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 = µ î 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______________.
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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 n̂ -direction) so there is a clockwise
induced current to oppose that change. Therefore
the magnetic dipole vector points in the
  
negative n̂ -direction. The torque τ = µ × B ext is therefore in the positive z -direction.
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Question 3 (5 points)

A coil of wire with resistance R defines an open surface whose normal dA 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)
(c)
(b)
(d)
The correct answer is ________c________________.
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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_______________.
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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_______________.
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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 = α r k̂ (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 k̂ 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 ,
a

2πα 3
3 I
J
⋅
n̂
da
=
I
=
o
∫∫S
∫0 2π r dr (α r ) = 3 a ⇒ α = 2π ao3
For current through b < r < c ,
c

2
2
J
⋅
n̂
da
=
−
I
=
o
∫∫
∫ 2π r dr ( − β ) = − βπ c − b ⇒ β =
S
b
(
)
(
Io
π c 2 − b2
)
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
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r
 

2π 3
B
⋅
d
s
=
2
π
rB
=
µ
J
⋅
n̂
da
=
µ
2π r ′ dr ′ (α r ′ ) =
αr
θ
0 ∫∫
0∫

∫
3
closed
S
0
path
αr
B = θˆ µ0
3
2
where θ̂ is a unit vector oriented counterclockwise.
For a<r < b, loop is circle of radius a<r < b, and
∫
closed
path
 

B ⋅ d s = 2π rBθ = µ0 ∫∫ J ⋅ n̂ da = µ0 I 0
S
µI
B = θˆ 0 o
2π r
where θ̂ is a unit vector oriented clockwise.
For b<r < c, loop is circle of radius b<r < c, and
r
⎛
⎞
 
⎛
⎞
β
β
B
⋅
d
s
=
2
π
rB
=
µ
I
1−
2π r ′ dr ′ ⎟ = µ0 I o ⎜ 1− π r 2 − b2 ⎟
θ
0 o⎜

∫
∫
Io b
Io
⎝
⎠
⎝
⎠
closed
(
)
path
⎞
µI ⎛ β
B = θˆ 0 o ⎜1 − π ( r 2 − b2 ) ⎟
2π r ⎝ I o
⎠
where θ̂ is a unit vector oriented counterclockwise. This can be written using the results
above as
B = θˆ
2
2
µ0 I o ( c − r )
2π r ( c 2 − b 2 )
For c<r, loop is circle of radius c<r, and
∫
 
B ⋅ d s = 2π rBθ = 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.
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The graph is a concave upward parabola from 0 to a, rising to a value of
Then it goes as inverse r from a to b, decreasing to
µ0 I o
at r = a.
2π a
µ0 I 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.
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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 ydirection. 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 ) k̂; 0 ≤ t ≤ T ,
⎪
T ≤t
⎩ J e k̂;
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 = −wBx = µ0 ∫∫ J ⋅ n̂ da =µ0 wy(J e t / T ) ⇒ B = − x̂µ0 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
 

d
d
B ⋅ d s = −wBx ( y ) = µ0 ∫∫ J ⋅ n̂ da =µ0 w (J e t / T ) ⇒ B = − x̂µ0 (J e t / T )
2
2
closed
S
∫
path
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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.
dΦ d
d
d
= B s 2 = µ0 ( J e t / T ) s 2
dt
dt
dt
2
I =
1 dΦ
d Je s2
.
= µ0
R dt
2 RT
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.
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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.
dΦ d
dr
= B π r 2 = B 2π r
dt
dt
dt
I =
1 d Φ 2π rBv
=
R dt
R
b) What is the rate at which energy is dissipated (Joule heating) during the time
interval 0 ≤ t ≤ T ?
2
4π 2 r 2 B 2 v 2
⎛ 2π rBv ⎞
.
I R=⎜
R
=
⎟
R
⎝ R ⎠
2
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?
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 
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
dFagents
ds

2π rB 2 v
= −I × B ext = r̂IB = r̂IB = r̂
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
 
⎡ 2π rB 2 v ⎤  2π rB 2 v 2 ds
.
F ⋅ v = ds ⎢r̂
⎥⋅v =
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.
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