Physics 1304/1404 - Spring 1999

Homework Assignment #5

Due: Thursday 18 March 1999


Chapters 31 and 32.


Chapter 31 - 1, 4, 5, 6, 9, 14, 16.
Chapter 32 - 7, 8, 10, 15.


Chapter 31 - 3, 8, 20, 21, 27, 30, 41, 46, 49.
Chapter 32 - 6A, 6, 10, 17, 25, 30A, 34, 42A, 48A, 60A.


These are the ANSWERS only, not the SOLUTIONS. It is not sufficient to copy these and turn them in as homework. You must show your work.

Q 31-1) Magnetic flux is the integral of the magnetic field over some area. The MKS units of magnetic field are tesla (T); the MKS units of magnetic flux are weber (Wb) = tesla.meter2.

Q 31-4) No. There is neither an electric nor a magnetic force to oppose the motion, so the bar will continue at the same speed in a straight line forever, as long as the B field into the page remains uniform and constant.
Q 31-5) By Lenz' law the induced current is clockwise so that the induced magnetic flux is into the page; this is opposite to the direction of the external magnet flux which is out of the page.

If the bar were moving to the left, the induced current would be counter-clockwise.

Q 31-6) As the bar moves to the right, its mechanical energy is converted into the electrical energy of the current flowing in the circuit, and finally this energy is then converted into thermal energy in the resistor. Since energy is being lost, work must be done to keep the bar moving, so a force that can do positive work must be applied to the bar.
Q 31-9) Dropping a magnet down a copper tube will produce one current (say clockwise) ahead of the magnet and another (say counter-clockwise) behind the magnet. The net current will be zero, but the two eddy currents will slow the fall of the magnet through the pipe.
Q 31-14) When the switch is closed, the magnetic field in the solenoid goes from zero to some finite value in a short time. The rate of change of the magnetic field, and hence also the rate of change of the magnetic flux through the metal ring is large. This changing flux induces a current in the ring by Faraday's Law. The current flows in the ring in such a direction as to oppose the change in flux, by Lenz' Law. If the solenoid's north pole is pointing up, then the ring's north pole will point down. If the solenoid's south pole is pointing up, then the ring's south pole will point down. The net result is repulsion in either case. This is the mechanism behind diamagnetism, which you remember is repulsive.
Q 31-16) Yes, Maxwell's equations can be made consistent with the existence of magnetic monopoles by adding two terms which are currently set to zero.
Q 32-7) The inductor attempts to keep the current flowing at the same rate that it had before the switch was opened. The charge piling up at the contacts of the open switch can jump the gap if the distance is small enough.
Q 32-8) quadrupled, U=1/2 LI2.
Q 32-10) Lequiv = L1 + L2
Q 32-15) Yes, the oscillations would persist forever without resistance. In an ideal LC circuit, energy is passed back and forth from the electric field energy stored in the capacitor to the magnetic field energy stored in the inductor. The only place that energy can leave the system (as heat) is in the resistance.
P 31-3) 160 A
P 31-8) 0.67 V
P 31-20) Negative.
P 31-21) 2 mV; the west end is positive.
P 31-27)
  1. to the right
  2. out of the page
  3. to the right
  4. into the page

P 31-30) 2.37 mV
P 31-41)
  1. 7540 V
  2. the plane of the loop is parallel to B

P 31-46)
  1. 1.60 V
  2. 0
  3. both would remain unchanged
  4. (graph required)
  5. (graph required)

P 31-49)
  1. N2B2w2v/R to the left
  2. 0
  3. N2B2w2v/R to the left

P 32-6A) LI/N
P 32-6) 2.4 x 10-7 Wb
P 32-10)
  1. 360 mV
  2. 180 mV
  3. 3 s

P 32-17) through inductor: (0.5 A)[1 - exp(-10 t)]
through switch: 1.5 A - (0.25 A)exp(-10 t)
P 32-25)
  1. 5.66 ms
  2. 1.22 A
  3. 58.1 ms

P 32-30A) through L1: I1
through L2: 0
P 32-34) 18 J
P 32-42A) C V2 R2/E2
P 32-48A) (L1L2 - M2)/ (L1 + L2 - 2M)
P 32-60A)
  1. 1/[2 pi sqrt(LC)]
  2. CE
  3. E sqrt(C/L)
  4. 1/2 CE2

Please report any corrections to Professor Scalise.

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