PHY 142!
Assignment 6!
Summer 2013
Reading: Electrodynamics, Inductance
Key concepts: Faraday’s law; Lenz’s law; eddy currents; induced fields. Self-inductance; back
emf; inductors; energy in the B-field; L-R circuits.
1.!
A cylindrical bar magnet is moving to the
v
right. Poles of the magnet are indicated,
S
N
with field lines as shown emanating
a
from the north pole and returning at
the south pole. Points a and b are slightly below the symmetry axis.
a.!
What is the direction of the induced E-field at a?
b.!
Repeat for b.
b
!
[Imagine circular conducting loops around the symmetry axis, passing through
the points in question. What would be the directions of the induced currents?]
2*.!
Shown in cross-section is a long solenoid of radius R
with a B-field into the page inside its windings, and
negligible outside. A positive charge Q is at rest at a
point outside the solenoid. The solenoid current is
increasing at rate dI /dt .
3.!
B⊗
r
Q
a.!
What direction is the induced E-field at the
location of the charge?
b.!
What is the magnitude of the force on the charge ? Ans: F = 12 µ0nR 2
dI Q
⋅ .
dt r
c.!
How would your answers change if r < R ? Ans: F = 12 µ0n
dI
⋅ Qr .
dt
A bar magnet is near a circular loop of copper wire which is
suspended at rest by a thread as shown.
a.!
How does the loop move if the magnet is moved
toward it?
b.!
Away from it?
c.!
How would your answers change if the poles of the
magnet were reversed?
S
N
PHY 142!
Assignment 6!
Summer 2013
4.!
In the situation shown from above, a bar magnet is rotating
counter-clockwise as shown at a small distance above a
conducting disk, initially at rest but free to rotate about a
vertical axle through its center. Does the disk move? If so,
which way? Explain your answers.
5.!
A small disk magnet is dropped from rest vertically down
three equally long empty cylindrical tubes, in succession.
Tube A is made of a non-conductor, tube B of aluminum, and tube C of copper. It
takes time T for the magnet to emerge from the bottom of the tube. Comment on
the following statements.
6.!
7.!
a.!
T is shortest for A.
b.!
T is shorter for B than for C because aluminum has lower resistivity.
c.!
The retardation of the fall in B and C is caused by eddy current loops
induced in the tube by the changing B-field as the magnet passes by.
d.!
The forces exerted by the induced current loops above and below the
magnet are both upward.
To get a qualitative understanding, one can sometimes imagine electron “orbits”
in an atom to be like tiny current loops. Consider hydrogen, which has only one
electron, and imagine the orbit to be that of a classical particle of charge −e
running in a circle of radius a at speed v.
a.!
What is the magnitude of the current? [How much charge passes a point
in the orbit per second?] Ans: I = ev / 2π a .
b.!
What is the magnitude of the magnet moment? Ans: µ = eva/ 2 .
c.!
Show that the ratio of the magnetic moment to the angular momentum of
the electron is given by e / 2m , where m is the electron mass.
Suppose the atom in the previous question is in the B-field
of a bar magnet as shown. Assume the magnetic moment of
the electron orbit is the only magnetic property of the atom.
Atom
S
N
a.!
How will the magnetic moment of the atom be aligned relative to the field
of the magnet?
b.!
Which direction will the atom move (toward or away from the magnet)?
[This is an example of paramagnetism.]
PHY 142!
8.!
Assignment 6!
Summer 2013
Instead of a hydrogen atom, consider a helium atom, with two electrons. Assume
their orbits are similar, but that the electrons run in opposite directions so that the
magnetic moments cancel. Let this atom be located at rest in the B-field of the
magnet in the previous question.
a.!
Suppose the atom is pushed toward the magnet, so it experiences an
increasing B-field. This will produce changes in the currents of the electron
orbits (by Faraday’s law). What will be the direction of the resulting total
magnetic moment of the atom? Explain carefully.
b.!
What will be the direction of the force exerted by the magnet on the atom?
[This is an example of diamagnetism.]
9.!
An AC generator consists of a flat coil of N turns and area A
rotating in a uniform magnetic field B. It is shown (in side
view) at an instant when the normal to the coil makes angle
θ with the direction of the B-field. The coil rotates counter-
coil
A
θ
B
clockwise at constant angular speed ω , so θ = ω t .
a.!
Find the emf E(t) produced by the generator as a
function of time, in terms of B, A, N and ω . Ans: E(t) = ω NAB ⋅ sin ω t .
b.!
The generator is put across a resistor R. Find the magnitude of the
sin ω t
magnetic moment µ of the coil. Ans: µ = ω N 2 A 2B ⋅
R
c.!
At the instant shown µ is parallel to the vector A. What is the magnitude
τ of the torque exerted on the coil by the B-field? Ans:
τ = ω N 2 A 2B 2 ⋅
d.!
sin 2 ω t
.
R
An equal and opposite external torque must be applied to keep the coil
rotating. The power supplied to the generator by this torque is P = τω .
Show that this is the same as the power delivered to the resistor. Ans:
P = ω 2 N 2 A 2B 2 ⋅
sin 2 ω t E 2
=
.
R
R
PHY 142!
10.!
Assignment 6!
Summer 2013
Questions about energy in the fields.
2
a.!
Show that the ratio of the energy densities is
um ⎛ cB ⎞
=
, where c is the
ue ⎜⎝ E ⎟⎠
speed of light.
b.!
Evaluate this ratio for the field of a large electromagnet, with B = 1 T and
an E-field large enough to ionize air, E = 3 × 106 V/m.
11.!
Consider an ordinary piece of wire, a cylinder of radius a and resistivity ρ . It
carries current I distributed uniformly across its cross-section.
a.!
Express the magnitudes of the E and B fields at the surface of the wire in
terms of I, a, ρ and universal constants. [Use E = ρ j , where j is the current
density in the wire.]
b.!
Calculate the ratio um /ue at the surface of the wire. [Use the result of the
above question.]
c.!
12*.!
Evaluate the ratio for typical numbers: a = 3 mm, ρ = 1.6 × 10−8 (SI units).
The toroidal coil shown has N closely wound turns.
Current I comes in at A and leaves at B.
a.!
Use Ampere’s law to find the formula for B in the
region inside the windings, as a function of R, the
distance from the center of the form.
b.!
Let the average B-field within the windings be
given by B = 0.02 I (SI units), the average radius
R
R = 0.1 m, and the cross section area of one
A B
winding A = 0.001 m 2 . Assume the B-field is
approximately uniform across the winding. Find
the total magnetic energy stored in the coil, in terms of I. [What is the
energy density?] Ans: 0.1 ⋅ I 2 J.
c.!
Use this to find the self inductance. Ans: 0.2 H.
d.!
The resistance of the wires is 1 Ω. At a particular time I = 1 A, increasing
at rate dI /dt = 10 A/s. Find the terminal voltage VB − VA . Ans: –3 V.
PHY 142!
13*.!
Assignment 6!
Summer 2013
A solenoid has N1 total turns, cross-section area A, length  , and negligible
resistance. The field inside its windings is (approximately) B(t) = µ0 (N1 /)I(t) ,
where I(t) = I0 cos ω t .
a.!
Find the total induced emf E1 (t) in the solenoid, assuming the B-field is
uniform over its whole length. [What is the induced emf in each turn?]
14.!
15.!
!
µ0 N12 A
Ans: E1 =
⋅ ω ⋅ I0 sin ω t .

b.!
µ0 N1 2 A
Find its self inductance. Ans: L =

c.!
Now a shorter “secondary” solenoid of slightly greater cross-section is
wrapped around the “primary” one near its center. This secondary
solenoid has N 2 turns. Find the total induced emf E 2 (t) in this solenoid.
!
Ans: E2 =
d.!
Find the ratio E 2 (t)/E1 (t) . (This is the law of the transformer).
µ0 N1 N 2 A
⋅ ω ⋅ I0 sin ω t .

In the circuit shown the switch is closed at t = 0 . At a
later time the current has risen to ½ its maximum
value. The dotted rectangle indicates an inductor
which has resistance 2 Ω in its windings.
12 V
10 Ω
a
b
a.!
What is the rate of increase of the current,
dI /dt , at that time? Ans: 6 A/s.
b.!
What is the terminal voltage Vb − Va at that time? Ans: 7 V.!
2Ω
1H
A coil has self inductance L and resistance R. At a time when the current is 1 A,
increasing at a rate 2 A/s, the potential drop across the coil is 120 V. At a later time
the current is again 1 A, but decreasing at a rate 2 A/s; the potential drop is 80 V.
a.!
What are L and R? Ans: R = 100 Ω, L = 10 H .
b.!
At an instant when the current is zero but increasing at a rate 4 A/s, what
is the potential drop? Ans: 40 V.