11/24/2016Ch 29 HW Ch 29 HW Due: 11:59pm on Friday, November 25, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy The AmpèreMaxwell Law Learning Goal: To show that displacement current is necessary to make Ampère's law consistent for a charging capacitor Ampère's law relates the line integral of the magnetic field around a closed loop to the total current passing through that loop. This law was extended by Maxwell to include a new type of "current" that is due to changing electric fields: ∮ ⃗ ⃗ B ⋅ dl = μ (Icharge + Idisplacement ) 0 . The first term on the righthand side, Icharge , describes the effects of the usual electric current due to moving charge. In this problem, that current is designated I (t) as usual. The second term, Idisplacement = ϵ0 dΦ E dt , is called the displacement current; it was recognized as necessary by Maxwell. His motivation was largely to make Ampère's law symmetric with Faraday's law of induction when the electric fields and magnetic fields are reversed. By calling for the production of a magnetic field due to a change in electric field, this law lays the groundwork for electromagnetic waves in which a changing magnetic field generates an electric field whose change, in turn, sustains the magnetic field. We will discuss these issues later. (Incidentally, a third type of "current," called magnetizing current, should also be added to account for the presence of changing magnetic materials, but it will be neglected, as it has been in the equation above.) The purpose of this problem is to consider a classic illustration of the need for the additional displacement current term in Ampère's law. Consider the problem of finding the magnetic field that loops around just outside the circular plate of a charging capacitor. The coneshaped surface shown in the figure has a current I (t) passing through it, so Ampère's law indicates a finite value for the field integral around this loop. However, a slightly different surface bordered by the same loop passes through the center of the capacitor, where there is no current due to moving charge. To get the same loop integral independent of the surface it must be true that either a current or an increasing electric field that passes through the Ampèrean surface will generate a looping magnetic field around its edge. The objective of this example is to introduce the displacement current, show how to calculate it, and then to show that the displacement current Idisplacement (t) is identical to the conduction current Icharge (t). Assume that the capacitor has plate area A and an electric field E(t) between the plates. Take μ0 to be the permeability of free space and ϵ0 to be the permittivity of free space. Part A ⃗ ⃗ First find ∮ R B⃗ ⋅ dl , the line integral of B around a loop of radius R located just outside the left capacitor plate. This can be found from the usual current due to moving charge in Ampère's law, that is, without the displacement current. Find an expression for this integral involving the current I (t) and any needed constants given in the introduction. ANSWER: ∮ R ⃗ ⃗ B ⋅ dl = μ0 I (t) Correct Part B ⃗ ⃗ Now find an expression for ∮ R B⃗ ⋅ dl , the same line integral of B around the same loop of radius R located just outside the left capacitor plate as before. Use the surface that passes between the plates of the capacitor, where there is no conduction current. This should be found by evaluating the amount of displacement current in the AmpèreMaxwell law above. Express your answer in terms of the electric field between the plates E(t), dE(t)/dt, the https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706981 1/36 11/24/2016 Ch 29 HW Express your answer in terms of the electric field between the plates E(t), dE(t)/dt, the plate area A, and any needed constants given in the introduction. Hint 1. Find the electric flux What is the electric flux ΦE (t) through this surface? Express your answer in terms of E(t) and other variables given in the introduction. Hint 1. The definition of electric flux The electric flux ϕE through a surface is defined by ⃗ ⃗ ϕE = ∫ E ⋅ dA, where the integral is taken over the entire surface. ANSWER: ΦE (t) = AE(t) ⃗ Hint 2. Express ∮R B⃗ ⋅ dl in terms of ΦE (t) ⃗ Find ∮ R B⃗ ⋅ dl in terms of the electric flux ΦE (t) between the plates. Express your answer in terms of ΦE (t), dΦE (t)/dt , and any needed constants given in the introduction. Hint 1. How to approach the problem Use the extension to Ampère's law and the definition of displacement current given in the problem introduction. ANSWER: ∮ R ⃗ ⃗ B ⋅ dl = μ0 ϵ0 dΦ E (t) dt ANSWER: ∮ R ⃗ ⃗ B ⋅ dl = μ0 Aϵ 0 dE(t) dt Correct A necessary consistency check Part C https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706981 2/36 11/24/2016 Ch 29 HW We now have two quite different expressions for the line integral of the magnetic field around the same loop. The point here is to see that they both are intimately related to the charge q(t) on the left capacitor plate. First find the displacement current Idisplacement (t) in terms of q(t). Express your answer in terms of q(t), dq(t)/dt , and any needed constants given in the introduction. Hint 1. Find the flux using Gauss's law Assign the capacitor an arbitrary plate area A. If we apply Gauss's law for electric flux to the the left plate of the capacitor, we find that ⃗ ⃗ ∮ E ⋅ dA = qencl ϵ0 , where the electric field is entirely inside the capacitor. What is the electric flux ΦE (t)? Express your answer in terms of q(t) and any needed constants given in the introduction. Hint 1. Gauss's law Gauss's law states that the electric flux through a closed surface equals the enclosed charge divided by ϵ0 . ANSWER: ΦE (t) = q(t) ϵ0 Hint 2. Find the displacement current What is the displacement current Idisplacement (t)? Express your answer in terms of ΦE (t), dΦE (t)/dt , and any needed constants given in the introduction. ANSWER: Idisplacement (t) = ϵ0 dΦ E (t) dt ANSWER: Idisplacement (t) = dq(t) dt Correct Part D Now express the normal current Icharge (t) in terms of the charge on the capacitor plate q(t). Express your answer in terms of q(t), dq(t)/dt , and any needed constants given in the introduction. ANSWER: Icharge (t) = dq(t) dt Correct Using Gauss's law, you have shown that the displacement current from the changing electric field between the plates equals the current from the flow of charge through the wire onto that plate. This means that the AmpèreMaxwell law can consistently treat cases in which the normal current due to the flow of charge is not continuous. This realization was a great boost to Maxwell's confidence in the physical validity of his new displacement current term. The Magnetic Field in a Charging Capacitor When a capacitor is charged, the electric field E , and hence the electric flux Φ, between the plates changes. This change in flux induces a magnetic field, according to Ampère's law as extended by Maxwell: ⃗ ⃗ https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706981 ( ) 3/36 where the electric flux changes but the conduction current I is zero. and B(r). ΦL . Express your answer in terms of μ0 and given quantities. ANSWER: f = r 2 a2 Hint 5. Such a case is given in this problem. Part A A parallelplate capacitor of capacitance C with circular plates is charged by a constant current I . ANSWER: dΦ tot dt = I ϵ0 Hint 4. How to approach the problem The AmpèreMaxwell law allows you to find the magnetic field in cases of high symmetry. which reduces to the ratio of the areas. ⃗ ⃗ B ⋅ dl evaluated about the Amperian loop defined in the first hint? Remember that the magnetic field is everywhere tangent Express your answer in terms of π. Φtot . What is the value of ∮ to this loop. is a fraction of the change of flux across the whole area of the capacitor. You should convince yourself that the magnetic field vector is everywhere tangential to circles around an imagined axis joining the centers of the two circular plates. Therefore. Find the scaling factor f ≡ ΦL Φ tot . the only component of the magnetic field that matters is Bθ . The radius a of the plates is much larger than the distance d between them. But the change of electric flux will be the same everywhere between the plates. You will calculate this magnetic field in the space between capacitor plates. We call the radius of this circle r. the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates. ANSWER: Φtot q = ϵ0 Hint 3. What is the rate of change of the electric flux? What is the rate of change of electric flux dΦ tot dt between the capacitor plates? Recall that dq dt = I . Hint 2. Scaling for the area within the Ampèrian loop The change of flux through the Ampèrian loop of radius r.com/myct/assignmentPrintView?assignmentID=4706981 4/36 . What is the electric flux between the plates? Express the electric flux Φtot charge q on each plate.11/24/2016 Ch 29 HW ⃗ ⃗ ∮ B ⋅ dl = μ (I + ϵ 0 0 dΦ dt ). Calculate B(r). ANSWER: ⃗ ⃗ ∮ B ⋅ dl = B(r)⋅2πr ANSWER: https://session. 2 = πa E between the plates of the capacitor in terms of the parameters of the problem and the magnitude of the Express your answer in terms of ϵ0 and given quantities. so fringing effects are negligible. Hint 1. all we need to do is to calculate the change of flux across all the space between the capacitor plates and then multiply by a scaling factor. Hence we can conveniently integrate around such a circle. the radius r. and realize that only the change of flux through the area within the circle will contribute to the magnitude of the magnetic field vector B on its perimeter.masteringphysics. which is constant around the loop. Evaluate the integral → Since we take the scalar product of B⃗ with dl . com/myct/assignmentPrintView?assignmentID=4706981 5/36 .00×10−2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. Part A What is the total magnitude of the magnetic flux ( Φinitial ) through the coil before it is rotated? Express your answer numerically. since the plane of the coil is parallel to the magnetic field and the area vector vector is normal to the plane of the coil. where ΔΦ1 is the total change in magnetic flux through each loop. in webers.11/24/2016 B(r) Ch 29 HW = μ0 I r 2πa2 Correct ± Magnetic Flux and Induced EMF in a Coil In a physics laboratory experiment. You did not open hints for this part.10×10−5 T . to at least three significant figures. Formula for the average emf induced in a coil (Faraday's law) The formula for the average emf E induced in a coil is E = −N dΦ 1 dt = −N ΔΦ 1 Δt . The magnitude of Earth's magnetic field at the lab location is 5. The angle between the magnetic field and the area vector The angle between the magnetic field and the area vector after rotation is 90 degrees. to at least three significant figures. Hint 1. and Δt is the time interval over which this change occurs. N is the total number of loops. in webers.masteringphysics.7 cm2 is rotated during the time interval 4. Hint 1. ANSWER: https://session. a coil with 250 turns enclosing an area of 12.25⋅10 −5 Wb Incorrect; Try Again; 10 attempts remaining Part B What is the magnitude of the total magnetic flux Φf inal through the coil after it is rotated? Express your answer numerically. ANSWER: |Φf inal | = 0 Wb Correct Part C What is the magnitude of the average emf induced in the coil? Express your answer numerically (in volts) to at least three significant figures. ANSWER: |Φinitial | = 1. Part A When the switch is open. There is a positive flux through the wire loop.com/myct/assignmentPrintView?assignmentID=4706981 6/36 . which of the following statements about the magnetic flux through the wire loop is true? Assume that the direction of the vector area of the wire loop is to the right. you will use Lenz's law to explore what happens when an electromagnet is activated a short distance from a wire loop. You will need to use the righthand rule to find the direction of the induced current. The induced current is counterclockwise.05×10−4 V Correct Understanding Changing Flux In this problem. ANSWER: There is no magnetic flux through the wire loop. the electromagnet does not produce a magnetic field. there is no current flowing in the circuit. Thus. What is the direction of the induced current in the wire loop immediately after the switch is closed (as seen from the left)? Hint 1. Correct Part C Now the switch on the electromagnet is closed. Consider the arrangement shown in . and the flux is zero. Part B What is the direction of the induced current in the wire loop (as seen from the left) when the switch is open? ANSWER: There is no induced current. The induced current is clockwise. There is a negative flux through the wire loop. Correct When the switch is open.masteringphysics.11/24/2016 Ch 29 HW average induced emf = 4. How to approach the problem https://session. C.com/myct/assignmentPrintView?assignmentID=4706981 7/36 . The figure shows some of the magnetic field lines produced by the electromagnet. How to approach the problem When the switch is closed. B.masteringphysics. https://session. a counterclockwise) current induced in the loop (as seen from the left). Enter the letters corresponding to the responses that correctly complete the statement above. The induced current is clockwise. the current in the electromagnet is clockwise as seen from the left. a current flows through the coil of the electromagnet. Hint 2. and the magnetic field disappears. The field produced by the electromagnet The figures show the magnetic field produced by the electromagnet when the switch is closed and then the field when the switch is open. zero. a clockwise.11/24/2016 Ch 29 HW When the switch is open. and a magnetic field is produced. there is a nonzero magnetic field produced by the electromagnet. remains constant). Determine how this changes the flux through the wire loop and use Lenz's law to determine the direction of the induced current. Determine how this field changes the flux through the wire loop. B. the magnetic field produced by the electromagnet is zero. and there is _______ (A. Hint 2. For example. type A. ANSWER: There is no induced current. Hint 3.C Hint 1. increases. When the switch is closed. when the switch is closed. Use the righthand rule to find the direction of the induced current that would generate a field opposing the change in flux. When the switch is opened. Then use Lenz's law to determine the direction of the induced current. Therefore. The field produced by the electromagnet The current flows from the positive battery terminal to the negative one. C. current stops flowing through the electromagnet. decreases. The magnitude of the external magnetic flux through the wire loop ______ (A. Correct Part D Finally. if the correct answers are A and C. The induced current is counterclockwise. Apply Lenz's law Lenz's law states that the current induced in the wire loop has the direction such that the magnetic flux created by the current opposes the change in the external magnetic flux that caused the current. the switch on the electromagnet is reopened. a current flows through the electromagnet. is closed. and a magnetic field is produced.11/24/2016 Ch 29 HW Hint 3. How to approach the problem When the switch is open. The field produced by the electromagnet The current flows from the positive battery terminal to the negative one. What is the direction of the induced current in the wire loop (as seen from the left)? Hint 1.com/myct/assignmentPrintView?assignmentID=4706981 8/36 .B Correct Now consider the new arrangement shown in . Note that the orientation of the battery is reversed with respect to the first arrangement you considered. initially open. the magnetic field produced by the electromagnet is zero. Apply Lenz's law Lenz's law states that the current induced in the wire loop has the direction such that the magnetic flux created by the current opposes the change in the external magnetic flux that caused the current. Hint 2. https://session. when the switch closes. Use the righthand rule to find the direction of the induced current that would generate a field opposing the change in flux. Answer the following questions related to the arrangment with the new battery orientation. Determine how this field changes the flux through the wire loop and use Lenz's law to determine the direction of the induced current. Therefore. the current in the electromagnet is counterclockwise as seen from the left. The figure shows some of the magnetic field lines produced by the electromagnet (recall that the battery has the opposite orientation to that of the first arrangement). ANSWER: B.masteringphysics. Part E The switch on the electromagnet. When the switch is closed. C.C Hint 1. Use the righthand rule to find the direction of the induced current that would generate a field opposing the change in flux. and use Lenz's law to determine the direction of the induced current. and the magnetic field disappears. a counterclockwise) current induced in the loop (as seen from the left. Hint 2. ANSWER: There is no induced current. Apply Lenz's law Lenz's law states that the current induced in the wire loop has the direction such that the magnetic flux created by the current opposes the change in the external magnetic flux that caused the current. Enter the letters corresponding to the responses that correctly complete the statement above. Apply Lenz's law Lenz's law states that the current induced in the wire loop has the direction such that the magnetic flux created by the current opposes the change in the external magnetic flux that caused the current. a clockwise. C. B. For example: A. The induced current is clockwise. The field produced by the electromagnet The figures show the magnetic field produced by the electromagnet when the switch is closed and then when the switch is open. How to approach the problem When the switch is closed. there is a nonzero magnetic field produced by the electromagnet. B. and there is _______ (A. remains constant). The induced current is counterclockwise. zero. Correct Part F Now the switch on the electromagnet is reopened.11/24/2016 Ch 29 HW Hint 3. current stops flowing through the electromagnet. Hint 3. The magnitude of the external magnetic flux through the wire loop ______ (A. increases.com/myct/assignmentPrintView?assignmentID=4706981 9/36 . decreases. Determine how this changes the flux through the wire loop. Use the righthand rule to find the direction of the induced current that would generate a field opposing the change in flux. When the switch is opened.masteringphysics. https://session. ANSWER: https://session. the electromagnets are switched on. The brake consists of a set of electromagnets that are held just above the rails. looking down at the rail through the electromagnet. Which way must the induced magnetic field point in the segments of rail under the leading and trailing edges of the moving electromagnet? The righthand rule will tell you the direction a given eddy current must have to induce the magnetic field predicted by Lenz’s law. launch the video below. Then. creating a magnetic field that induces eddy currents in the metal rails passing beneath them. which of the choices correctly represents the eddy currents induced in the rails? The diagrams represent a view from above. Part A Electric rail cars often use magnetic braking.com/myct/assignmentPrintView?assignmentID=4706981 10/36 .masteringphysics. According to Lenz’s law. this induced magnetic field points in the direction that opposes the change in the external magnetic flux. You can watch the video again at any point. Is the flux the same under the leading and trailing edges of the electromagnet? Are there places where the flux is increasing or decreasing? The eddy currents induced in the rail by the changing external flux will themselves induce a magnetic field. and the magnetic field points into the screen. First. This problem asks you to use Lenz’s law. You will be asked to use your knowledge of physics to predict the outcome of an experiment. think about the magnetic flux through a given portion on the rail as the electromagnet passes over that portion. How to approach the problem.11/24/2016 Ch 29 HW ANSWER: B. Hint 1. In the figure. The electromagnet moves to the right. close the video window and answer the question at right.C Correct Video Tutor: Eddy Currents in Different Metals First. To brake the train. 8 A flat.11/24/2016 Ch 29 HW A B C D Correct The magnetic flux increases under the leading edge of the electromagnet and decreases under its trailing edge.masteringphysics.122e V Correct Part B When is the induced emf equal to 1 20 of its initial value? ANSWER: t = 52.4T)e )t . ANSWER: E = −0. as viewed from above the loop. according to the righthand rule. The field is changing with time. by Lenz’s law. the induced magnetic field will point out of the screen under the leading edge and into it under the trailing edge. Part A Find the emf induced in the loop as a function of time (assume t is in seconds). as shown in an edgeon view in the figure . Therefore.057t 0. steel loop of radius 75 cm is at rest in a uniform magnetic field. circular. Express your answer in terms of the variable t.6 s Correct Part C Find the direction of the current induced in the loop. according to B(t) −1 −(0.057s = (1. Exercise 29. The eddy currents in choice B have the right directions to induce such magnetic fields. ANSWER: https://session.com/myct/assignmentPrintView?assignmentID=4706981 11/36 . ANSWER: I = 0.8Ω.28 A 1.4 V Correct Part B Find the direction of the current induced in the circuit.11/24/2016 Ch 29 HW Counterclockwise Clockwise Correct Exercise 29.745T magnetic field. ANSWER: E = 3.masteringphysics. ANSWER: clockwise counterclockwise Correct Part C Calculate the current through the resistor. so the apparatus makes a complete circuit. Express your answer using two significant figures. Part A Calculate the magnitude of the emf induced in the circuit. as shown in the figure . The bar rides on parallel metal rails connected through R = 24.com/myct/assignmentPrintView?assignmentID=4706981 12/36 . You can ignore the resistance of the bar and the rails.05mlong metal bar is pulled to the right at a steady 4.14 A Correct https://session. Express your answer using two significant figures.3 m/s perpendicular to a uniform. 0. The circular plates have radius 4.11/24/2016 Ch 29 HW Exercise 29.60 cm . and at a particular instant.masteringphysics.278 A . Part A What is the displacement current density jD in the air space between the plates? ANSWER: j D = 54.0 A/s .00 cm from the axis of the solenoid? ANSWER: E2 = 3. thin solenoid has 930 turns per meter and radius 2.65×10−4 V/m Correct Part B What is the magnitude of the induced electric field at a point 1.51×10−4 V/m Correct Exercise 29.36 A long. Part A What is the magnitude of the induced electric field at a point 0.16×1012 V/m ⋅ s https://session. the conduction current in the wires is 0.5 A/m2 Correct Part B What is the rate at which the electric field between the plates is changing? ANSWER: dE dt = 6.03 cm .42 A parallelplate airfilled capacitor is being charged as in the figure .com/myct/assignmentPrintView?assignmentID=4706981 13/36 .470 cm from the axis of the solenoid? ANSWER: E1 = 1. The current in the solenoid is increasing at a uniform rate of 60. Express your answer in terms of the variables B. ANSWER: x = mRv0 2 2 B L Correct https://session. R.63×10−7 T Correct Problem 29.com/myct/assignmentPrintView?assignmentID=4706981 14/36 . ANSWER: F = 2 B L 2 v R Correct Part B Find the distance x that the wire moves before coming to rest.masteringphysics.91 cm from the axis? ANSWER: B1 = 6.11/24/2016 Ch 29 HW Correct Part C What is the induced magnetic field between the plates at a distance of 1. the magnitude of the force exerted on the wire while it is moving at speed v . Express your answer in terms of the variables B. v 0 . R.54×10−7 T Correct Part D What is the induced magnetic field between the plates at a distance of 1.06 cm from the axis? ANSWER: B2 = 3. L. The slide wire is given an initial speed of v 0 and then released. There is no friction between the slide wire and the loop.69 A rectangular loop with width L and a slide wire with mass m are as shown in the figure . A uniform magnetic field B⃗ is directed perpendicular to the plane of the loop into the plane of the figure. and the resistance of the loop is negligible in comparison to the resistance R of the slide wire. and m. v . L. and m. Part A Obtain an expression for F . The current is clockwise in the first case. Correct Part B Find the magnitude of the current induced in the circuit as it is going into the magnetic field . clockwise in the second case and zero in the third case. ANSWER: The current is zero in the first case.00 m/s into. Part A Find the direction (clockwise or counterclockwise) of the current induced in the circuit as it is going into the magnetic field (the first case). zero in the second case and clockwise in the third case. as shown in the figure . totally within the magnetic field but still moving (the second case).masteringphysics. and then out of a uniform 1. ANSWER: I1 = 0.0 cm . The current is clockwise in the first case.com/myct/assignmentPrintView?assignmentID=4706981 15/36 . The magnetic field region is considerably wider than 50. and moving out of the field (the third case).25 T magnetic field. ANSWER: https://session.35 A rectangular circuit is moved at a constant velocity of 3. The current is counterclockwise in the first case. counterclockwise in the second case and zero in the third case. through.counterclockwise in the second case and clockwise in the third case . zero in the second case and counterclockwise in the third case. ANSWER: I2 = 0 A Correct Part D Find the magnitude of the current induced in the circuit as it is moving out of the field . The current is counterclockwise in the first case.225 A Correct Part C Find the magnitude of the current induced in the circuit as it is totally within the magnetic field but still moving .11/24/2016 Ch 29 HW Exercise 29. The formula for the magnetic flux through a loop The formula for the magnetic flux Φ through a wire loop is ⃗ ⃗ ⃗ ⃗ Φ = B ⋅ A = |B||A | cos(ϕ) . where Φ is the magnetic flux through the wire loop. A is the area vector associated with the loop. Note that the area vector is normal to the plane of the wire loop and can be chosen to be parallel or antiparallel to the magnetic field in this case. Hint 1. https://session. That is. as in this problem. and ϕ is the angle between the magnetic field and the area vector. Hint 2. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. c. ⃗ B ⃗ where is the magnetic field.masteringphysics.11/24/2016 I3 Ch 29 HW = 0. there can be contributions to the induced emf in a wire loop both from a changing magnetic field through the loop (about which you may have studied earlier) and from the change in the area of the loop (within the magnetic field region). Which of the following gives the proper value for d dx [a(x)b(x)] ? ANSWER: d(ab) dx d(ab) dx d(ab) dx d(ab) dx = da dx db =a = = dx da dx 1 b b 2 b+a ( da dx db dx b−a db dx ) Hint 3. How to take the derivative of the product of two functions Let a(x) and b(x) be two functions of x. What is the magnitude B of the magnetic field that the loop was in? Express your answer in terms of some or all of the variables A.225 A Correct A Simple Way to Measure Magnetic Fields A loop of wire is at the edge of a region of space containing a uniform magnetic field B⃗ . with c > 0 . This ambiguity as to which area vector to choose does not affect the problem as long as the vector chosen stays the same throughout the problem. and V . dA dt = −c. The plane of the loop is perpendicular to the magnetic field. Part A The induced emf in the loop is measured to be V .com/myct/assignmentPrintView?assignmentID=4706981 16/36 . ANSWER: B = V c Correct So you see that in general. The formula for the emf induced in a loop (Faraday's law) The formula for the emf E induced in a wire loop (Faraday's law) is E=− dΦ dt . 7 ms .masteringphysics. or. you will learn. If you have already studied this. that the "motional emf" E associated with a rod of length L moving through a uniform magnetic field of magnitude B with speed v is given by E = vLB . This is another way of thinking about the result derived above. can you see which sides of the square loop contribute to the motional emf and which do not. what is the rate of change of area c = − dA dt ? Express your answer in terms of L and v . which is upward. Part A If the loop is removed from the field region in a time interval of 2.5 T is directed along the positive zdirection.com/myct/assignmentPrintView?assignmentID=4706981 17/36 . Express your answer using two significant figures. equivalently.11/24/2016 Ch 29 HW Part B For the case of a square loop of side length L being pulled out of the magnetic field with constant speed v (see the figure). if you have not already.0 cm and oriented in the horizontal xyplane is located in a region of uniform magnetic field. find the average emf that will be induced in the wire loop during the extraction process. and why? Exercise 29. A field of 1. ANSWER: E = 29 V Correct Part B https://session.5 A circular loop of wire with a radius of 13. B= E vL . How to approach the problem Think about the following questions: What is the shape of the region of the coil in the magnetic field? What is the formula for the area of such a shape? For the part of the coil that is in the magnetic field. is the length changing? Is the width changing? ANSWER: c = Lv Correct Later. Hint 1. 122e V Correct Part B 1 When is the induced emf equal to 18 of its initial value? ANSWER: t = 50. ANSWER: E = −0. The field is changing with time.4T)e )t .com/myct/assignmentPrintView?assignmentID=4706981 18/36 . Express your answer in terms of the variable t.7 s Correct Part C Find the direction of the current induced in the loop. as shown in an edgeon view in the figure .masteringphysics. circular.8 A flat. according to B(t) −1 −(0.11/24/2016 Ch 29 HW If the coil is viewed looking down on it from above. ANSWER: Counterclockwise Clockwise https://session. is the induced current in the loop clockwise or counterclockwise? ANSWER: Clockwise Counterclockwise Correct Exercise 29.057s = (1. as viewed from above the loop. Part A Find the emf induced in the loop as a function of time (assume t is in seconds).057t 0. steel loop of radius 75 cm is at rest in a uniform magnetic field. EXECUTE the solution as follows: 1. SET UP the problem using the following steps Part A Which of the following is the reason why the magnetic flux through the bracelet is changing? ANSWER: The bracelet is moving in the magnetic field. Draw the area vector A⃗ for the bracelet. what is the induced emf E ? Take the magnetic flux through an area to be positive when B⃗ crosses the area from top to bottom. Determine what is making the magnetic flux change. and be sure that you've correctly used the sign rules. you can calculate the magnitude of the induced current I using E = IR . IDENTIFY the relevant concepts There is a changing magnetic flux through the area enclosed by the loops of the bracelet due to the magnetic field produced by the metal detector. Use the black dot as the starting point of A⃗ . ANSWER: https://session. being mindful of the direction you chose for the area vector. Calculate the magnetic flux using ΦB = B⃗ ⋅ A⃗ = BA cos ϕ if B⃗ is uniform over the area and Φ B uniform.00 cm with the plane of the loops parallel to the ground. Is the conductor moving or changing orientation? Is the magnetic field changing? → 2. Suppose a metal detector that generates a uniform magnetic field perpendicular to its surface is held stationary at an angle of 15.0250 T/s. not its length. Faraday's law applies to this problem. Therefore. If the circuit resistance R is known. If the magnetic field increases at a constant rate of 0. A metal detector uses a changing magnetic field to detect metallic objects. Only the orientation of your vector will be graded. ProblemSolving Strategy 29. EVALUATE your answer: Does your answer make physical sense? Check over your units.1: Faraday's Law.1: Faraday's Law IDENTIFY the relevant concepts: Faraday’s law applies when there is a changing magnetic flux.masteringphysics. Choose a direction for the area vector A⃗ or dA that is perpendicular to the plane of the area.11/24/2016 Ch 29 HW Correct ± PSS 29. To use the law. Remember the sign for the positive direction of emf and use it consistently. identify an area through which there is a flux of magnetic field. Calculate the induced emf using E = −N dΦ B dt ⃗ ⃗ = ∫ B ⋅ dA = ∫ BdA cos ϕ if it isn't for a conductor that has N turns in a coil.1: Faraday's Law Learning Goal: To practice ProblemSolving Strategy 29. 2. The bracelet is depicted as a single loop parallel to the ground.com/myct/assignmentPrintView?assignmentID=4706981 19/36 . make certain to draw your vector so that its tail is on the dot. The magnetic field is changing direction with respect to the bracelet. that is. 3.0∘ to the ground. Whatever direction you choose. while just below the surface there lies a silver bracelet consisting of 6 circular loops of radius 5. use it consistently throughout the problem. SET UP the problem using the following steps: 1. Correct Part B The diagram below shows a side view of the situation described in this problem. The magnitude of the magnetic field is changing. its orientation is arbitrary.masteringphysics. Use this choice of direction for A⃗ to work through the next part of the problem.89⋅10 −4 V https://session. In this specific problem. what is the total induced emf E in the bracelet? Express your answer in volts to three significant figures.11/24/2016 Ch 29 HW Correct While the direction of A⃗ is fixed (by definition.com/myct/assignmentPrintView?assignmentID=4706981 20/36 . as shown in the diagram below. A⃗ is perpendicular to the plane of the area). However. so that the angle between B⃗ and A⃗ is less than 90 ∘ and its cosine is positive. we need to take A⃗ to be downward. ANSWER: E = 1. since the problem introduction states that the magnetic flux through an area is to be taken positive when B⃗ crosses the area from top to bottom. A⃗ can be upward or downward. EXECUTE the solution as follows Part C Based on the definition of A⃗ given in the previous part. You did not open hints for this part. com/myct/assignmentPrintView?assignmentID=4706981 21/36 . The current in loop C is zero. ANSWER: The current in loop A is counterclockwise. Part A Find the direction (clockwise or counterclockwise) of the current induced in loop A if I is steadily increasing. The current in loop A is clockwise. Correct Part B Find the direction (clockwise or counterclockwise) of the current induced in the loop B if I is steadily increasing. Exercise 29. Correct Part C While I is increasing. The current in loop C is clockwise.11/24/2016 Ch 29 HW Incorrect; Try Again; 24 attempts remaining EVALUATE your answer Part D This question will be shown after you complete previous question(s). what is the direction of the net force that the wire exerts on loop A? ANSWER: https://session. The current in loop A is zero. ANSWER: The current in loop C is counterclockwise.17 Two closed loops A and C are close to a long wire carrying a current I .masteringphysics. The wire exerts a downward force on loop A.com/myct/assignmentPrintView?assignmentID=4706981 22/36 .0 cm moves in a magnetic field B⃗ of magnitude 0.420 T directed into the plane of the figure. Induced currents in the loops and the forces exerted on them are directed so they oppose this change. Correct Part E Explain how you obtain your answer. The wire exerts no force on loop C . what is the direction of the net force that the wire exerts on loop C ? ANSWER: The wire exerts an upward force on loop C .11/24/2016 Ch 29 HW The wire exerts an upward force on loop A. The rod moves with speed v = 5.90 m/s in the direction shown. Drag the terms on the left to the appropriate blanks on the right to complete the sentences.25 In the figure a conducting rod of length L = 35. ANSWER: Reset Help decreasing When the current in the wire is increasing the increasing magnitude of the magnetic flux in the loops is oppose support increasing .masteringphysics. Correct Exercise 29. The wire exerts a downward force on loop C . Correct Part D While I is increasing. https://session. The wire exerts no force on loop A. 11/24/2016 Ch 29 HW Part A What is the potential difference between the ends of the rod? ANSWER: V = 0. is at higher potential? ANSWER: a b Correct Part C When the charges in the rod are in equilibrium.masteringphysics.867 V Correct Part B Which point.48 V/m Correct Part D What is the direction of the electric field within the rod? ANSWER: From a to b From b to a Correct https://session. a or b.com/myct/assignmentPrintView?assignmentID=4706981 23/36 . what is the magnitude of the electric field within the rod? ANSWER: E = 2. Part A At the instant shown. Faraday's law of induction The magnitude of the current in the light bulb is proportional to the magnitude of the emf induced in each scenario. Rank from largest to smallest. has an excess of positive charge? ANSWER: a b Correct Part F What is the potential difference across the rod if it moves parallel to ab? ANSWER: V = 0 V Correct Part G What is the potential difference across the rod if it moves directly out of the page? ANSWER: V = 0 V Correct Magnetic Light Switch Ranking Task Six identical vertical metal bars start at the positions shown below and move at constant velocities through identical magnetic fields. any change in flux must be due solely to a change in area. a or b.masteringphysics. overlap them. The bars make electrical contact with and move along frictionless metal rods attached to light bulbs. Hint 3.11/24/2016 Ch 29 HW Part E When the charges in the rod are in equilibrium. Hint 1. To rank items as equivalent. Change in magnetic flux Magnetic flux is the product of the magnetic field perpendicular to a given area and the area itself. rank these six scenarios on the basis of the magnitude of the current in the light bulb. Since the magnetic field is constant in this case. ANSWER: https://session. The emf induced is equal to the time rate of change of the magnetic flux through the closed loop formed by the sliding bar. Change in area The area of the closed loop formed by the sliding bar changes at a rate that is proportional to the velocity of the sliding bar. Hint 2.com/myct/assignmentPrintView?assignmentID=4706981 24/36 . which point. Correct Exercise 29. https://session.11/24/2016 Ch 29 HW Reset largest Help smallest The correct ranking cannot be determined. Express your answer with the appropriate units.30 A 0. Ignore the resistance of the bar and the rails.1 V Correct Part B Find the direction of the current induced in the circuit. Part A Calculate the magnitude of the emf induced in the circuit. so the apparatus makes a complete circuit.com/myct/assignmentPrintView?assignmentID=4706981 25/36 . The bar rides on parallel metal rails connected through a 25Ω. 0.550T magnetic field.650mlong metal bar is pulled to the right at a steady 6.masteringphysics. resistor . ANSWER: E = 2.0 m/s perpendicular to a uniform. Hence charges are moving perpendicular to the magnetic field. The arrangement is shown in the figure.11/24/2016 Ch 29 HW ANSWER: clockwise counterclockwise Correct Part C Calculate the current through the resistor. There is a uniform magnetic field of magnitude B.6×10−2 A Correct Rail Gun This problem explores how a currentcarrying wire can be accelerated by a magnetic field.com/myct/assignmentPrintView?assignmentID=4706981 26/36 . The rod is released from rest. Part A A conducting rod is free to slide on two parallel rails with negligible friction. and point your middle finger in the direction of L⃗ . The force on a conducting rod due to a magnetic field There is a force on the rod because a current is flowing through it. The rod ⃗ ⃗ experiences a force F . The direction of the magnetic field Use the righthand rule with the cross product: Take your right hand and point with your index finger in the direction of the rail's motion. The rails extend to infinity on the left. a voltage source of strength V in series with a resistor of resistance R makes a closed circuit together with the rails and the rod. The rails and the rod are taken to be perfect conductors. You will use the ideas of magnetic flux and the EMF due to change of flux through a loop. In what direction does the magnetic field point? Hint 1. and it is observed that it accelerates to the left. Hint 2. ANSWER: into the plane of the figure out of the plane of the figure https://session. At the right end of the rails. Your thumb will then point in the direction of the magnetic field. Note that there is an involved followup part that will be shown once you have found the answer to Part B.masteringphysics. ANSWER: I = 8. Express your answer with the appropriate units. pervading all space. perpendicular to the plane of rod and rails. which is given by F = ⃗ ⃗ IL × B . com/myct/assignmentPrintView?assignmentID=4706981 27/36 . Hint 1. L. . ANSWER: I (t) = V −BLvr (t) R https://session. and reduces it. Correct Part C What is the acceleration ar (t) of the rod? Take m to be the mass of the rod. Express your answer as a function of V . ANSWER: E(t) = −BLv r (t) Hint 2. The induced current works against that source current. and the magnetic field B. Hint 3. R. where E is the induced EMF. Alternatively. But the force accelerating the rod is proportional to the current. B. Lenz's law An EMF is induced in the circuit due to the change in magnetic flux through it.masteringphysics. and the mass of the rod m.11/24/2016 Ch 29 HW Correct Part B Assuming that the rails have no resistance. and the lower the current flowing through the loop. The velocity of the rod The higher the velocity of the rod. But will this EMF increase the current through the loop or decrease it by opposing the voltage coming from the source? Lenz's law states that induced currents will always flow in such a way that they oppose the change in flux that caused them. we need the EMF E(t) induced by the change in magnetic flux Φ(t) Faraday's law states that E(t) = − dΦ(t) dt ⃗ ⃗ = B ⋅ A (t) through the current loop of area A(t). The velocity cannot increase beyond the point at which the induced EMF is equal and opposite to V . what is the most accurate qualitative description of the motion of the rod? Hint 1. What is the Express your answer in terms of v r (t) and other given quantities. Find the induced EMF To determine the current. The change in flux that induced this current is caused by the motion of the rod. Under these idealized conditions the rod will experience constant acceleration and the velocity of the rod will increase indefinitely. the higher the induced EMF. the velocity of the rod v r (t). What is the EMF E(t) ? Express E in terms of the velocity v r (t). Hint 2. Hence the acceleration goes down as the velocity goes up. The rod will accelerate indefinitely with acceleration proportional to its (increasing) velocity. one could say that the induced EMF opposes the voltage from the source. The motion of the rod will change A(t) according to dΦ(t) dt = dA(t) dt B . ANSWER: The rod will accelerate but the magnitude of the acceleration will decrease with time; the velocity of the rod will approach but never exceed a certain terminal velocity. Appyling Lenz's law to this problem Let us apply Lenz's law here. which in turn is caused by the current flowing around the circuit due to the voltage source. We want to find the direction of the induced current. Find the current in the rod We can find the current through the rail by using Kirchhoff's rule for a closed circuit: I R current I (t)? = V +E . the separation of the rails L. masteringphysics.00 cm2 and are separated by a 2. What is the condition for this? Write down this condition. but lowers the terminal velocity: a tradeoff for rail gun engineers! Exercise 29. Write down an expression for ar (t).i.50mmthick sheet of dielectric that completely fills the volume between the plates. a magnetic field of large magnitude and a high voltage are advantageous.43 Suppose that the parallel plates in the figure have an area of 3. which you can solve to find the velocity of the rod as a function of time: vr (t) = V BL (1 − e − 2 2 B L mR t ). We have already determined its direction. Find an expression for the terminal velocity The terminal velocity is reached when the acceleration is zero. expressing the applied voltage V as a function of the terminal velocity v t . ANSWER: ar (t) = I (t)BL m ANSWER: ar (t) = (V −BLvr (t))LB Rm Correct Making the substitution ar (t) = dvr (t) dt . you obtain the dfferential equation dvr (t) dt = BL mR (V − BLvr (t)). To achieve a high acceleration. B.com/myct/assignmentPrintView?assignmentID=4706981 28/36 .11/24/2016 Ch 29 HW Hint 3. using Newton's second law. (You can ignore fringing effects. the potential difference between the plates is 120 V and the conduction current iC equals 6.70.) At a certain instant.00 mA. current I (t). and L. The dielectric has dielectric constant 4. ANSWER: V = BLv t ANSWER: vt = V BL Correct A larger magnetic field increases the acceleration of the rod. The magnitude of this force is given in Part A. Part D What is the terminal velocity v t reached by the rod? Hint 1. https://session. Find the acceleration of the rod To find the acceleration ar (t). we need the force on the rod. Expres your answer in terms of mass m. which is necessary for a useful gun. all the magnetic vectors are parallel to the xaxis. Part A At temperatures near absolute zero.11/24/2016 Ch 29 HW Part A At this instant. by symmetry. Bc approaches 0.com/myct/assignmentPrintView?assignmentID=4706981 29/36 . what is the charge q on each plate? ANSWER: |q| = 5.142 T for vanadium. Consider a long. The normal phase of vanadium has a magnetic susceptibility close to zero. what is the displacement current in the dielectric? ANSWER: id = 6.00×10−3 A Correct Part C At this instant. thin vanadium cylinder with its axis parallel to an external magnetic field B⃗ 0 in the +xdirection.00×10−3 A Correct Exercise 29. At points far from the ends of the cylinder. what is the rate of change of charge on the plates? ANSWER: dq dt = 6. a typeI superconductor.99×10−10 C Correct Part B At this instant.130T)^i ? = ( ANSWER: B = 0 T https://session. what is the magnitude of the resultant magnetic field B⃗ inside the cylinder for B⃗ 0 0.45 At temperatures near absolute zero.masteringphysics. 11/24/2016 Ch 29 HW Correct Part B What is the direction of the resultant magnetic field B⃗ inside the cylinder for this case? ANSWER: in the +xdirection in the −xdirection perpendicular to the xaxis the field is zero Correct Part C What is the magnitude of the resultant magnetic field B⃗ outside the cylinder (far from the ends) for this case? ANSWER: B = 0.com/myct/assignmentPrintView?assignmentID=4706981 30/36 .masteringphysics.03×105 A/m Correct Part F What is the direction of the magnetization M⃗ inside the cylinder for this case? ANSWER: https://session.130 T Correct Part D What is the direction of the resultant magnetic field B⃗ outside the cylinder (far from the ends) for this case? ANSWER: in the +xdirection in the −xdirection perpendicular to the xaxis the field is zero Correct Part E What is the magnitude of the magnetization M⃗ inside the cylinder for this case? ANSWER: M = 1. 260T) i ? ANSWER: B = 0.com/myct/assignmentPrintView?assignmentID=4706981 31/36 .masteringphysics.260 T Correct Part J What is the direction of the resultant magnetic field B⃗ inside the cylinder for this case? ANSWER: in the +xdirection in the −xdirection perpendicular to the xaxis the field is zero Correct https://session. what is the magnitude of the resultant magnetic field B⃗ inside the cylinder for B⃗ 0 ^ = (0.11/24/2016 Ch 29 HW in the +xdirection in the −xdirection perpendicular to the xaxis the magnetization is zero Correct Part G What is the magnitude of the magnetization M⃗ outside (far from the ends) the cylinder for this case? ANSWER: M = 0 A/m Correct Part H What is the direction of the magnetization M⃗ outside the cylinder (far from the ends) for this case? ANSWER: in the +xdirection in the −xdirection perpendicular to the xaxis the magnetization is zero Correct Part I At temperatures near absolute zero. 260 T Correct Part L What is the direction of the resultant magnetic field B⃗ outside the cylinder (far from the ends) for this case? ANSWER: in the +xdirection in the −xdirection perpendicular to the xaxis the field is zero Correct Part M What is the magnitude of the magnetization M⃗ inside the cylinder for this case? ANSWER: M = 0 A/m Correct Part N What is the direction of the magnetization M⃗ inside the cylinder for this case? ANSWER: in the +xdirection in the −xdirection perpendicular to the xaxis the magnetization is zero Correct Part O What is the magnitude of the magnetization M⃗ outside the cylinder for this case? ANSWER: M = 0 A/m Correct https://session.masteringphysics.11/24/2016 Ch 29 HW Part K What is the magnitude of the resultant magnetic field B⃗ outside the cylinder (far from the ends) for this case? ANSWER: B = 0.com/myct/assignmentPrintView?assignmentID=4706981 32/36 . 60 A metal rod with a length of 23.4 turns of wire per centimeter. Part A What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.7 ∘ with the positive xaxis and an angle of 51.142 tesla. by symmetry. consider an applied magnetic field B⃗ 0 ^ = (0. solid vanadium cylinder with its axis parallel to an external magnetic field B⃗ 0 in the +x direction. The rod is moving in the +xdirection with a speed of 6. At points far from the ends of the cylinder. ANSWER: = 1.masteringphysics.3 ∘ with the positive yaxis.45×10−5 V |E| Correct A Superconducting Cylinder At temperatures near absolute zero.160T) i − (0. straight solenoid with a crosssectional area of 1. The rod is in a uniform magnetic field B⃗ = ^ ^ ^ (0.0 cm lies in the xy plane and makes an angle of 38.0200T )k . what is the magnitude of the resultant magnetic field B⃗ inside the cylinder (far from the ends)? Express your answer in teslas. a typeI superconductor. Part A At temperatures near absolute zero. such that the secondary winding has the same crosssectional area as the solenoid. all the magnetic vectors are parallel to the x axis.84 cm2 is wound with 86.49 A very long. approaches 0.com/myct/assignmentPrintView?assignmentID=4706981 33/36 .11/24/2016 Ch 29 HW Part P What is the direction of the magnetization M⃗ outside the cylinder for this case? ANSWER: in the +xdirection in the −xdirection perpendicular to the xaxis the magnetization is zero Correct Exercise 29.165 A/s2 )t2 . https://session. thin.130 T) i . Part A What is the magnitude of the emf induced in the rod? ANSWER: E = 0. the current in the solenoid is increasing according to i(t) = ( 0.80 m/s .200T) j − (0. Consider a long. Starting at t = 0. The normal phase of vanadium has a magnetic susceptibility close to zero.2 A ? Express your answer with the appropriate units. A secondary winding of 5 turns encircles the solenoid at its center. the critical field Bc for vanadium.0467 V Incorrect; Try Again; 22 attempts remaining Problem 29. First. Because the applied magnetic field is parallel to the x axis. Do you recall what the magnetic field is outside of a solenoid? ANSWER: ∣ ⃗ ∣ Bcyl ∣ ∣ = 0 ANSWER: ⃗ |B| = 0. What is the critical field? Choose the correct definition of the criticial field Bc . We know that the magnetic field inside the cylinder is zero. Current flowing around the circumference produces a magnetic field inside that is parallel to the x axis (which is also the axis of the cylinder). Let's recap: We have a cylindrical object behaving as though a current were flowing along its surface in such a way that there is a magnetic field inside it parallel to its axis. Find the magnetic field due to the cylinder Think of the magnetic field outside the cylinder as an applied field B⃗ 0 plus a magnetic field B⃗ cyl due to the cylinder. Hint 1.masteringphysics.130 T Correct Part C What is the magnitude of the magnetization M⃗ in inside the cylinder? What is the magnitude of the magnetization M⃗ out outside the cylinder? Express your answers in amperes per meter separated by a comma. the induced magnetic field must also be parallel to the x axis. you may be able to solve it without performing any calculations. What is the magnitude of B⃗ cyl ? Hint 1. ANSWER: ⃗ |B| = 0 T Correct Part B What is the magnitude of the resultant magnetic field B⃗ outside the cylinder (far from the ends) for the situation described in the problem introduction? Express your answer in teslas. How to approach the problem ⃗ ⃗ ⃗ ⃗ ⃗ https://session. Hint 1.11/24/2016 Ch 29 HW Hint 1. ANSWER: the maximum magnitude of magnetic field into which an object can be placed and still exhibit superconductivity (at a given temperature below the critical temperature) the magnetic field due to coulomb (or conservative) forces. A solenoid analogy If you give some thought to this problem. We know that the object is behaving as though there were a current flowing on its surface creating a magnetic field that exactly cancels the applied magnetic field.com/myct/assignmentPrintView?assignmentID=4706981 ⃗ 34/36 . Does this sound familiar? The cylinder is acting like a solenoid. masteringphysics. In other words.260 T Correct Part E What is the magnitude of the resultant magnetic field B⃗ outside the cylinder (far from the ends)? Hint 1. an object will not exhibit superconductivity. More on superconductivity When an object is not in the superconducting phase. How to approach the problem Recall that when ∣∣B⃗ 0 ∣∣> ∣∣B⃗ c ∣∣. ANSWER: ⃗ M in .0 A/m. the magnetic dipoles of which it is made cannot align so as to cancel the applied magnetic field. the magnetization of free space is always zero.11/24/2016 Ch 29 HW Magnetization M⃗ is defined by the equation B⃗ = B⃗ 0 + μ0 M⃗ . Part D At temperatures near absolute zero. Value of μ0 Recall that μ0 = 4π × 10 −7 T⋅m A in standard SI units. Hint 2. ANSWER: ⃗ |B| = 0. Hint 2. How to approach the problem This part is conceptually similar to Part E. A/m Correct Recall that there needs to be a material present to induce magnetization. All of the vectors point along the x axis. M⃗ out = 1.260 T Correct Part F What is the magnitude of the magnetization M⃗ in inside the cylinder? What is the magnitude of the magnetization M⃗ _out outside the cylinder? Express your answers in amperes per meter separated by a comma.com/myct/assignmentPrintView?assignmentID=4706981 35/36 . so solving this equation is straightforward. making the material no longer superconducting.260 T) i .03×105. When the critical field is exceeded. the magnetization depends on the magnetic susceptibility of the material in its normal phase. what is the magnitude of the resultant magnetic field B⃗ inside the cylinder (far from the ends)? Hint 1. Now consider an applied magnetic field B⃗ 0 ^ = (0. ANSWER: ⃗ |B| = 0. https://session. in that any magnetic field caused by the cylinder will be based on its normal phase magnetic susceptibility. where B⃗ is the total magnetic field and B⃗ 0 is the applied magnetic field. 11/24/2016 Ch 29 HW ANSWER: ⃗ M in . M⃗ out = 0.com/myct/assignmentPrintView?assignmentID=4706981 36/36 .0 A/m. https://session.9 out of a possible total of 25 points. A/m Correct Score Summary: Your score on this assignment is 91.masteringphysics. You received 22.6%.