Solutions and Explanations for HW Chapter 27 BBPrepared by Prof Omar Chmaissem and TA Abhilash Sajja Question 1 A circular loop of wire of radius 0.50 m is in a uniform magnetic field of 0.30 T. The current in the loop is 2.0 A. What is the magnetic torque when the plane of the loop is perpendicular to the magnetic field? 0.47 m∙N 0.52 m∙N 0.59 m∙N zero 0.41 m∙N Explanation: Magnetic Torque is given by = B sin where is the angle between the magnetic moment and the Magnetic field. Since the direction of is perpendicular to plane of the loop and the direction of B is perpendicular to the plane, should be parallel to B which results in to be 0°. = Bsin0° = 0 m.N Question 2 A stationary proton is in a uniform magnetic field of 0.20 T. What is the magnitude of the magnetic force on the proton? zero 3.2 × 10-20 N 1.6 × 10-20 N 1.6 × 10-21 N 3.2 × 10-21 N Explanation: Force on a particle moving in a magnetic field is given by F = q(v x B) where q is the charge of the particle, v is the velocity of the particle entering magnetic field, and B is the magnetic field strength. The magnitude of this magnetic force can be written as F = qvBsin where is the angle between the velocity vector and the magnetic field. In this case, the proton velocity is 0m/sec so the magnetic force will also be 0N. Question 3 A proton travels through a potential of 1.0 kV and then moves into a magnetic field of 0.040 T. What is the radius of the proton's resulting orbit? 0.11 m 0.19 m 0.17 m 0.14 m 0.080 m 1).2 × 10-19 N 2.04 T 0.2 × 10-18 N 2. Other given parameters are q = 1.01 T Explanation: The mass m experiences an acceleration a due to a net force F given by F = ma = 0. we get v = 4. Hence. Hence.004 T 0. q = 1.38 x 105 m/s. Question 5 A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg.2). Initially the proton is accelerated by a potential V = 1000V. What is the minimum magnetic field that would produce such an acceleration? 0.Explanation: Remember the equation we derived in class for the cyclotron radius. we get r = 0. the angle between the two vectors is indeed 45 degrees. r = mv/Bq [eq(25.11m 5 Question 4 An electron traveling due north with speed 4.3 × 10-19 N 2.6 x 10-19C in eq(25. the Electric potential energy gained by the proton is qV. Hence.06 N .0 × 10-5 T and is directed downward at 45° below the horizontal. This energy is converted into kinetic energy resulting in the proton moving with some velocity v.1 T 0. The mass experiences an acceleration a = 3 × 104 m/s2.3 × 10-20 N 3. v =4.67 x 10-27Kg.04T. Draw a sketch for the direction of the electron’s velocity due north and the earth’s magnetic field downward and tilted.0 × 10-5 T Substituting these parameters in the equation. This was obtained by setting the net force which is centripetal (mv2/r) to equal the magnetic force qvb.6 x 10-27kg. and speed v = 5 × 106 m/s enters a uniform magnetic field.3 x 10-18N.38 x 105 m/s. V = 1000V . the proton enters the magnetic field with this velocity v.1)]. Substituting m= 1. q = 1.6 x 10-19 C in eq(25. B = 5.0 × 105 m/s. B = 0. KE = PE (½)mv2 = qV eq(25. v = 4. To solve this equation we need to figure out the proton’s velocity.3 × 10-18 N Explanation: Straightforward.2) Substituting the values of mproton = 1. we get F = 2. The magnitude of the magnetic force on a particle moving in a magnetic field is given by F = qvB sin.001 T 0. We just need to clearly identify the angle .6 x 10 -19C. What force acts on the electron? 3.0 × 10 m/s enters a region where the Earth's magnetic field has the magnitude 5. 4 A is placed inside a magnetic field B = (0.3 × 103 V/m 5. Hence. Initially proton is accelerated by a potential difference V = 10.9 × 105 V/m 7.5 × 10-4 J +9.1 x 10-31Kg. KE = qV (½)mv2 = Vq eq(27.0 × 10-4 J -9.010 T.6 x 10-19C in eq(27.01 E = 5. to produce this force.1) Substituting the values of m = 9.2 T) iˆ + (0.93 x 107 x 0.5 × 10-4 J -2. It’s obvious that I need to figure out the velocity in order to solve for the electric field magnitude.0 × 105 V/m 2. Minimum magnetic field is obtained when sin is maximum because it’s in the denominator. sin = 1 and Bmin= F/qv Plug F= 0. we have the magnitude (IA) and the direction which is the unit vector nˆ . the magnetic field B should be B = F/qvsin. If the B-field of the velocity selector has a value of 0. we get v = 5.93 x 107 m/s Hence.06N.3 × 10-4 J Explanation: Magnetic loop potential energy is given by U = – μ .4 T) ˆj .6 iˆ – 0.2 × 106 V/m 6. we get B = 0.93 x 105 V/m Question 7 A loop of diameter d =12 cm. V = 10000 V. q = 1. the proton enters and exits undeflected the velocity selector with this velocity v. Applied to the magnetic moment vector μ . B (dot product of magnetic moment vector and magnetic field vector B) A vector is defined as having two quantities: magnitude and direction.9 × 103 V/m Explanation: We already know that in a velocity selector we obtain v = E/B. The normal to the loop is parallel to the unit vector nˆ = 0. v = E/B E = vB E = 5. what value of the E-field is required if the particles are to be undeflected? 7.0 × 10-4 J -4. I know the magnetic field but v and E are unknown. Therefore.The only force acting significantly on this charge is the magnetic force.000V Hence. carrying a current I = 0. q = 3 x 10-6C.8 ˆj . Therefore the net force found above is the same as F = qvBsin.004T Question 6 A beam of electrons is accelerated through a potential difference of 10 kV before entering a velocity selector. v = 5x106 m/s into the above equation. What is the potential energy of the loop? +4.1). 6 iˆ – 0.6m. 0.6 x 10-3)(0.0 × 103 m/s into a magnetic field of 0.2 iˆ + 0.2 × 10-16 N Explanation: Magnetic force on a particle moving in a magnetic field is given by F = qvB sin where is the angle between the direction of the particle motion and the magnetic field.8 ˆj ) μ = (2. =90°. Substituting the parameters I = 2A.6 x 10-3 ˆj ) . B = 0. What is the force that acts on the proton? 0N 6.4) = 0.4( * (0. Substituting the parameters as v = 7000m/s. F = 0.7 x 10-3 iˆ – 3.3T. (0.4 ˆj ) = (–2.m2 B = (0.2 iˆ + 0.60 m in length.60 T perpendicular to the motion of the proton.9 x 10-3 J = 9 x 10-4 J Question 8 A wire.6 x 10-3 ˆj ) A.18 N.0 A and is placed at a certain angle with respect to the magnetic field of strength 0.7 x 10-3)(0. μ = 0.4 ˆj ) T U=–μ.2)+( 3. what angle does the wire make with respect to the magnetic field? 35° 20° 25° 60° 30° Explanation: Magnetic force on a current carrying wire is given by F = ILB sin() where is the angle between the wire length vector taken in the direction of the current (I) and the magnetic field (B).4 ˆj ) U = (2.6 sin() sin() = 0. Question 9 A proton is projected with a velocity of 7. B = 0.4 × 10-16 N 4. we get 0.3 x 2 x 0.7 x 10-3 iˆ – 3.7 x 10-16 N .18 = 0. we get F = 6. If the wire experiences a force of 0.6 * 10-19 (charge of proton).5 = 30° Note that if the angle were to be zero (meaning the wire is parallel to the field) then there would be no magnetic force present.2 iˆ + 0.7 x 10-3 iˆ + 3. is carrying a current of 2. L = 0.18N in the above equation.06)2) * (0. (0.6T q = 1.30 T.6 x 10-3 ˆj ) .B U = –(2.7 × 10-16 N 13 × 10-16 N 3. 8 ˆj .8 × 10-4 Nm 4. L = 1.m . Or.8 ˆj ) = (–9.05)2 * (– 0.3 kˆ ) Please review the cross product rules = (+2.42 x 10-4 iˆ –12. The normal to the loop is parallel to a unit vector nˆ = . magnetic moment vector can be obtained by multiplying its magnitude (IA) with the unit vector representing its direction nˆ = – 0. To get the minimum value of the current flowing in the wire we have to set to be 90o ( maximizing the denominator – there are no other variables to change).10 T. you do cross product according to the torque equation.83) 2 (3. we get I = 4.8 A 2.0 m long has a mass of 0.6 iˆ – 0.m2 (cross product) = xB = = (–9.42 x 10-4 iˆ –12.6 iˆ – 0.3 T kˆ .6 iˆ – 0.5 A Explanation: Magnetic force on a current carrying wire is given by F = ILBsin where is the angle between the current direction (I) and the magnetic field (B).6 × 10-4 Nm 2. Substituting the parameters . =90° in the above equation. What minimum current in the rod is needed in order for the magnetic force to cancel the weight of the rod? 7.56 x 10-4 ˆj ) A.1T. Question 11 ) A loop of diameter d = 10 cm. (This is a nice but tedious problem to solve. Calculate the magnitude of the torque on the loop. You will need to determine vector A.Question 10 A thin copper rod 1. carrying a current I = 0.56 x 10-4 ˆj ) x (0.77) 2 = 4.050 kg and is in a magnetic field of 0.2 × 10-4 Nm 0 0.2 A is placed inside a magnetic field B = 0. mg = ILBsin.7 × 10-4 Nm Explanation: As described in the hint. B = 0.9A 4.8 ˆj Hence vector = (I*A) *unit vector n = 0. This force has to be canceled by gravity.2**(0.7 x 10-4 N.0m.2 A 4. This can be done by multiplying the magnitude of A by the unit vector direction because it is in the same direction. Once you figure it out.77 x 10-4 iˆ ) Nm Magnitude of =10-4 (2.0. Solving for I.) 1.9 A is the minimum current required to generate a magnetic force which could cancel the force due to gravity. we get I = mg/ LBsin.83 x 10-4 ˆj – 3. The answer is in a vector notation form.9 A 9.6 A 1. You will need then to calculate the magnitude of the torque vector. Hence. and A is the area of the loop. Question 13 An electron moving perpendicular to a magnetic field of 3. It enters a magnetic field of 0.2 × 10-2 T moves in a circle of radius 0. 1. It is the same as the direction of the vector A representing the loop area A.Question 12 What is the magnetic moment of a rectangular loop of 120 turns that carries 6.40 cm.3 × 107 m/s 1.9 × 10-53 m/s Explanation: The centripetal force is balanced by the magnetic force resulting in a circular motion.3 A. mv2/r = qvBsin() (with =90°) v = qBr/m -2 v = (3.6 x 10-19 x 0.0 cm? 2.0 × 109 m/s 1.3 x 107 m/s. Question 14 A proton is accelerated from rest through 500 V.80 × 107 m/s 2. I is the current in the loop. Determine the radius of the path it follows.30 T oriented perpendicular to its direction of motion.23 A∙m2 Explanation: Magnetic moment of a rectangular loop is given by = NIA where N is the number of turns.1 m 11 m 11 cm 1.2 × 10 x 1.9 × 10-30 m/s 0.004) / (9.m2 The magnetic moment is a vector quantity with a direction perpendicular to the current loop according to the righthand rule.1 mm 1.023 A∙m2 0.3 A∙m2 230 A∙m2 23 A∙m2 0.0 A if its dimensions are 4.1 cm Explanation: .0 cm × 8. How fast is this electron moving? 3. = 120*6*32x10-4 = 2.1 x 10-31) v = 2. 6 x 10-19 C.52 m∙N Explanation: Magnetic Torque is given by = Bsin where is the angle between the magnetic moment and the Magnetic field. we get r = 1. and e is electron charge. n is the carrier density.50 m is in a uniform magnetic field of 0.5 × 1028/m3 in this material? 0.67 x 10-27Kg.5 x 1028 /m3. Remember the equation.2 μV Explanation: Hall voltage is given by VH = IB/ned where I is the current through the strip. KE = qV (½)mv2 = qV eq(25.3 μV 6.3 x 10-6 V (approx). B is the magnetic field. we get VH = 1. q = 1. We need the speed. we get v = 3.3T.1).0 A. r = mv/qB eq(25. v = 3. the centripetal force balancing the magnetic force From that we derive. e = 1. What is the magnetic torque when the plane of the loop is parallel to the magnetic field? 0. should be perpendicular to B which results in to be 90°. q = 1.We know this is going to be a circle. d=10-4 m in the above equation.59 m∙N zero 0.2).1 x 105 m/s Hence.05 T.72 μV 4. the proton enters the magnetic field with this velocity v.30 T.05T. The magnetic field is perpendicular to the direction of the strip along which there is a 10-A current.6 x 10-19C in eq(25.47 m∙N 0.5 μV 1. Substituting I = 10A.1) Substituting the values of m = 1. = Bsin90° ( = IA) . V = 500V . The current in the loop is 2.2 μV 2.1 x 10 m/s.6 x 10-19C in eq(25. B = 0. Since the direction of is perpendicular to the plane of the loop and the direction of B is parallel to the plane. n = 2.1 cm Question 15 In a Hall experiment a conducting strip of thickness d = 100 μm is placed in a magnetic field B = 0.2) -27 5 Substituting m= 1. We can calculate the incoming kinetic energy since we know that the proton is accelerated in a potential difference V = 500V Hence. d is the strip thickness. B = 0. What is the Hall voltage measured across the strip if the charge carrier density is 2.67 x 10 kg.41 m∙N 0. Question 16 A circular loop of wire of radius 0. 4A.52) x (0.4 × 10 T.4 in the above equation.0 × 10-4 T) = + (2. L = 0. B = 11.0 × 10-4 T) = . is in a uniform magnetic field of 0. carrying a current of 4 A.0 × 10-4 T) = .3) x sin90° = 0.2m. B = 0.2 N Question 18 A wire carries a current of 11.4° with a magnetic field of -3 magnitude 11. The electric field is through.(2.6 T.0 × 10-4 T) = + (5.14 x0.4mT.104 .93 × 10-3 N.6 x 4 x 0.N Question 17 A straight wire 20 cm long.47 m.114m.4 x 0. we get F = 0. 0. 2. Explanation: Magnetic force on a current carrying wire is given by F = ILBsin where is the angle between the current direction (I) and the magnetic field (B).5 N 0.4 × 10-3 N. 11.114x sin 11.0 × 10-4 T) =.(2.4 A in a direction that makes an angle of 11.4x10-3 x 11.4 cm of this wire is 1.48 × 10-2 N. Question 19 An electron moving with a velocity = 5.4 N 0. What is the force on the wire when it is at an angle of 30° with respect to the field? 0.2 N Explanation: Magnetic force on a current carrying wire is given by F = ILBsin where is the angle between the current direction (I) and the magnetic field (B).4 F = 2. undeflected? = + (2.6 N 0.93 x 10-3 N. Substituting the parameters I = 4A. 0.130 N. = 11. What magnetic field will allow the electron to go . we get F = 11. Substituting the parameters I = 11.6T.2 sin 30° F = 0. The magnitude of the force on 11.013 N. = 30° in the above equation. = IABsin90° = (2) x (3.0 × 107 m/s enters a region of space where perpendicular electric and a magnetic fields are present. L = 0.3 N 0. 0 × 107 m/s ) x (Bx +By +Bz kˆ ) = (5.0 × 10-4 T) kˆ 2 Question 20 A circular loop of wire of cross-sectional area 0.) .12x0. Lorentz force = Electric force + Magnetic force When the charge goes undeflected it’s because the electric force is canceled out by the magnetic force and vice versa. It is placed in a magnetic field of 0. each carrying 0. Hence. We get the following relationship: (-5.52 m.Electric force Using vector notations because we also need to figure out the magnetic field direction: q v B qE v B E (where the x sign represents the cross product) v B = (5.050 T oriented at 30° to the plane of the loop.Explanation: This is the velocity selector device.25 m∙N 2. B = .5 m∙N 25 m∙N Explanation: Since the magnetic field is oriented 30° to the plane of the loop. 0 = Electric force + Magnetic force Magnetic force = .5x0.2 × 10 m/s at an angle of 80° to a uniform magnetic field of strength 1.2.0 × 107)Bz = (104) or Bz = 104/(-5. This is essentially saying that the Lorentz force is zero. What torque acts on the loop? 5. Draw a sketch and check the actual angle.12 m consists of 200 turns.) E 10 4 ˆj so E 10 4 ˆj Equating the two equations v B E show that (By) must be zero. v = E/B We did not use this name in class but the sum of the electric and magnetic forces is called the Lorentz force.0 × 107)Bz (. = 60° Magnetic Torque is given by =NIABsin = 200x0.N 6 Question 21 A proton is moving at a speed of 3. What is the distance moved by the proton along the spiral path as it completes one revolution? (This distance is called the "pitch" of the helix.0 × 107) = .0 x 10-4 T.50 A.2 m∙N 0.52 m∙N 0. B should be 60° (90° – 30°) to the magnetic moment .05xsin60° = 0.5 × 10-4 T.(2. Bz obviously represents a magnetic field that is fully pointing in the negative z direction with the magnitude we just calculated.0 × 107) By kˆ + (5. T is the time required to make a full revolution.1 × 10-16 N eastwards 0N 1.37 x 10-4 seconds Notice that the period (or frequency) is actually independent of the velocity. To emphasize: the forward distance is solely due to the parallel velocity component as the distance equation shows. we can calculate two components for the velocity: Parallel to the magnetic field v// = v cos = 0.555× 106)( 4.5 x 10-4)]= 4.602 x 10-19 ) (1.672 x 10-27 kg)/[(1. In vector notation v = 5000 iˆ m/s. It’s obvious that we need to calculate the period T. In vector notation B = 0.20 T pointing northward. This means that the distance travelled around the circular path (due to the perpendicular component of the velocity) is 2r = v┴T .4 × 10-16 N westwards Explanation: The proton is moving eastwards and can be considered to be moving in the positive x direction ( iˆ ). The pitch distance is p = v//T. F = q(v x B) F = 1.2 ˆj T. Plug and cancel the v┴ terms to get: 2 m/qB = T T = 2 (1. Now that we know the time. When the proton completes one full revolution in one period T due to the perpendicular velocity component.555 × 106 m/s and Perpendicular to the field v┴ = v sin = 0. 90o or 180o should produce a spiral path for a charge moving in a magnetic field.240 m 1390 m 1370 m 220 m cannot be determined Explanation: We already know that any travel angle different than 0o. We just mentioned that it is the time for the time to complete a full revolution.6x10-19 (5000 iˆ X 0. The magnetic field is pointing in the north direction and can be considered to be in the positive y direction ( ˆj ). let’s calculate the pitch distance: p = (0. r is not known so we have to use r = mv┴/qB.6 × 10-16 N downwards 1.0 × 103 m/s enters a magnetic field of 0.985 × 106 m/s. In this problem.2 ˆj ) F = 1.2( iˆ X ˆj ) (cross product of iˆ x ˆj = kˆ ) . What is the magnitude and direction of the force that acts on the proton? 1.6x10-19 x 5000 x 0.6 × 10-16 N upwards 4. it finds itself propelled forward by the pitch distance due to the parallel velocity component.37 x 10-4) = 242 m Question 22 A proton moving eastward with a velocity of 5. (8 × 10-3 N) kˆ is acting on a charge q = 2x10-9 C moving inside a uniform magnetic field with a velocity v = (2x106 m/s) iˆ + (3x106 m/s) ˆj .(8 × 10-3) kˆ .58 ˆj + 0.24 kˆ . this would give a thumb direction pointing up.0.(3T) + (2T) = (2 T) .(2 T) + (1 T) = (4 T) . . The force on the wire is 0.5 T) = . =.(1 T) Explanation: F = q*(v x B) F is known in a vector form F = (6 × 10-3) iˆ .F = 1.(0.(4 × 10-3 N) ˆj .(4T) + (0. Question 23 A force F = (6 × 10-3 N ) iˆ .78 iˆ . Question 24 A current is flowing in a wire in direction 3 iˆ + 4 ˆj where the direction of the magnetic field is 5 ˆj + 12 kˆ . All we need is to calculate the right side vector and compare the components from which we can get the magnetic fields components.6x10-16 kˆ ( kˆ represents the positive z direction or upwards) Alternately.5 T) =.(4 × 10-3) ˆj . magnetic field vector is (-2 ˆj + 1 kˆ ) Tesla. curl them towards the north. We have a positive charge so this is the direction of the magnetic force. we can determine the magnitude F = qvBsin90o (field and velocity vectors are perpendicular to each other) and figure out the direction of the force using the right hand rule. Consider the general magnetic field B = Bx iˆ +By ˆj +Bz kˆ but the x.component is zero so: B = By ˆj +Bz kˆ iˆ -9 q(v B) = 2 10 2 10 6 0 ˆj 3 10 6 By kˆ 0 iˆ 6 10 3 Bz ˆj 4 10 3 Bz kˆ 4 10 3 B y Bz Now equate the two sides: 6 10 iˆ 4 10 ˆj 8 10 kˆ iˆ6 10 3 3 3 3 Bz ˆj 4 10 3 Bz kˆ 4 10 3 By This equality leads to: 6 10 6 10 B or B 1 4 10 4 10 B same solution 8 10 4 10 B or B 2 3 3 z 3 z 3 z 3 3 y y Hence. Calculate the components of the magnetic field knowing that the x-component of the field is equal to zero. Fingers towards east. Question 26 An electron enters a region of space where there is a magnetic field B = 7 mT kˆ with a velocity v = ( 9 × 106 m/s) iˆ + ( 7 × 106 m/s) ˆj + ( 5 × 106 m/s) kˆ . What is the mass of the particle? 1.020 m.2T.5 × 10-27 kg 3.5 cm 2.02m.58 ˆj + 0.7 cm 1. r= 0.4 x 10-28 kg. we get m = 6.58 ˆj + 0.4 × 10-28 kg Explanation: Mass spectrometer is an application which uses the magnetic force on a moving particle.6 0 ˆj kˆ 0.8 0 0.7 × 10-27 kg 3.92) iˆ –(0. L is a scalar quantity.78 iˆ + 0. v = 106 m/s in the above equation.20 T. the direction of the force is determined by the current and magnetic fields (cross product using the right hand rule).78 iˆ + 0.78 iˆ – 0.0 × 106 m/s and enters a uniform magnetic field of 0. What is the spacing of the circles of the helix on which the electron moves? 7. Remember the equation. iˆ Direction of force is given by directions of L x B = 0. The radius of the circular orbit is 0.1 × 10-31 kg 6.92 L x B = (0.ˆj .38 0.24 kˆ 0.92) ˆj +(0. F = I(L x B) Since. direction of B = (5 ˆj + 12 kˆ )/ (5)2 (12)2 . Explanation: The direction of force on the wire can be obtained using the unit vectors of the current and magnetic field.6 x 10-19C. direction of L = (3 iˆ + 4 ˆj )/ (3) 2 (4) 2 .38) kˆ L x B = 0.58 ˆj . 0. a single-charged particle (charge e) has a speed of 1.24 kˆ (approx) Question 25 In a mass spectrometer. q=1.24 kˆ .71( iˆ + ˆj ).0.6*0.2 × 10-28 kg 4. 0.8*0.6 cm . L = (3/5) iˆ + (4/5) ˆj . B = (5/13) ˆj + (12/13) kˆ .6*0.1 cm 5. the centripetal force balancing the magnetic force From that we derive m = qBr/v Substituting B = 0. the pitch of the spiral is Pitch (p) = (v//)(2m/qB) p = (5 x 106 * 2 *9.025 m∙N.0019 m. The magnitude of the torque on the loop is 0. the magnetic field is in the positive z direction ( B = 7 mT kˆ ) so the parallel velocity component is ( 5 × 106 m/s) kˆ .037. = 90° = 0. 0. Explanation: Draw a sketch and convince yourself that the angle between vectors and B is 90o.037 T.5A.6 cm Question 27 A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.N . Two sides of the loop are oriented parallel to a uniform magnetic field of strength 0.1 x 10-31) / (1. Here.02. T = 2m/qB So. B = 0.050 m∙N. 2The time registered during travel is the period.9. 0.0019 m∙N.5 A flowing through it. 0.0093 m∙N. 0. the other two sides being perpendicular to the magnetic field.5 cm Explanation: Please check problem #21 for important conclusions we made there: 1Pitch depends on the velocity component that is parallel to the magnetic field direction. 3The period is independent of the perpendicular velocity component(s). A = 0.038 m∙N.6 x 10-19 * 7 x 10-3) x = 2. Magnetic Torque is given by = Bsin where is the angle between magnetic moment and Magnetic field =IABsin Substituting the values I = 2.