Problem 1 (25 points)

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2012 Exam Three Solutions Problem 1 (25 points) Question 1 (5 points) Consider two circular rings of radius R, each perpendicular to the axis of symmetry, with their centers located at z = ± l / 2. There is a steady current I flowing in the same direction around each coil, as shown in the figure below. A magnetic dipole, with dipole moment µ = µî where µ is a positive constant with units A m 2, is placed on the symmetry axis, at the position z = l / 4. The dipole will a) experience no force and no torque. b) align itself to point in the positive z -direction and experience a force in the positive z -direction. c) align itself to point in the positive z -direction and experience a force in the negative z -direction. d) align itself to point in the negative z -direction and experience a force in the positive z -direction. e) align itself to point in the negative z -direction and experience a force in the negative z -direction. f) align itself to point in the positive z -direction but feel no force. g) align itself to point in the negative z -direction but feel no force. The correct answer is b and f accepted as correct. 1

Question 2 (5 points) A square wire loop rotates in the direction shown (see sketch) in a magnetic field directed to the right. At the instant shown, when 0 < θ < π / 2, which of the figures below best describes the direction of current in the square wire loop and the direction of the magnetic torque on the square wire loop? The correct answer is c. At the instant shown the flux is increasing (in the ˆn -direction) so there is a clockwise induced current to oppose that change. Therefore the magnetic dipole vector points in the negative ˆn-direction. The torque τ = µ B ext is therefore in the positive z -direction. 2

Question 3 (5 points) A coil of wire with resistance R defines an open surface whose normal d A points upward, as shown in the sketch. The coil is below a magnet whose magnetic field lines and directions are shown in the figure above. If positive current is defined as counterclockwise as viewed from the top, and if we ignore any self-magnetic field generated by the induced current, then as the coil moves from well below the magnet to well above that magnet, the induced current through the coil will look like (a) (b) (c) (d) The correct answer is c. 3

Question 4 (5 points) The figure above on the left shows a side view of a section of a very long solenoid with radius R carrying current I with magnetic field pointing up at time t. The figure above on the right shows a top view of the electric field E inside the solenoid at a radius r and the direction of the magnetic field B at time t. In the solenoid, the current I is a) increasing in time. b) constant. c) decreasing in time. d) cannot tell without more information. The correct answer is c. 4

Question 5 (5 points) A very long solenoid consisting of n turns per unit length has radius R and length d ( d >> R ). Suppose the current running through the solenoid is doubled keeping all the other parameters fixed. You may neglect edge effects. Which of the following is true? a) The energy stored in the magnetic field and the self-inductance remain the same. b) The energy stored in the magnetic field doubles and the self-inductance remains the same. c) The energy stored in the magnetic field is four times as large and the selfinductance remains the same. d) The energy stored in the magnetic field remains the same and the self-inductance doubles. e) The energy stored in the magnetic field remains the same and the self-inductance is four times as large. f) None of the above. The correct answer is c. 5

Problem 2 (25 points) NOTE: YOU MUST SHOW WORK in order to get any credit for this problem. Make it clear to us that you understand what you are doing (use a few words!) A very long coaxial cable consists of a solid cylindrical inner conductor of radius a, surrounded by a concentric cylindrical conducting shell of inner radius b and outer radius c. The inner conductor has a non-uniform current density J inner = αr ˆk (pointing to the left in the figure just below) where α is a positive constant with units A m -3. The outer conductor has a uniform current density J outer = β ˆk where β is a positive constant with units A m -2. The conductors carry equal and opposite currents of magnitude I 0. a) Find expressions for α and β in terms of a, b, c, and I 0. For current through 0 < r < a, For current through b< r < c, a J ˆn da = I o = 2πr dr αr S 0 ( ) c J ˆn da = I o = 2πr dr ( β ) = βπ c 2 b 2 S b = 2πα 3 a3 α = 3 I o 2π a 3 ( ) β = I o π c 2 b 2 ( ) b) Determine the magnitude and direction of the magnetic field for the regions (i) r < a, (ii) a < r < b, (iii) b < r < c, (iv) and r > c. For each region, redraw the coaxial cable clearly indicating your choice of Amperian loop and associated parameters. For r < a, loop is circle of radius r < a, and 6

B d s = 2πrB θ = µ 0 J ˆn da closed path = µ 0 2π S r 0 r d r α r ( ) = 2π 3 αr 3 2 ˆ αr B = θ µ where ˆθ is a unit vector oriented counterclockwise. 0 3 For a<r < b, loop is circle of radius a<r < b, and B d s = 2πrB θ = µ 0 J ˆn da closed path where ˆθ is a unit vector oriented clockwise. = µ 0 I 0 S ˆ µ B = θ 2π r 0I o For b<r < c, loop is circle of radius b<r < c, and closed path B d s = 2πrB θ = µ 0 I o 1 β 2π I o ˆ µ 0I o β B = θ 1 π 2π r Io r b r d r = µ I 1 β π r 2 b 2 0 o ( ) I o 2 2 ( r b ) where ˆθ is a unit vector oriented counterclockwise. This can be written using the results above as For c<r, loop is circle of radius c<r, and 2 2 ( c r ) 2 2 ( ) ˆ µ 0Io B = θ 2π r c b B d s = 2πrB θ = 0 closed path c) Make a graph of the magnitude of the magnetic field as a function of the distance r from the central axis of symmetry. Clearly label each axis with any relevant values. 7

µ 0I The graph is a concave upward parabola from 0 to a, rising to a value of o at r = a. 2π a µ 0I Then it goes as inverse r from a to b, decreasing to o at r = b. Then it decreases from 2π b its value at r = b to 0 as we move from b to c. It is zero thereafter. 8

Problem 3 (25 points) NOTE: YOU MUST SHOW WORK in order to get any credit for this problem. Make it clear to us that you understand what you are doing (use a few words!). Consider a slab that is infinite in the x and z directions that has thickness d in the y- direction. The slab has a time varying current with the current density as a function of time given by the following expression: 0; t 0 J = (J e t / T ) ˆk; 0 t T, J e ˆk; T t where J e is positive constant with units of amps per square meter and T is a constant with units of seconds. a) Find the direction and magnitude of the magnetic field for the interval 0 t T in the regions: (i) 0 y d / 2 ; (ii) y d / 2. Clearly show all your work. Answers without justification will receive no credit. 0 y d / 2 : By symmetry we argue that the field is zero at y = 0. We take an Amperean loop whose bottom is at y = 0 and whose top is at 0 y d / 2, of width w. We have closed path B d s = wb x = µ 0 J ˆn da = µ 0 wy(j e t / T ) B = ˆxµ 0 y(j e t / T ) S d /2 y: We take an Amperean loop whose bottom is at y = 0 and whose top is at d /2 y, of width w. We have closed path B d s = wb x ( y) = µ 0 J ˆn da = µ 0 w d 2 (J e t / T ) B = ˆxµ 0 S d 2 (J e t / T ) 9

Suppose a square conducting loop with resistance R, and side s is placed in the region y d / 2, at a height h above the top of the slab oriented as shown in the figure below. What is the induced current in the square loop for the time interval 0 t T? Draw the direction of the induced current on the figure. The direction of the current is counterclockwise when looking from the right. d Φ d d d = = dt dt dt µ 2 1 dφ d Js = =. R dt 2 RT 2 B 2 2 e s 0 ( Je t/ T) s I µ 0 b) What is the direction and magnitude of the force due to the induced current on the square loop during the time interval 0 t T? What is the direction and magnitude of the torque due to the induced current on the square loop during the time interval 0 t T? Since the loop is sitting in a uniform field, the force is zero. Since the loop has a magnetic dipole moment anti-parallel to the magnetic field, the torque τ = µ B ext is also zero. 10

Problem 4 (25 points) NOTE: YOU MUST SHOW WORK in order to get any credit for this problem. Make it clear to us that you understand what you are doing (use a few words!). A stretchable and flexible conducting band in the shape of a circle with radius r(t) has constant resistance R. It sits in a uniform magnetic field B that is directed out of the page (see figure). External agents distributed uniformly over the circumference of the ring exert radial outward forces that cause the ring to expand at a constant speed from radius a to a larger radius b over a time interval 0 t T, where T is a constant with units of seconds. Let v = dr / dt be the constant speed at which the ring expands. Express your answers to the following questions in terms of r, v, a, b, R, B = B, and T as needed. Note that in this problem R is a resistance, not a radius. a) Give an expression for the induced current I in the ring. Draw the direction of the induced current on the figure above. You may ignore any magnetic field generated by the induced current. The current flows clockwise in the band. d Φ d 2 r 2 r dr 1 dφ 2π rbv = B π = B π I = = dt dt dt R dt R b) What is the rate at which energy is dissipated (Joule heating) during the time interval 0 t T? 2 2 2 2 2 2 2πrBv 4π r B v I R= R= R R c) What is the direction and magnitude of the force per unit length that the external agents must apply to overcome the magnetic force per unit length on the conducting band due to the induced current?. 11

At any given point on the band, the Id s B ext force is radially inward, and therefore at that point the external agents must exert a force per unit length given by df agents ds = I B ext = ˆrIB = ˆrIB = ˆr 2πrB2 v R d) Based on your result for the force per unit length in part c), what power do the external agents provide during the time interval 0 t T? Is this the same as your answer to part b)? If yes, explain why; if no, explain why not. Be sure to give your reasoning. An external agent at a given point on the band exerting a force on that ds section of the band does work at a rate given by F v = ds ˆr 2πrB2 v v = 2πrB2 v 2 ds. R R The power from all the agents is found by integrating the above over the circumference, giving 4π 2 r 2 B 2 v 2 / R, the same as above. They are the same because of conservation of energy. 12