divided by the surface area of the light bulb: Ivis = Pvis 0.05Ptot (0.05)(60.0) = = = 353 W/m2 . A 4π(d/2)2 4π(0.026)2 (b) The intensity at the surface of the bulb calculated above is just the magnitude S¯ of the time averaged Poynting vector, which is related to the electric field amplitude E0 by S¯ = 12 0 cE02 . Using this expression we can solve for the amplitude of the electric field: √ √ 2S¯ 2Ivis E0 = = = 516 V/m. 0 c 0 c (c) The amplitudes of the electric and magnetic fields are related by B0 = E0 = 1.7 µT. c We could have found the magnetic field amplitude directly ¯ from S¯ using an alternate expression for S: √ c 2µ0 S¯ S¯ = 12 B02 =⇒ B0 = . µ0 c 32-41 A small helium-neon laser emits red visible light with a power of 3.40 mW in a beam that has a diameter of 2.00 mm. (a) What is the amplitude of the electric field of the light? (b) What is the amplitude of the magnetic field of the light? (c) What is the average energy density associated with the electric field? (d) What is the average energy density associated with the magnetic field? (e) What is the total energy contained in a 1.00 m length of the beam? (a) The amplitude of the electric field of the light is related to the time average of the Poynting vector S¯ by 1 S¯ = 0 cE02 . 2 The time average of the Poynting vector is in units of power per unit area; it can be calculated by dividing the power given by the area of the beam: .0034 S¯ = W/m2 = 1082 W/m2 . π0.0012 Solving the above equation for electric field yields √ √ 2S¯ 2 × 1082. = = 903 V/m E0 = 0 c 0 c (b) One can find the amplitude of the magnetic field in a similar manner, using another relation from the equation sheet: √ ¯ 0 1 2Sµ c 2 S¯ = B0 =⇒ B0 = = 3.01 × 10−6 T. 2 µ0 c One could also use B = E/c. (c) The energy density associated with the electric field is uelec = 12 0 E 2 . However, this expression gives the instantaneous energy density at the time when the electric field magnitude is E. To find the average energy density associated with the electric field, you must use the time-averaged value of E 2 . Because the electric field is sinusoidal, E(t) = E0 cos(ωt) and E(t)2 = E02 cos2 (ωt). The time average of cos2 ωt = 21 , so huelec i = 12 0 hE 2 i = 41 0 E02 = 1.80 × 10−6 J/m3 (d) Similarily, the average density associated with the magnetic field is 1 2 1 hB 2 i = B = 1.80 × 10−6 J/m3 . humag i = 2µ0 4µ0 0 Note that the average energy in the electric field is the same as the average energy in the magnetic field. (e) To find the total energy contained in a 1.00 m length of the beam, use the total (average) energy density multiplied by the volume of that length. The total energy density is hutot i = huelec i + humag i = 3.60 × 10−6 J/m3 The volume of a 1.00 m length of beam is V = πr2 L = π × .0012 × 1.0 = 3.14 × 10−6 m3 , and the total energy is hutot iV = 1.13 × 10−11 J. December 2, 2011 32-18 A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.500 m2 . At the window, the electric field of the wave has an rms value 2.60 × 10−2 V/m. How much energy does this wave carry through the window during a 30.0 s commercial? The Poynting vector gives the energy per unit area per unit time carried by the electromagnetic wave. To get the energy carried through the window in 30 s, we just multiply the magnitude S¯ of the time-averaged Poynting vector by the area of the window and then by the length of the time interval. (Since the wave propagates perpendicular ∫ to the win¯ dow, the surface integral over the window is S · dA = SA.) ¯ S is given on the equation sheet in terms of the electric field amplitude by The fact that we get a negative result means that the image is inverted. Mastering Physics only asks for the size of the image in this part, so enter the absolute value of the result. (c) The image is inverted. (d) The fact that we got a positive result in part (a) means that the image is in front of the mirror and it is a real image. (This image could be focussed on a screen.) (e) Here is the ray diagram for this situation: S¯ = 12 0 cE02 . Since we are given Erms , it is convenient to √ replace the peak amplitude E0 by Erms , using Erms = E0 / 2. The result is 2 . S¯ = 0 cErms The energy is then 2 ¯ E = SA∆t = 0 cErms A∆t −12 = (8.85 × 10 )(3.00 × 108 )(0.026)2 (0.500)(30.0) = 2.69 × 10−5 J. 34-5 An object 0.550 cm tall is placed 17.0 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Determine the position of the image. (b) Determine the size of the image. (c) Determine the orientation of the image. (d) Determine the nature (real or virtual) of the image. (e) Make a ray diagram and bring it to recitation. Since we are dealing with mirrors, the equations we will use are 1 1 1 + = do di f and m= hi di =− ho do The focal length f is positive since the mirror is convex, and f= R 22.0 cm = = 11.0 cm 2 2 We take do = 17.0 cm to be positive since it is in front of the mirror. We take ho = 0.550 cm to be positive since the object is always assumed to be upright. (a) 1 1 1 = − =⇒ di f do f do 11.0 cm · 17.0 cm di = = = 31.2 cm do − f 17.0 cm − 11.0 cm (b) hi = −ho di 31.2 cm = −0.550 cm = −1.01 cm do 17.0 cm YF 34-8 An object is a distance of 25.0 cm from the CENTER of a silvered spherical glass Christmas tree ornament which has a diameter of 5.70 cm. (a) What is the position of its image? Use the mirror equation to answer this question, but draw a ray diagram and bring it with you to recitation. Hint: be careful to determine do (the distance to the SURFACE of the mirror) correctly. (b) What is the magnification of its image? 6 cm from the center of the mirror? (b) What is the distance di to the image? (c) Is the image real or virtual? Use the mirror equation to answer this question.5 cm behind the mirror. .49.25 cm.25 11.YF 34-14 A spherical. do 11. 1 1 1 1 1 1 = − = − = di f do 16. or −16.5 = = 3. concave. (a) The magnification is m=− di 40.6 (c) The image is virtual.6 −40. Work part (b) first: (b) The focal length is half the radius of curvature. (a) What is the magnification of a person’s face when it is a distance do = 11. (distances in cm.) The sign is positive for a concave mirror.5 The image is 40. shaving mirror has a radius of curvature of 32. but draw a ray diagram and bring it with you to recitation.5 cm. Physics 21 Fall. The index of refraction of air is 1. 2011 . use the magnification m=− do = Inverting both sides of the equation.00.41 cm = 7. Find the speed of light in the plastic. The image is 80. 33-9 Light traveling in air is incident on the surface of a block of plastic at an angle of 61.7 − = f R1 R2 ∞ −13 18.57 1 25 = 72. f do di do |m|do do |m| f= 1 − 1 18. with n = 1.0 The image height hi is hi = mho = −2. then R1 = 13 cm and R2 = ∞.00 mm tall is placed 25. |m| 80 do = 7.57 25 di We solve these two equations by substituting the second into the first. |m| + 1 80 + 1 (b) In the last part we found that m was negative.57 cm The value of f determined can be substituted into the lens equation: 1 1 1 = + f do di 600 cm L = = 7. Using this in the equation for the speed of the wave we obtain vplas = c/nplas = c sin(θplas ) sin(47. Thus the index of refraction of the plastic is nplas = sin(θair )/ sin(θplas ).00 mm) = −8. do + di = L di = |m|do (a) Using the formula for the focal length of a lens with two curved surfaces (the Lensmakers’ Equation). so the lens is converging.41 cm. (d) If the lens is reversed. The lens projects an image of the slide onto a wall 6.00 m to the right of the slide.89(3. we have ( ) ( ) 1 1 1 1 1 1 = (n − 1) − = 0. Since we know do is also positive. All the results are the same. and R2 = −13 cm. The left surface of this lens is flat.57 cm The value of 1/f is exactly the same as we had before. We can write two equations for do and di from the values of L and |m| given. (c) Is the image real or virtual? Erect or inverted? 34-28 A photographic slide is to the left of a lens. R1 = ∞. December 2.53 × 108 m/s.3◦ to the normal and is bent so that it makes a 47. we can conclude that the magnification m = −di /do will be negative.7. therefore. The speed of light in the material is given by the equation v = c/n. we find The result is: 1 1 1 = + =⇒ di = 18.0 cm.7) − = f 13 ∞ 18.67mm (c) The image is real because the image distance is positive.00 × 108 ) sin(θair ) sin(61. The focal length is therefore: ( ) 1 1 1 1 = (0. the right surface has a radius of curvature of magnitude 13. (a) Calculate the location of the image this lens forms of the insect.70. 2011 Solution to HW-26 34-23 An insect 3. which leads to 72.3◦ ) = 2. (b) Calculate the size of the image.0 cm to the left of a thin planoconvex lens. di is positive. The image is inverted because the magnification is negative. and the index of refraction of the lens material is 1.8◦ ) = (3.2 cm b) To calculate the size of the image. (c) We get the focal length of the lens from the lens equation: ( ) 1 1 1 1 1 1 1 = + = + = 1+ . (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging? lens do di do + di = L (a) The image is on the opposite side of the lens. Snell’s Law states nair sin(θair ) = nplas sin(θplas ).0 times the size of the slide. which implies an inverted image.89 do 25.8◦ angle with the normal in the plastic.2 di =− = −2.32 cm |m| + 1 80 + 1 (d) The focal length f is positive. the image is upright.69 =− = −0. (b) Determine the size of the object. the sign of di is negative. the lens is moved away from or toward the film. ray 1 F (b) Obtain the image size from the magnification: hi 8. In order to find the image the rays must be extended back behind the lens (shown by dotted lines) in order to create a virtual image. Therefore.3 mm m −(−16. (d) The object and image are on the same side of the lens.00 mm tall.69 cm The distance di from the film to the lens is 8. do 7.90 cm = −39.5 cm to the right of the lens. (c) Is the image erect or inverted? (d) Are the object and image on the same side or opposite sides of the lens? Use the lens equation to answer this question. 17. we see that the image will not fit upright on the film.03 cm m (c) The sign of ho is positive. . If you take a picture of your friend. its distance di will be negative.5 390 8. you could tilt the camera and fit the 39 √mm image along the film’s diagonal. In a lens system a virtual image is produced on the same side as the object that creates the image.00 mm = = 12. using a camera with a lens with an 85-mm focal length.4) (c) From the ray diagram and since m > 0.4 cm The ray diagram shows the relative positions of the image and object. fit on film that is 24 × 36 mm? lens ho hi do di (a) Use the Lens Equation with do = 390 cm and f = 8. This result can also be determined by a ray diagram.) Comparing the height of the image to the size of the film. We can calculate the object distance from the Lens Equation.03 cm do f di 12 cm −17 cm (b) The size of the object is related to the size of the image by the magnification. 34-32 A converging lens with a focal length of 12.5/25.69 cm.00 mm tall.0223 × 175 = −3. 2. Of course.5 cm: 1 1 1 1 1 1 1 1 1 + = ⇒ = − = − = do di f di f do 8. hi = mho . who is 175 cm tall.8 cm =− = 2. (b) Determine the size of the object.0 mm (The height is negative because the image is inverted.331 cm. This situation occurs when the object is closer than the focal point. who is standing 3. since that is the way we’ve usually done it in clsss. The position of the object can be found by using the thin lens equation: 1 1 1 1 1 = − = − =⇒ do = 7. then draw a ray diagram and bring it to recitation.5 25.0 cm forms a virtual image 8. we can infer that it is on the same side of the lens as the object. 1 1 1 1 1 1 = − = − = do f di −47 −16. The rays are numbered 1. and 3 to correspond to those described on the optics handout.YF 34-33 A diverging lens with a focal length of −47. (c) Is the image erect or inverted? (d) Are the object and image on the same side or opposite sides of the lens? (a) Since the image is a virtual image. 34-36 When a camera is focused.42 ⇒ ho = = 0. (a) Determine the position of the object. (d) The image and object are on the same side of the lens.90 m from the lens.3 mm. therefore it is erect.) (b) We determine the size of the object from the magnification m = −di /do : hi = mho =⇒ ho = m=− di 8. (We drew the image and object on the left.0 cm to the right of the lens. (a) Determine the position of the object.0 cm forms a virtual image 8. ray 2 do di F ray 3 (a) Since the image is virtual.0223 do 390 Then the height of the image is hi = mho = −0. the length of which is 242 + 362 = 43. Notice that on the left of the lens the rays diverge. (a) how far from the film is the lens? (b) Will the whole image of your friend. where m is given by m=− di −17 cm 0. The situation is illustrated by the ray diagram shown below. 16. This means working out a little geometry.75.50◦ .0 cm. (a) Find the index of refraction of the prism for each of the two wavelengths. (c) What is the ABSOLUTE VALUE of the angular size of the final image as viewed by an eye very close to the eyepiece? Give your answer in RADIANS.0 cm lens is used as the objective. letting nair = 1 be the index in air. When it emerges at face AB.5◦ Now we can write Snell’s Law for these situations as nred sin 62◦ = nair sin 74◦ and nblue sin 62◦ = nair sin 82. and the halos have an December 8. We can relate θc to the index of refraction n of the glass using nair 1 = . Now. sin(74◦ ) nair sin(θ1 ) = = 1.02 mm.5◦ ) = = = 1.60 mm. but we expect the light that is bent less (more) will be more to the red (blue) end of the spectrum. n n Using the geometry shown in the picture we see that we can determine θc using ( ) R R/2 −1 =⇒ θc = tan tan θc = t 2t sin θc = (a) In order to use Snell’s Law to find the index of refraction.12 ◦ sin(62 ) sin(62◦ ) nred = nblue 33-45 Old photographic plates were made of glass with a light-sensitive emulsion on the front surface. and the 90. (b) Find the ABSOLUTE VALUE of the height of the image formed by the objective of a building 60. Both the object being viewed and the final image are at infinity. is totally internally reflected (at the critical angle θc ) at A. where nred (nblue ) is the index for the red (blue) light. 2011 Solution to HW-27 33-42 A light ray in air strikes the right-angle prism shown in the figure (6 B = 28◦ ). This emulsion was somewhat transparent.00 km away.0 m tall. it has been split into two different rays that diverge from each other by 8. = sin θc = √ n (R/2)2 + t2 θ1 = 90◦ − 28◦ + 12◦ = 74◦ . inner radius. we can find the angle between the upper refracted ray and the normal: Using R and t given we find that θc = 35◦ . Some of it is then totally reflected at the back surface of the plate and returns to the front surface. The index of refraction of the glass will be n= 1 = 1. When a bright point source is focused on the front of the plate. A ray leaves from the point source at O. The angle between the normal and the other refracted ray is θ2 = θ1 + 8. If the glass plate has a thickness. the developed photograph will show a halo around the image of the spot.5◦ . it is necessary to determine the angle each of these light rays make with a line normal to the surface of the prism. we solve for each index: YF 34-57 A telescope is constructed from two lenses with focal lengths of 90. 3. sin θc One can derive a simple expression for n by direct evaluation of sin θc using the sides of the right triangle ABC: √ 1 R/2 =⇒ n = 1 + (2t/R)2 . and then reaches the inner radius of the halo at point B. 2011 . what is the index of refraction of the glass? (Hint: Light from the spot on the front surface is scattered in all directions by the emulsion.5◦ = 82. This ray consists of two different wavelengths. t = 3. (a) Find the angular magnification for the telescope.) B R/2 t C R/2 O R θc θc A The figure above shows the rays that produce the the inner radius of the halo.Physics 21 Fall. We aren’t told the exact colors. The angle θ between the incident ray and the normal to the hypotenuse of the prism is θ = 90◦ − 28◦ = 62◦ Since the 12◦ angle is given.0 cm and 20. R = 5.09 ◦ sin(62 ) sin(62◦ ) nair sin(θ2 ) sin(82. 3 cm 30. It is an upright. The quadratic equation for do has two roots.5.75 − 5.00 cm 22.75 Mastering Physics wants height (absolute value of hi ) in units of cm. So the distance between the object and final image is 5 cm + 12 cm + 315 cm = 332 cm. that means it is 12 cm + 3. The image from the first lens becomes the object for the second lens. because we get a positive image distance (di ) for the second lens. but do must be positive because the object is on the side of the lens from which the light comes. We need to determine do : because the image from the first lens is 3.00 cm. The location of the image formed by the first lens can be found using do = 5 cm. (1) and (2) equal to each other and use the quadratic equation to solve for do : 1 1 1 1 + = + do 30. Now do and the original di can be used to solve for the focal length f : 1 1 1 1 1 = + = + ⇒ f = 10. (a) In a two lens system.75 cm and f = +15 cm. solve for di = 315 cm.75 cm from the second lens. For the first set up. and we can tell from the ray diagram. We know that it is virtual because di is a negative number.0 cm In the second set up the object distance is increased by 4. the image distance is di = 30. Therefore the new image distance is d0i = di − 8.75 cm = 15. The image distance is decreased by 8.00 cm 22.00 cm = 30.00 mm tall is placed 5. thus d0o = do + 4. When the lens is moved 4. We know that it is upright because the magnification is a positive number.00 cm for the screen moving to the left). Determine the focal length of the lens. (1) (2) We can set the expressions for 1/f given by Eqs. f do di do + 4. the equation for the focal length is 1 1 1 1 1 = . Therefore we used the plus sign in the quadratic formula to ensure do > 0. f = −15 cm.0 cm to the right of the lens. An object 5. and solving the lens equation for di . ][ ] [ 315 −3.0 cm. (b) To find the distance from the final image to the object.0 cm (c) The final image is real. Using do = 15. the first diverging and the second converging.00 mm = −75 mm hi = m1 m2 ho = − 5 15.6 cm. A ray diagram confirms this. because the total magnification is negative. + = + f do di do 30.0 cm.00 cm = 22. These values can be used in another equation for the focal length 1 1 1 1 1 = 0 + 0 = + .00 cm from the lens moving to the right and another 4. Thus.34-89 Two thin lenses with focal lengths of magnitude 15. (d) We can find the height of the final image by multiplying the magnification of each lens.3 cm. are placed 12. so enter 7. serves as the object of the second lens. we need to determine the location of the final image. the image from the first lens.0 cm and let the object distance be the variable do . virtual image.75 cm to the left of the first lens.0 cm do + 4. That means it the image is be 3. (e) The final image is inverted. the screen must be moved 4.0 cm .00 cm to the left to refocus the image.75 cm to the left of the lens.00 cm to the right. and the ray diagram confirms this.00 cm. (a) Where is the image formed by the first lens located? (b) How far from the object is the final image formed? (c) Is the final image real or virtual? (d) What is the height of the final image? (e) Is the final image erect or inverted? 34-91 An object to the left of a lens is imaged by the lens on a screen 30.00 cm (4. 1 1 1 = + f do di Then di = −3. f do di 16.0 cm 0 = d2o + 4do − 330 do = 16.75.00 cm apart. on the same side as object.00 cm to the left of the first (diverging) lens.0 cm − 8. This means that the final image is 315 cm to the right of the second lens. for m = 1) and find the difference (x1 − x0 ): d sin θ ≈ d 3Lλ x1 = (1 + 12 )λ =⇒ x1 = . (a) What is the width of the central interference maximum? (b) What is the width of the first-order bright fringe? (a) The position x = x0 of the first dark spot is given in terms of the angle it makes with with central bright spot (at x = 0) by d sin θ ≈ d x0 Lλ = (m + 12 )λ =⇒ x0 = . Fringes are measured carefully on a screen 1. L 2d Clearly x1 = 3x0 = 12 mm.00 m)(400 × 10−9 m) = 0.6 mm from the center of the central bright fringe. L 2d where we invoked small-angle approximations and set m = 0 since we are interested in the central peak.0 mm. (b) The width of the first order bright fringe will be the distance between the first and second dark spots. .200 mm and the interference pattern is observed on a screen L = 4. We already know the position of the first dark spot. and the center of the twentieth fringe (not counting the central bright fringe) is found to be 10. All we must do now is calculate the position of the next dark spot (at x1 . What is the separation of the two slits? 35-12 Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by d = 0.20 m away from the double slit.00 m from the slits. We have x0 = (4. which is 2x0 = 8.004 m = 4 mm 2(2 × 10−4 m) The width of the central interference maximum will be double the distance from the bright central spot to the first dark spot.35-8 Young’s experiment is performed with light from excited helium atoms (λ = 502 nm). so the width of the first order bright fringe is x1 − x0 = 12 mm − 4 mm = 8 mm. Since the axis of the first polarizer is at 24. (r2 − r1 ). the objective and the eyepiece. Note that in this problem. and interference maxima occur when d sin θ = mλ. We call the phase difference between these arriving waves φ. Give the number m of the first interference maxima that will be missing in the pattern. 34-58 Saturn is viewed through the Lick Observatory refracting telescope (objective focal length 18 m). We can find θ by using the right triangle formed by the center of the objective lens. so we enter 5.44 × 10−5 = 9. the base of the image. φ = π radians. 35-21 Coherent light with wavelength 450 nm passes through narrow slits with a separation of 0. Saturn’s average distance from the earth is 1. or as shown in the figure.350 mm. what is the phase difference (in radians) in the light from the two slits at an angle of 22. 2011 . a d 3a a 3a 3 The lowest solution for m and n is m = 3 and n = 1. and the axis of the second polarizer is at 65. Mastering Physics is expecting an answer in degrees and to two significant figures. If these coincide at the same θ. tan θ ≈ θ. Both angles are marked as θ in the figure. so we can use do = ∞. and so on. they define the angles subtended by the object and by the image.0◦ counterclockwise from the vertical.8◦ counterclockwise from the vertical (viewed in the direction the light is traveling) and the axis of the second filter is at 65. If the diameter of the image of Saturn produced by the objective is 1.7 mm = = 9. and the other from the tip of the object to the tip of the image. What is the intensity of the light after it has passed through the second polarizer? When the unpolarized light (intensity I0 ) passes through the first polarizer. one wave has gone through one complete cycle more than the other.433 × 1012 m. which is very large compared to the other distances in this problem.350 mm) sin θ = sin 22.0 W/cm2 is incident on two polarizing filters. December 9. or 40. For instance.2◦ .44 × 10−5 fo 18 m ( ) θ = tan−1 9. we are only dealing with the objective lens. then sin θ = mλ mλ nλ mλ m nλ = = ⇒ = ⇒n= .4 × 10−3 . where I2 is the intensity after the second polarizer.Physics 21 Fall. its intensity decreases by half and it will be polarized in the same direction as the axis of the first polarizer. The light is now polarized. the light that is emitted from each slit is in phase with each other. 2011 Solution to HW-28 33-31 Unpolarized light of intensity 26. This fact follows by considering two straight rays: one from the base of the object to the base of the image. and φ is the angle between the polarization of the light and the axis of the polarizer. Combining these equations gives φ= 2πd 2π (0. it is shown that a refracting telescope is made of two lenses.6◦ = 1878 rad λ 450 nm YF 36-24 An interference pattern is produced by two identical parallel slits of width a and separation (between centers) d = 3a.44 × 10−5 rad tan θ = Note that we could have used the small angle approximation. At a distance from the slits which is large compared to their separation. so when it passes through the second polarizer its intensity decreases by another factor of cos2 φ. Therefore the intensity after the first polarizer is I1 = I0 /2 = 13 W/cm2 . the path difference is given by r2 − r1 = d sin θ where θ is the angle measured from the centerline. and the tip of the image. The axis of the first filter is at an angle of 24. hi 1.7 mm. when the path difference is one wavelength. the objective lens creates a real image at its focal point.0◦ in the same direction. When the path difference is λ/2. what angle does Saturn subtend from when viewed from earth? We now have all the relationships needed to answer the question.8◦ .2◦ ) = 7. When the light waves arrive at the same point far from the slits. Note that the angle subtended by the object is the same as the angle subtended by the image. Diffraction minima occur when a sin θ = nλ. the angle between them is the difference. and φ = 2π radians. In the figure above (taken from page 1193 in the textbook). We can express this proportionality in the equation φ = 2π r2 − r1 λ If the point where the waves meet is far from the slits in comparison to their separation d.58 W/cm2 . according to Malus’ Law: I2 = I1 cos2 (φ). Both rays pass through the center of the lens and are not bent. they will be out of phase by an amount proportional to the difference in their path lengths.6◦ from the centerline? Since the light source is coherent. Then I2 = (13 W/cm2 ) cos2 (40. This allows us to calculate di = f = 18 m. f 1250 Hz To find the angles. A bat that is preparing to leave the cave emits a 31. for m = 1. ±3. Here m = 2. we solve the first equation above for θ. Setting m = 3 leads to d= mλ 3 × 700 nm = = 2281. otherwise the sine becomes greater than one. The wavelength of the sound can be found from the stated frequency and speed of sound: λ= vsound 344 m/s = = 0.2752) = 33. the condition for constructive interference is d sin θ = mλ From the information given in the problem.).2752) = 16. we would have obtained 15.0◦ from the central maximum. K 22-19 The opening to a cave is a tall crack 43. How wide is the ”sound beam” 110 m outside the cave opening? Use exact formulas. ±2.0 kHz ultrasonic chirp.53 mm center to center (except for the missing spots) on a screen 2. The bright spots are equally spaced at 1. respectively.6◦ for m = 2 for m = 3 θ = sin−1 Had we used the small angle approximation sin−1 θ ∼ θ. the slit spacing of the diffraction grating.2752) = 55. is given by mλ sin θ = .0◦ Then it is possible to find the angle for a second-order bright spot for violet light. it is possible to determine d. .2752 m.4◦ = sin−1 (3 × 0. Note that the maximum possible value of m is 3. and 3. From the optics handout. and a is the width of the door. 31. Use vsound = 340 m/s. The condition for a dark fringe.1◦ θ = sin d 2281. Even though this example involves sound waves. or in this case a region with no sound.40 m from the slits. don’t make small angle approximations. At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply.4 nm sin θ sin 67. You can ignore effects of reflections. without making small angle approximations: mλ a = sin−1 (1 × 0. 36-26 A diffraction experiment involving two thin parallel slits yields the pattern of closely spaced bright and dark fringes shown in the figure. the basic phenomenon is still diffraction from a single slit.36-9 Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1. Only the central portion of the pattern is shown in the figure.4 nm .00 m.. (a) How far apart are the two slits? (b) How wide is each one? 36-30 If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 nm) at 67. 2.. and 47.5◦ . and converted radians to degrees. The light source was a He-Ne laser producing a wavelength of 632. a where m is the order of the region of no sound (m = ±1.3◦ .8◦ .8 nm. and ( ( ) ) 2 × 430 nm mλ −1 −1 = sin = 22.0◦ for m = 1 = sin−1 (2 × 0. at what angle will the second-order bright spot be for violet light (of wavelength 430 nm)? (a) The bright spots are caused by constructive interference.0 cm wide. • In a magnetic field: What is the force on a moving charged particle? What is the force on a current? How does a charged particle move in a magnetic field? How does a velocity selector work? • What are the total force and torque on a current loop? What is the magnetic moment? September 28. Know how to evaluate cross products. • Draw the electric field lines and the equipotential lines of a point charge. • Know the vector form of Coulomb’s Law. Find σ or ρ for two. Subject Areas: The emphasis of the exam will be the material covered in lecture or recitation or that has been on the homework. • What is Gauss’s Law? Be able to use it to find the field of a spherical. planar. PA416 and PA466. Cell phones. What is an equipotential surface? • What is the electric field of an infinite sheet of charge? Use this result to find the electric field between the parallel plates of a capacitor. Do the same for a dipole. find λ. because the online homework system usually provides the units for you. and resistance? Know how to indicate the + and − terminals of a battery. • Know how to integrate over a linear charge distribution to get the electric field or potential at an arbitrary point. 5.or three. Units: In order to receive full credit. or linear charge distribution. The following is a list of topics and questions you should be familiar with.Physics 21 Fall. HW-10. For some questions you may need to solve three equations in three unknowns. Extra-time students will start in CU248. music players and headphones are prohibited and must not even be visible during the exam. 2011 Information about Exam-1 First Hour Exam: There will be an exam Wednesday. Any physical constants and integrals you need will be given on the exam. 2011 . Oct. what are the symbols for battery. • What is the energy density in an electric field? How much energy is stored in a charged capacitor? • What does it mean to do a surface integral E · dA? What is the convention for the direction of dA if the surface is closed? • What is current? • What is an electron volt? • What is the dielectric constant? What happens to the capacitance and electric field if an insulator with dielectric constant K is placed between the capacitor plates? • Be able to use the right hand rule to get the direction of magnetic fields or of cross products. Know how to find the potential difference between any two points on an electric circuit. you must include the correct units with all numerical answers. cylindrical. • What is the relation between the electric field and the electric potential? Know how to evaluate the gradient. • For a distribution of charges on a line. PA416 or PA466 and at 5:10 pm will move to LL221. You must solve such equations by hand and show the solutions for full credit. • Know how to write the loop equation for an RC circuit. and L-10.dimensional distributions. This list is not necessarily complete but is representative. Coverage: You are responsible for all the reading assignments. • Know how to account for the energy stored in the capacitor or lost in the resistor of an RC circuit. • What is the electric field inside a conductor? • In a circuit diagram. You may use a calculator to check your hand solution. lectures. capacitor. and homework problems up to and including R-10. how does a dielectric (insulator) respond to an electric field? • What is Ohm’s Law? • How much power is dissipated when a current passes through a resistor? • How does a current divide when it branches to flow through two resistors in parallel? • How do the charges arrange themselves on two capacitors in parallel? in series? What is the effective capacitance for two capacitors in series or parallel? What about resistors? • What are Kirchhoff’s Rules? Know how to write the loop and junction (node) equations that we discussed in class for any circuit. the equation sheet posted on the course web site will be included on the exam. • Know how to verify that a given solution satisfies the loop equations for a circuit. Use of Calculators and Other Electronic Devices: You should bring a calculator. 2011 at 4:10 pm in CU248. The exam will be closed book and closed notes. What is the time constant for such a circuit? Know what the function exp(−t/RC) looks like. Know how to find the electric field or potential of several point charges. All reading assigments throught the end of chapter 27 in the text are covered. With what time dependence does charge build up on a capacitor? With what time dependence does a capacitor discharge? • Know the common prefixes femto through Giga. • On a microscopic level. Be careful about this point. What is E · dl? Know how to evaluate it for a circuit. see the class web site for room assignments. I2 . You should have three equations for three unknowns. Show your work. All problems count 20 points.Physics 21 Fall. using the path through corner point b. and I3 (including the correct sign) by explicit solution of the equations you determined in part (a). but show every number that you use to get numerical results. Give units for all final answers.) Use the currents you calculated in part (b) to find the potential difference Vd − Va between the corner points labelled d and a on the diagram. You must show your work. Show your work. 2010 1 2 Hour Exam #1 3 4 Recitation Time 5 Total Recitation Leader Name: October 6. You may use a calculator. (b) (3 pts. Make it easy for the grader to identify your final answers to each question or part of a question. Problem 1.) Determine the currents I1 . but take the path through corner point c.) Write Kirchhoff’s loop and junction (or node) equations needed to determine the currents I1 . You must write these equations as defined in Physics 21. (d) (2 pts. (c) (2 pts. You must show enough work on each problem to convince the grader you understand how to solve the problem. . including the potential change across the resistor and battery separately. Consider the following circuit: I2 a I3 b I1 21 V 3Ω 7Ω 2Ω 11 V c d (a) (13 pts.) Repeat part (c). I2 . Use the currents and their directions specified by the arrows in the diagram. Draw clearly and label on the diagram the loop used to determine each loop equation. The penalty for arithmetic errors is small if the grader can tell what you intended to do. 2010 This exam is closed notes and closed book. You should get the same value for Vd − Va that you found in part (c). There is an equation sheet on the last page. and I3 . ) Assuming that the circuit in panel (a) has been connected for a very long time. For the first capacitor.) At time t = 0 the switch S is flipped so that the circuit appears as shown in panel (b). C1 C2 (d) (4 pts. S (a) V0 C1 R Q1 Q2 C2 V0 R (b) (6 pts. What is the time constant of the circuit? (c) (2 pts. C1 = 1. The voltage of the battery is V0 = 12 volts. Two parallel plate capacitors are connected to a battery and a resistor as shown in the circuit.) At what time t will the total charge on the two capacitors diminish to 60% of its initial value? What fraction of the initial energy will remain at time t = t ? . and the resistance R = 3500 Ω.24 nF.) What is the current that starts to flow in the circuit of panel (b) just after t = 0? S (b) (a) (8 pts. C2 = 3. what are the charges Q1 and Q2 on each capacitor? What is the energy stored in each capacitor? Give numerical answers. for the other.72 nF.Problem 2. and the capacitors discharge through the resistor. dQ -b x' 2b (a) (3 pts.Problem 3.y. b. Your answer should be in terms of Q. and y. Write an expression for the vector (r − r ) that indicates its components.0) (b) (3 pts.) On the diagram shown. . b. draw the vector (r − r ) (as defined in class) from the charge dQ on the rod at x = x to the point P on the y axis.) What is the linear charge density λ on the rod? (d) (7 pts. x .) Write an expression for the electric field dE at the point P due to the element of charge dQ at the point x on the rod. For this problem you are to find the electric field at the point P on the y axis due to the charge distribution on the x axis. y.) Integrate to find ONLY THE y COMPONENT of the electric field E at the point P due to the rod. P = (0. (c) (7 pts. dx . The expression for dE should be in terms of Q. A total charge of 3Q is uniformly spread out from x = −b to x = +2b. and NOT dQ. Sketch the Gaussian surface that you use and explain your result.00 mm made of an insulating material has a uniform volume charge density of ρ = 4. A solid sphere of radius R = 2. R (a) (3 pts. .Problem 4.) Use Gauss’s Law to find the magnitude of the electric field at a distance r = 1.) What is the total charge on the sphere? (b) (3 pts. (d) (7 pts.) Write Gauss’s Law.) Find the magnitude of the electric field at the surface of the sphere. (c) (7 pts.00 mm from the center of the sphere.80 × 10−12 C/m3 . 00 × 10−19 C moving along the z axis away from eight identical point charges with q = 3. and total energy of the ion. All charges are the same distance (b) from the origin. Assume that the particle is constrained to stay on the z axis. all eight charges are the same distance b = 0.50 × 10−26 kg and charge Q = −2.) At the instant shown in the figure.5 nm from the origin.2 nm. These eight charges are fixed in the xy plane as shown in panel (b). (a) PERSPECTIVE VIEW: (b) VIEW LOOKING DOWN z AXIS: z Q (v = v0 k) y b The charges are in the xy plane at the vertices of a regular octagon centered at the origin. (b) (6 pts. Find the potential energy. the z coordinate of the negative ion is 1.) The negative ion will slow down and reach a point z = z where its speed is zero. Panel (a) shows a negative ion with mass m = 1. Show all work and give numerical answers in joules. . Hint: Use symmetry.3 km/s. (c) (7 pts. kinetic energy.) Write an expression for the potential at an arbitrary point on the z axis due to the fixed charge distribution in the xy plane. b. Give your answer in terms of q.Problem 5.25 × 10−19 C. Show explicitly the equation that must be solved to find z . y q q q b x q x q q q q (a) (7 pts. and its speed is v0 = 15. and standard physical constants. and then solve that equation for z . z. = Vp Np Ip Ns µ χm = −1 µ0 τ =µ×B µ = IA ξi =Ci (parallel) or Ri (series): sin(a ± b) = sin a cos b ± cos a sin b B · dA = 0 F = qv × B.6605 × 10−27 kg 8.626 × 10−34 J s 1.11 × 10−31 kg 1.4 m/s) T /300 v = λf = ω/k ω = 2πf k = 2π/λ T = 1/f (T =period) P = 12 ρA2 ω 2 v 2 1 ωC RC time constant = RC LR time constant = L/R Q(t) for RLC circuit Q0 exp(−Rt/2L) cos ωt ω2 = = sin θ cos π 2 1 R2 − LC 4L2 ± cos θ sin E · dA u du = a2 + u 2 XR = R.38 × 10−23 J/K 1.055 × 10−34 J s 9.85 × 10−12 C2 /(N m2 ) 4π × 10−7 T m/A 9.02 × 1023 mol−1 1.Physics 21 Fall. dF = Idl × B µ0 Idl × (r − r ) dB = 4π |r − r |3 1 1 1 = + ξeffective ξ1 ξ2 circ.99 × 109 N m2 /C2 h h ¯ = h/2π me mp mn u k ξi =Ci (series) or Ri (parallel): Eline = F = qE dE (at r) = Planck’s constant Planck’s constant/(2π) electron rest mass proton rest mass neutron rest mass atomic mass unit 1/(4π0 ) a−b a+b sin a + sin b = 2 cos sin 2 2 du u = √ (a2 + u2 )3/2 a2 a2 + u 2 e du = ln u u 2π cos2 θ dθ = 0 ∂ 2D 1 ∂ 2D = 2 2 ∂x v ∂t2 1 S= (E × B) µ0 √ c = 1/ 0 µ0 c 2 S= = B0 µ0 E0 B0 Erms Brms = = 2µ0 µ0 λ = h/p 1 2 1 un+1 n+1 du 1 = ln(a + bu) a + bu b 1 du = eau a ln u du = u ln u − u 1 cE02 2 0 un du = u du 1 = −√ (a2 + u2 )3/2 a2 + u 2 au ax2 + bx + c = 0 ⇒ √ −b ± b2 − 4ac x= 2a E×B∝v ∆x∆p > ¯ h ∼ − 2π sin2 θ dθ = π 0 KE = p2 /(2M ) (plane wave) p=¯ hk (¯ h = h/2π) eiθ = cos θ + i sin θ E=¯ hω = hf (de Broglie) ¯ 2 ∂ 2ψ h ∂ψ = i¯ h 2M ∂x2 ∂t August 29. Eplane = r 20 σ for plate 0 capacitor A A Q = CV . XC = solenoid B = µ0 nI solenoid L = µ0 N 2 A/l C = 2πr C = πd A = πr2 A = 4πr2 V = 43 πr3 π 2 circumference of circle circumference of circle area of circle surface area of sphere volume of sphere cos(a ± b) = cos a cos b ∓ sin a sin b du √ = ln u + a2 + u2 2 2 a +u π ) 2 long wire: B = ξeffective = ξ1 + ξ2 = ± cos θ B · dA d B · dl = µ0 I + µ0 0 dt √ sin(θ ± µ0 I 2πR center loop: B = µ0 I/2R dQ I= I = −neAvd dt Vs Ns Is Np = . C = 0 K = d d 2 Q 2 Ucap = 12 CV = 12 C Uind = 12 LI 2 ρL V = IR R= A P = IV P = I 2R 1 2π0 Q E= = 0 A 1 dQ (r − r ) 4π0 |r − r |3 E = −∇V ∂V ∂V ˆ ∂V = − ˆi + ˆj +k ∂x ∂y ∂z V f − Vi = − f E · dl i 1 Q dQ 1 .00 × 108 m/s 6. dV = 4π0 r 4π0 |r − r | 1 uelec = 12 0 E 2 . XL = ωL.67 × 10−11 N m2 /kg2 6. 2010 . umag = B2 2µ0 V= Work = R = mv⊥ /(qB) A × B = F · dl E · dA = Q 0 d dt E · dl = − u du = a2 + u 2 a2 + u 2 du 1 = tan−1 a2 + u 2 a v= T /ρ u a ln a2 + u 1 2 (T =tension) v = (347.807 m/s2 c G NA kB e 0 µ0 g 1 q1 q2 (r1 − r2 ) 4π0 |r1 − r2 |3 λ σ .6726 × 10−27 kg 1.6749 × 10−27 kg 1.60 × 10−19 C 8. 2010 Equation Sheet speed of light in vacuo Gravitational constant Avogadro’s Number Boltzmann’s constant charge on electron free space permittivity free space permeability gravitational acceleration F2 on 1 = 3. orbit ˆj ˆ k Ay Az By Bz ˆi Ax Bx 6. I2 . and I3 (including the correct sign) by explicit solution of the equations you determined in part (a). Show your work. 3 Malenda.Physics 21 Fall.) Determine the currents I1 .) Repeat part (c). You must show your work. see the grader by Oct. (b) (3 pts. You must write these equations as defined in Physics 21. 2 Faust. 4 and 5 Beels For questions about the grading. (c) (2 pts. including the potential change across the resistor and battery separately. 27. You should get the same value for Vd − Va that you found in part (c). I2 . and I3 .) Use the currents you calculated in part (b) to find the potential difference Vd − Va between the corner points labelled d and a on the diagram. but take the path through corner point c. (d) (2 pts. 2010 Solution to Hour Exam #1 The graders for the problems were: 1 Jones. You should have three equations for three unknowns. . Consider the following circuit: I2 a I3 I1 loop 1 b loop 2 21 V 3Ω 7Ω 2Ω 11 V c loop 3 d (a) (13 pts. Use the currents and their directions specified by the arrows in the diagram. using the path through corner point b. Problem 1. Draw clearly and label on the diagram the loop used to determine each loop equation. Show your work.) Write Kirchhoff’s loop and junction (or node) equations needed to determine the currents I1 . ) What is the current that starts to flow in the circuit of panel (b) just after t = 0? S (b) (a) (8 pts.) Assuming that the circuit in panel (a) has been connected for a very long time. what are the charges Q1 and Q2 on each capacitor? What is the energy stored in each capacitor? Give numerical answers. The voltage of the battery is V0 = 12 volts.72 nF. and the resistance R = 3500 Ω. What is the time constant of the circuit? (c) (2 pts. C2 = 3. C1 = 1. For the first capacitor. for the other. C1 C2 (d) (4 pts.24 nF. S (a) V0 C1 R Q1 Q2 C2 V0 R (b) (6 pts. and the capacitors discharge through the resistor.) At time t = 0 the switch S is flipped so that the circuit appears as shown in panel (b).Problem 2. Two parallel plate capacitors are connected to a battery and a resistor as shown in the circuit.) At what time t will the total charge on the two capacitors diminish to 60% of its initial value? What fraction of the initial energy will remain at time t = t ? . ) Integrate to find ONLY THE y COMPONENT of the electric field E at the point P due to the rod.) On the diagram shown.r' dQ -b x' 2b (a) (3 pts. and NOT dQ.0) (b) (3 pts. draw the vector (r − r ) (as defined in class) from the charge dQ on the rod at x = x to the point P on the y axis. y.) Write an expression for the electric field dE at the point P due to the element of charge dQ at the point x on the rod. dx . b.y. and y. P = (0. Write an expression for the vector (r − r ) that indicates its components. A total charge of 3Q is uniformly spread out from x = −b to x = +2b. b. The expression for dE should be in terms of Q.Problem 3. . (c) (7 pts. x .) What is the linear charge density λ on the rod? (d) (7 pts. For this problem you are to find the electric field at the point P on the y axis due to the charge distribution on the x axis. Your answer should be in terms of Q. r . Sketch the Gaussian surface that you use and explain your result.) Write Gauss’s Law. (c) (7 pts.Problem 4.00 mm from the center of the sphere.) Find the magnitude of the electric field at the surface of the sphere.) What is the total charge on the sphere? (b) (3 pts. R/2 R (a) (3 pts. A solid sphere of radius R = 2. .00 mm made of an insulating material has a uniform volume charge density of ρ = 4. (d) (7 pts.80 × 10−12 C/m3 .) Use Gauss’s Law to find the magnitude of the electric field at a distance r = 1. y q q q b q x q q q q (a) (7 pts. (b) VIEW LOOKING DOWN z AXIS: (a) PERSPECTIVE VIEW: z Q (v = v0 k) y b x The charges are in the xy plane at the vertices of a regular octagon centered at the origin. Find the potential energy.) Write an expression for the potential at an arbitrary point on the z axis due to the fixed charge distribution in the xy plane. These eight charges are fixed in the xy plane as shown in panel (b). Panel (a) shows a negative ion with mass m = 1.) At the instant shown in the figure. Give your answer in terms of q.00 × 10−19 C moving along the z axis away from eight identical point charges with q = 3. Hint: Use symmetry. the z coordinate of the negative ion is 1. Show explicitly the equation that must be solved to find z . and standard physical constants.2 nm. (c) (7 pts.5 nm from the origin.50 × 10−26 kg and charge Q = −2. b. Show all work and give numerical answers in joules.Problem 5. and its speed is v0 = 15. kinetic energy. and then solve that equation for z .3 km/s.) The negative ion will slow down and reach a point z = z where its speed is zero. All charges are the same distance (b) from the origin.25 × 10−19 C. (b) (6 pts. z. Assume that the particle is constrained to stay on the z axis. and total energy of the ion. . all eight charges are the same distance b = 0. Show your work.) Write Kirchhoff’s loop and junction (or node) equations needed to determine the currents I1 . (d) (2 pts.) Use the currents you calculated in part (b) to find the potential difference Vc −Vb between the corner points labelled c and b on the diagram. You must show your work. Problem 1. You should get the same value for Vc − Vb that you found in part (c).) Repeat part (c). and I3 . but take the path through corner point d. Draw clearly and label on the diagram the loop used to determine each loop equation. 2011 Solution to Hour Exam #1 The graders for the problems were: 1 Tupa. You should have three equations for three unknowns. Use the currents and their directions specified by the arrows in the diagram. . Show your work. Consider the following circuit: I2 a 8V 7Ω loop 1 c I3 b I1 2Ω 12 V 6Ω loop 2 d (a) (13 pts. of the equations you determined in part (a). 26. including the potential change across any circuit elements on this path. I2 . using the path through corner point a. 3 Beels. You must write these equations as defined in Physics 21. I2 .Physics 21 Fall. 4 Malenda.) Determine the currents I1 . and I3 (including the correct sign) by explicit solution. by hand. (b) (3 pts. including the potential change across any circuit elements on this path. (c) (2 pts. see the grader by Oct. 5 Glueckstein For questions about the grading. 2 Faust. (a) (3 pts.) What is the total charge enclosed by S? S (d) (6 pts.) Find the magnitude of the electric field at distance r = 5.6 cm from the center of the cavity. rcav = 3.5 × 10−3 C/m3 .) What is the direction of the electric field? Justify your answer. the charge density in the solid is ρ = 1. and you are to use Gauss’ Law to calculate the electric field inside the solid at a distance r = 5. (e) (2 pts. q = −6.6 cm from the point charge q. spherical charged solid: For this problem. r q rcav (c) (6 pts.) What is the shape and location of the Gaussian surface S that you will use? Draw and label S neatly on the diagram to the left.Problem 2. .) Write Gauss’ Law.3 cm. (b) (3 pts.1 µC. A point charge q is located at the center of a spherical cavity of radius rcav inside an insulating. Problem 3. A battery with V = 12 V is connected to a circuit with capacitors C1 = 3.0 mF, C2 = 3.5 mF, and C3 = 2.5 mF as shown. C2 C1 C3 V R (a) (2 pts.) What is the equivalent (or effective) capacitance of the three capacitors shown? (b) (12 pts.) Assume the circuit has been connected for a very long time. Find the charge on each capacitor and the potential difference across each capacitor. Identify the charges as Q1 , Q2 , and Q3 , and the potential differences as V1 , V2 , and V3 . Explain carefully the steps you take to determine your answer. (c) (6 pts.) Suppose it takes the capacitors 15 s after the battery is connected to become 99% charged. What is R? Problem 4. For this problem you are to find the electric field at the point P on the x axis due to the charge distribution on the y axis. A total charge of Q is uniformly spread out on a thin wire of length L. The lower end of the wire is at the origin. y (b) (4 pts.) On the diagram shown, draw the vectors r and r that correspond to the field point (P ) and the charge point (where dQ is), respectively. Give the vector (r − r ) in terms of its components. (c) (6 pts.) Write an expression for the electric field dE at the point P due to the element of charge dQ on the wire. Show how to write dQ in terms of the variable of integration. charge Q length L dQ (0,y',0) r' r P = (x,0,0) x (a) (3 pts.) What is the linear charge density λ on the wire? (d) (7 pts.) Integrate to find ONLY THE y COMPONENT of the electric field E at the point P due to the wire. Your answer should be in terms of Q, L, x, and physical constants. Problem 5. A particle moves in a circular orbit in a uniform magnetic field B in the −z direction (into the page). The orbit is confined to the xy plane. The charge and mass of the particle are q = 3.20 × 10−19 C and m = 6.75 × 10−26 kg, respectively, and the magnitude of B is B = 0.500 T. (a) (5 pts.) At some time t0 the particle is at the point shown on the diagram, and its instantaneous velocity is v = (3.42 × 104 m/s) ˆi + (3.08 × 104 m/s) ˆj. Give the components of the magnetic force F on the particle at the time t0 . Draw an arrow on the diagram that shows the direction of F. (b) (5 pts.) Find the radius R of the circular orbit. y x +z out of page (c) (5 pts.) How much time does it take for the particle to make one revolution? (d) (5 pts.) Through what potential difference would the particle have to be accelerated from rest to acquire the speed it has? recitation or homework. This list is not necessarily complete but is representative. The solution to the practice exam will be posted on the web and also discussed in a review session Tuesday. check the class website for the schedule. what is the force on a moving charged particle? What is the force on a current? • Be able to use the right hand rule to get the direction of magnetic fields or of cross products. What is the displacement current? • What do “transient” and “steady state” mean with respect to dc and ac circuits? • Know how to analyze the transient behavior of an LR circuit. The exam will emphasize the material covered in lecture. • What are the magnetic and electric flux? Know how to evaluate them. The following is a list of topics and questions you should be familiar with. Beels (10 a. Coverage: The exam will cover material starting with our initial discussion of magnetism (the beginning of Chapter 27).) Use of Calculators and Other Electronic Devices: You should bring a calculator. except for §31. 2011 at 4:10 pm. 2011 . Some of the problems on the practice exam are taken from previous exams in Physics 21. What does it mean if. because the online homework system usually provides the units for you. November 3. L. Be careful about this point. for example. what is the relation between peak values and rms values? • In a magnetic field. You should know the phase relations between current and voltage in the prototypical ac circuits with only R. What happens if you add a resistor to the circuit? • What are Maxwell’s Equations? • Know how to use phasor diagrams. Beels (9 a. all numerical answers must include the correct units. Other review sessions may also be held. the exam will cover the lectures numbered 9–18 inclusive and homeworks 10–20 inclusive. November 8 at 7:10 pm in LL270.) and all extra time students Malenda (all). • Know the common prefixes femto through giga. November 9.m. There is some overlap with the material on the magnetic field covered on the first exam. Any physical constants and integrals you need will be given on the exam. The exam will be closed book and closed notes. • How does a velocity selector work? • What is the magnetic dipole moment of a current loop? What are the forces and torques on a current loop in a uniform magnetic field? • How does a simple dc motor work? • What is Faraday’s Law? What is Lenz’s Law? Know how to apply them. or C elements besides the power supply. Here are the room assignments: Chandler Ullman 248 Packard 466 Packard 416 Glueckstein (all).Physics 21 Fall. The equation sheet posted on the course web site will be included on the exam. 2011 Information about Exam 2 Second Hour Exam: There will be an exam on Wednesday. Units: In order to receive full credit. What do the exponential functions exp(−t/τ ) and 1 − exp(−t/τ ) look like? What is τ ? • What is the energy stored in a magnetic field? • Why can there be oscillations in an LC circuit? What is the resonant frequency? Be able to track where the energy is during the oscillations. Cell phones. you should be well prepared for the real exam.m. as well as all the reading assignments in Chapters 27–31 of the text. Know how to evaluate cross products. • What does the magnetic field of a current loop look like? • Be able to use the Biot-Savart Law. music players and headphones are prohibited and must not even be visible during the exam. the voltage leads the current across a circuit element? What is the resonant frequency of an RLC circuit with an ac power supply? • What is the power delivered to an ac circuit? • In an ac circuit. You should be familiar with the common prefixes femto through giga. If you can work the problems on the practice exam within the time alloted. Faust (all) Tupa (all). • What is hysteresis? • What is ferromagnetism? diamagnetism? netism? What is a magnetic domain? paramag- • What is mutual inductance? What is self inductance? Why does a solenoid exhibit self inductance? • What is Ampere’s Law (final form)? Know how to use it. Specifically. Practice Exam and Review Session: A practice exam is available from the class web site. • Know why a charged particle can exhibit circular or spiral motion in a magnetic field.6 on transformers. Problem 31-37 on hw20 is about transformers and won’t be on the test. and ω = 800.) What is the time interval between a maximum in voltage across the inductor and the next maximum in voltage across the power supply? . You may use a calculator. Does the voltage lead or lag the current? (a) (8 pts. For the following circuit: (b) (3 pts. C = 10.) What is the average power delivered to the circuit by the power supply? (e) (3 pts. Draw a mark on your diagram to indicate the phase angle φ. draw a phasor diagram that is approximately to scale.) Evaluate the phase angle φ.0 rad/s. 2010 1 2 Recitation Time Hour Exam #2 3 4 5 Total Recitation Leader Name: November 10.0 Ω. (d) (3 pts. Problem 1. You must show enough work on each problem to convince the grader you understand how to solve the problem. There is an equation sheet on the last page. L = 25. The penalty for arithmetic errors is small if the grader can tell what you intended to do. All problems count 20 points.0 µF.00 V. but show every number that you use to get numerical results. 2010 This exam is closed notes and closed book.) If R = 22. (c) (3 pts.) Evaluate the peak value of the current I if the peak voltage supplied by the power supply is 8.Physics 21 Fall. Include a phasor for the ac voltage V . and give the length of each phasor. Give units for all final answers. Show all work.0 mH. carefully draw a plot that shows the current i(t) in the large loop as a function of time.) What is the current in the large loop a long time after the switch has been closed? (e) (2 pts. but the small wire inner loop is replaced with one that has half the diameter.0 Ω.Problem 2. When the switch S is closed. R = 30. (d) (3 pts.) Now this experiment is repeated. At t = 0 the switch S is closed to complete the circuit. The values of the circuit elements are L = 15.) What is the time constant of this circuit? (b) (5 pts. would the induced emf in the wire loop be smaller or larger than it was in the first experiment? Briefly explain your answer. starting at t = 0. The large loop remains the same. and V0 = 12.0 H.) Using the box printed below. Fill in numerical values in seconds and amperes for several tic marks on the horizontal and vertical axes to set reasonable scales for these axes. (c) (5 pts. A small circular wire loop is inside a larger loop that is connected to the circuit as shown. Draw the graph for part (c) in this box: .0 V.) Is the direction of the induced current in the small circular wire loop soon after the switch is closed clockwise or counterclockwise? (a) (5 pts. (e) (4 pts.) Calculate the current through the resistor. Find the magnitude and direction of that force. The distance between the rails is 0.Problem 3. (a) (4 pts.) Because of the induced current in the circuit.00 m/s through a uniform magnetic field of magnitude B = 1. A metal bar moves to the left with constant speed v = 8.) Is the direction of the current induced in the circuit clockwise or counterclockwise? (d) (4 pts.5 T as shown in the diagram. (b) (4 pts.) What is the magnitude of the emf induced in the circuit (before the metal bar hits the resistor)? (c) (4 pts.5 m. .) Give Faraday’s Law (as an equation). the magnetic field exerts a force on the moving metal bar. The only resistance in the circuit may be taken to be the resistance R = 24 Ω shown. y = 7. (Use the unit vectors ˆi and ˆj.500 A.) Write a vector expression for the current element I dl. z = 3. z = 0 . (a) (6 pts.0 m (c) (7 pts. The positive z axis points out of the page.) Find dB at x = 4. For this problem you are to find the contribution dB to the magnetic field at several points due to the dark segment dl of the wire centered at the origin. y = 0.) (b) (7 pts. The long wire shown in the diagram lies in the xy plane and carries a current I = 0. The dark segment has a length 3.2 m.3 m.0 mm and makes an angle of 60◦ with the x axis.) Find dB at x = 0.Problem 4. (a) (4 pts. C = 25 µF and L = 32 mH. (i gives the direction of positive current flow just after the switch is closed.) (c) (6 pts.Problem 5. (b) (3 pts.25 mC.) What is the total energy stored in the circuit before the switch is closed? (e) (3 pts. the charge on the capacitor is Q0 = 1. q(t).) Give the equation that relates the current i shown on the diagram and the charge q on the capacitor.) What is the energy stored in the electric field of the capacitor at an instant when the magnitude of the magnetic field in the inductor is 60% of its maximum value? . Just before the switch is closed at time t = 0.) Write the loop equation for this circuit and convert it to a differential equation for the charge on the capacitor.) Calculate the value of ω for this circuit. (d) (4 pts. For the following circuit. Verify that the solution to the differential equation is q(t) = Q0 cos ωt. = Vp Np Ip Ns µ χm = −1 µ0 τ =µ×B µ = IA ξi =Ci (parallel) or Ri (series): sin(a ± b) = sin a cos b ± cos a sin b B · dA = 0 F = qv × B. C = 0 K = d d 2 Q 2 Ucap = 12 CV = 12 C Uind = 12 LI 2 ρL V = IR R= A P = IV P = I 2R 1 2π0 Q E= = 0 A 1 dQ (r − r ) 4π0 |r − r |3 E = −∇V ∂V ∂V ˆ ∂V = − ˆi + ˆj +k ∂x ∂y ∂z V f − Vi = − f E · dl i 1 Q dQ 1 .6749 × 10−27 kg 1.6605 × 10−27 kg 8.4 m/s) T /300 v = λf = ω/k ω = 2πf k = 2π/λ T = 1/f (T =period) P = 12 ρA2 ω 2 v 2 1 ωC RC time constant = RC LR time constant = L/R Q(t) for RLC circuit Q0 exp(−Rt/2L) cos ωt ω2 = = sin θ cos π 2 1 R2 − LC 4L2 ± cos θ sin E · dA u du = a2 + u 2 XR = R. Eplane = r 20 σ for plate 0 capacitor A A Q = CV .00 × 108 m/s 6.85 × 10−12 C2 /(N m2 ) 4π × 10−7 T m/A 9.Physics 21 Fall. dF = Idl × B µ0 Idl × (r − r ) dB = 4π |r − r |3 1 1 1 = + ξeffective ξ1 ξ2 circ.807 m/s2 c G NA kB e 0 µ0 g 1 q1 q2 (r1 − r2 ) 4π0 |r1 − r2 |3 λ σ . orbit ˆj ˆ k Ay Az By Bz ˆi Ax Bx 6. dV = 4π0 r 4π0 |r − r | 1 uelec = 12 0 E 2 .38 × 10−23 J/K 1.02 × 1023 mol−1 1. 2010 .99 × 109 N m2 /C2 h h ¯ = h/2π me mp mn u k ξi =Ci (series) or Ri (parallel): Eline = F = qE dE (at r) = Planck’s constant Planck’s constant/(2π) electron rest mass proton rest mass neutron rest mass atomic mass unit 1/(4π0 ) a−b a+b sin a + sin b = 2 cos sin 2 2 du u = √ (a2 + u2 )3/2 a2 a2 + u 2 e du = ln u u 2π cos2 θ dθ = 0 ∂ 2D 1 ∂ 2D = 2 2 ∂x v ∂t2 1 S= (E × B) µ0 √ c = 1/ 0 µ0 c 2 S= = B0 µ0 E0 B0 Erms Brms = = 2µ0 µ0 λ = h/p 1 2 1 un+1 n+1 du 1 = ln(a + bu) a + bu b 1 du = eau a ln u du = u ln u − u 1 cE02 2 0 un du = u du 1 = −√ (a2 + u2 )3/2 a2 + u 2 au ax2 + bx + c = 0 ⇒ √ −b ± b2 − 4ac x= 2a E×B∝v ∆x∆p > ¯ h ∼ − 2π sin2 θ dθ = π 0 KE = p2 /(2M ) (plane wave) p=¯ hk (¯ h = h/2π) eiθ = cos θ + i sin θ E=¯ hω = hf (de Broglie) ¯ 2 ∂ 2ψ h ∂ψ = i¯ h 2M ∂x2 ∂t August 29.11 × 10−31 kg 1.055 × 10−34 J s 9. 2010 Equation Sheet speed of light in vacuo Gravitational constant Avogadro’s Number Boltzmann’s constant charge on electron free space permittivity free space permeability gravitational acceleration F2 on 1 = 3. XC = solenoid B = µ0 nI solenoid L = µ0 N 2 A/l C = 2πr C = πd A = πr2 A = 4πr2 V = 43 πr3 π 2 circumference of circle circumference of circle area of circle surface area of sphere volume of sphere cos(a ± b) = cos a cos b ∓ sin a sin b du √ = ln u + a2 + u2 2 2 a +u π ) 2 long wire: B = ξeffective = ξ1 + ξ2 = ± cos θ B · dA d B · dl = µ0 I + µ0 0 dt √ sin(θ ± µ0 I 2πR center loop: B = µ0 I/2R dQ I= I = −neAvd dt Vs Ns Is Np = . umag = B2 2µ0 V= Work = R = mv⊥ /(qB) A × B = F · dl E · dA = Q 0 d dt E · dl = − u du = a2 + u 2 a2 + u 2 du 1 = tan−1 a2 + u 2 a v= T /ρ u a ln a2 + u 1 2 (T =tension) v = (347.60 × 10−19 C 8.6726 × 10−27 kg 1. XL = ωL.67 × 10−11 N m2 /kg2 6.626 × 10−34 J s 1. ) Evaluate the phase angle φ.Physics 21 Fall. (d) (3 pts. Does the voltage lead or lag the current? (a) (8 pts.0 rad/s.) Evaluate the peak value of the current I if the peak voltage supplied by the power supply is 8.) If R = 22. Draw a mark on your diagram to indicate the phase angle φ. 2010 Solution to Hour Exam #2 If you want to discuss the grading. Show all work.) What is the average power delivered to the circuit by the power supply? (e) (3 pts. draw a phasor diagram that is approximately to scale. and ω = 800.) What is the time interval between a maximum in voltage across the inductor and the next maximum in voltage across the power supply? .0 µF. For the following circuit: (b) (3 pts.00 V. L = 25. Include a phasor for the ac voltage V .0 Ω. C = 10. and give the length of each phasor. 1: Beels 2: Jones 3: Faust 4: Beels 5: Glueckstein Problem 1.0 mH. (c) (3 pts. you must speak with the grader by Dec. 8. 0 V. starting at t = 0. but the small wire inner loop is replaced with one that has half the diameter. At t = 0 the switch S is closed to complete the circuit. Draw the graph for part (c) in this box: . (c) (5 pts. Fill in numerical values in seconds and amperes for several tic marks on the horizontal and vertical axes to set reasonable scales for these axes. The values of the circuit elements are L = 15.Problem 2.) What is the current in the large loop a long time after the switch has been closed? (e) (2 pts. A small circular wire loop is inside a larger loop that is connected to the circuit as shown.) Now this experiment is repeated.0 Ω. carefully draw a plot that shows the current i(t) in the large loop as a function of time. (d) (3 pts.0 H. The large loop remains the same. would the induced emf in the wire loop be smaller or larger than it was in the first experiment? Briefly explain your answer. and V0 = 12. R = 30.) What is the time constant of this circuit? (b) (5 pts.) Using the box printed below.) Is the direction of the induced current in the small circular wire loop soon after the switch is closed clockwise or counterclockwise? (a) (5 pts. When the switch S is closed. . A metal bar moves to the left with constant speed v = 8. Find the magnitude and direction of that force.) Calculate the current through the resistor.5 m. (a) (4 pts. (b) (4 pts.) Is the direction of the current induced in the circuit clockwise or counterclockwise? (d) (4 pts.) Because of the induced current in the circuit.) Give Faraday’s Law (as an equation). The distance between the rails is 0.) What is the magnitude of the emf induced in the circuit (before the metal bar hits the resistor)? (c) (4 pts. (e) (4 pts.00 m/s through a uniform magnetic field of magnitude B = 1.5 T as shown in the diagram.Problem 3. the magnetic field exerts a force on the moving metal bar. The only resistance in the circuit may be taken to be the resistance R = 24 Ω shown. 0 mm and makes an angle of 60◦ with the x axis.2 m. The dark segment has a length 3.500 A.) Find dB at x = 4.3 m. (a) (6 pts. The long wire shown in the diagram lies in the xy plane and carries a current I = 0. z = 3. (Use the unit vectors ˆi and ˆj.Problem 4.0 m (c) (7 pts.) (b) (7 pts.) Write a vector expression for the current element I dl. The positive z axis points out of the page.) Find dB at x = 0. For this problem you are to find the contribution dB to the magnetic field at several points due to the dark segment dl of the wire centered at the origin. z = 0 . y = 0. y = 7. ) (c) (6 pts.) Give the equation that relates the current i shown on the diagram and the charge q on the capacitor.) What is the total energy stored in the circuit before the switch is closed? (e) (3 pts.) What is the energy stored in the electric field of the capacitor at an instant when the magnitude of the magnetic field in the inductor is 60% of its maximum value? . q(t).25 mC. Verify that the solution to the differential equation is q(t) = Q0 cos ωt. For the following circuit. Just before the switch is closed at time t = 0. the charge on the capacitor is Q0 = 1.Problem 5. (b) (3 pts. (i gives the direction of positive current flow just after the switch is closed.) Calculate the value of ω for this circuit. C = 25 µF and L = 32 mH. (d) (4 pts. (a) (4 pts.) Write the loop equation for this circuit and convert it to a differential equation for the charge on the capacitor. . . . . . but you may be asked to verify that a given function satisfies a differential equation. If you are asked to solve simultaneous algebraic equations. Physics 19: The Physics 19 exam will consist of selected problems from the full Physics 21 exam. You should be familiar with all the topics and questions listed on the study guides for Hour Exams #1 and #2. • What is total internal reflection? What is the critical angle? Use of Calculators: You should bring a calculator. you might be asked for a short definition or an example. The equation sheet and the handout on optics (now posted on the course web site) will be included with the exam. Be able to draw a ray diagram for a magnifying glass or a telescope. You will not be asked to solve differential equations. extra time students will be in LL221 and will start 90 minutes early. • What does a polarizing filter do? • What is the paraxial approximation? • Be able to draw ray diagams for mirrors and lenses using the notes handed out. 2011 . December 9. 2011 Information about Final • What is refraction? What is the index of refraction? • What is disperson? Final Exam: The final exam will be Thursday. that is. What is a diffraction grating? • Be able to sketch the interference pattern for the two slit experiment if you are given the width and separation of the slits.Physics 21 Fall. The list is not necessarily complete but is representative. How does your analysis depend on whether the speakers are in phase or out of phase? • What is chromatic aberration? • What is an achromatic doublet? • Be able to give examples of and explain phenomena included under the heading of diffraction and interference. In general. Coverage: The exam will cover all the material presented this semester. Be able to analyize a two-lens system algebraically. December 15. There may be some shortanswer questions. The solution to the practice questions will be posted on the web. 2011 from 7:10–10:10 pm. however. The exam will be closed book and closed notes. • What is the difference between geometric and physical optics? What is the essential assumption of geometric optics? • How can you describe the path of a light ray using the principle of least time? • Why is the case of parallel light rays important for a lens or mirror? • What makes a rainbow? • What is the mathematical form of a traveling plane wave? • Know how to determine where the sound from two speakers will interfere constructively or destructively. The two previous hour exams and the associated practice exams provide representative questions on electricity and magnetism. including the sign conventions for the radii of curvature. About 50% of the exam will be on electricity and magnetism and about 50% on waves and optics. you must solve them by hand and show the solution to receive full credit. • What limits the ability of a telescope to resolve two closely spaced binary stars? What is the Rayleigh criterion? • What is the law of reflection (for a mirror)? • What determines the polarization of an E&M wave? What is the relation between E and B and the velocity of the wave? • Know how to use Snell’s Law. Almost everyone will be in PA 101. Physics 19 students may take the full three hours to work the exam. as well as with the items listed below that cover optics. What are missing orders and why do they occur? • What are the approximate wavelengths of visible light? • What is Huygens’ principle? • What is the f -number of a lens? • What is the difference between a real and a virtual image? • What is the difference between discrete and continuous spectra? Be able to give an example of each. Know how to use the lensmaker’s equation. recitation or homework. Some questions may come from material presented only in the lectures. Any physical constants and integrals you will need will be given on the equation sheet. at 5:30 pm. Practice Questions: Representative questions on waves and optics taken from previous exams in Physics 21 will be posted on the class web site. The emphasis will be on the material covered in lecture. • What is the Poynting vector? Know how to use it to describe the transport of energy by an E&M wave. • Know the sign conventions for converging and diverging lenses. setting up the problems and demonstrating the correct strategy for solving them are worth more than doing the arithmetic to get a numerical result. Two stars 10 light years away are barely resolved by a 90 cm (mirror diameter) telescope. (b) Determine the location and the height of the image using a ray diagram.14 m from an object of height 0. (a) Algebraically determine the location and the height of the image. Consider the polarizers in this problem as “perfect. the sun is directly overhead at a point on the equator. 2011 Practice Questions on Optics Some of these questions on optics have been adapted from questions on the Physics 21 final given at the end of the spring semester of 2004. what is the average intensity (power/meter2 ) of this sunshine? Problem 5. (a) What are the frequency and wavelength of this wave? (b) A second polarizer is oriented at 60◦ to the first. the electric field of a light wave is (in SI units) E = E0 cos[6..0 × 1015 t − 2. with E0 = 2.Physics 21 Fall. Let z be the direction straight down. Problem 1. What is the magnitude and direction of the electric field after passing through the third polarizer? Problem 2. t) = E0 cos 2π 5.0 × 10−7 z 14 ˆj B(z. wavelength and velocity of this wave? (b) If E0 = 1000 V/m.0 × 10 t − 6. i.0 × 10−4 m apart are illuminated with monochromatic light with a wavelength of 600 nm.0 × 10−7 14 (a) What are the frequency. the suns shines perpendicularly to the earth’s surface at that point.e. t) = B0 cos 2π 5.0 × 10 t − 6. .0 m from the slits? z ˆi E(z. L (a) What is the angular separation θ between the central fringe and the next bright fringe? (b) What is the spacing (in meters) between the bright fringes if the screen is L = 3. Try to work them using only the equation sheet and the notes on optics that will be provided with the final. Problem 3. A diverging lens with a focal length f = −0.” After passing through the first polarizer.0 × 107 x]ˆj.02 m. Two slits that are d = 2. Assume the electric and magnetic fields of this wave are (in SI units) 60O z (c) A third polarizer is oriented in the z direction. What is the magnitude and direction of the electric field after passing through this second polarizer? y (c) Is the image real or virtual? Problem 4. How far apart are the stars? Assume λ = 550 nm and that the resolution is limited by diffraction.06 m is 0. At noon on the first day of spring.0 × 104 V/m. 7 × 103 V/m.0 × 104 V/m. the electric field of a light wave is (in SI units) Problem 2. What is the magnitude and direction of the electric field after passing through the third polarizer? The projection of E1 on the axis of the third polarizer is E2 = E1 cos 30◦ = 8. Consider the polarizers in this problem as “perfect. E = E0 cos[6.0 × 1015 t − 2. (c) A third polarizer is oriented in the z direction.5 × 1014 Hz k = 2π/λ = 2. Problem 1.0 × 1015 ⇒ f = 9. (b) What is the spacing (in meters) between the bright fringes if the screen is L = 3.0 m from the slits? The spacing in meters is L∆θ = 3.0 × 10−4 m apart are illuminated with monochromatic light with a wavelength of 600 nm. What is the magnitude and direction of the electric field after passing through this second polarizer? y E0 L (a) What is the angular separation θ between the central fringe and the next bright fringe? The angular splitting (in radians) is given by O 60 E1 E2 ∆θ = 30O z The projection of E0 on the axis of the second polarizer is E1 = E0 cos 60◦ = 1 × 104 V/m. .Physics 21 Fall.” After passing through the first polarizer.0 m × 0.003 = 0.0 × 107 x]ˆj. with E0 = 2.009 m = 9 mm. Try to work them using only the equation sheet and the notes on optics that will be provided with the final.003 d 2 × 10−4 Note that this angular splitting is small enough that the small angle approximations used are justified. 600 × 10−9 λ = = 0.1 × 10−7 m (b) A second polarizer is oriented at 60◦ to the first.0 × 107 ⇒ λ = 3. 2011 Solution to Practice Questions Some of these questions on optics have been adapted from questions on the Physics 21 final given at the end of the spring semester of 2004. (a) What are the frequency and wavelength of this wave? ω = 2πf = 6. Two slits that are d = 2. (b) If E0 = 1000 V/m.02 m. We have do = 14 and f = −6. Problem 5. t) = B0 cos 2π 5. which is about 4 light minutes.0 × 10−7 1 object F image 2 3 F (The diagram above is not exactly to scale.22 = 7.14 m from an object of height 0. Let’s work the problem in cm. Let z be the direction straight down. the suns shines perpendicularly to the earth’s surface at that point. How far apart are the stars? Assume λ = 550 nm and that the resolution is limited by diffraction. That works out to 7 × 1010 m.00 × 108 )(1000)2 = 1330 W/m2 Using the fact that B0 = E0 /c for a plane wave. where L is 10 ly. The Rayleigh criterion gives θ = 1. wavelength and velocity of this wave? ω = 2πf = 2π(5. what is the average intensity (power/meter2 ) of this sunshine? The expression that uses only the electric field amplitude E0 is S = 12 0 cE02 = .0 × 10−7 z 14 ˆj B(z.06 m is 0.Problem 3. one can write equivalent formulas involving just E0 or B0 that give the same result.22 λ 550 × 10−9 m = 1.46 × 10−7 rad D 0.9 m The absolute distance is given by Lθ. t) = E0 cos 2π 5.) (c) Is the image real or virtual? The image cannot be focussed on a screen. so it is virtual.2 × 2 = 0. . At noon on the first day of spring. Two stars 10 light years away are barely resolved by a 90 cm (mirror diameter) telescope.2 So the image is 4. The height is hi = 4. (a) Algebraically determine the location and the height of the image.0 × 10 t − 6. The lens equation gives 1 1 1 1 1 1 − = = − = di f do −6 14 −4.6 cm 14 (b) Determine the location and the height of the image using a ray diagram. Problem 4. the sun is directly overhead at a point on the equator. A diverging lens with a focal length f = −0..0 × 1014 ) ⇒ f = 5 × 1014 Hz 2π ⇒ λ = 6 × 10−7 m k = 2π/λ = 6.2 cm in front of the lens.0 × 10 t − 6. but it’s close enough to show how the ray diagram looks. or about half the distance from the earth to the sun.e.5(8.0 × 10−7 14 (a) What are the frequency.85 × 10−12 )(3. Assume the electric and magnetic fields of this wave are (in SI units) z ˆi E(z. i.