MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2014
Exam 3 Equation Sheet




Force Law: Fq = q(Eext + v q × B ext )
Force on Current Carrying Wire:

F=
∫
 
Id s′ × B ext
wire
Magnetic Dipole

µ = IAn̂ RHR
Torque on a Magnetic Dipole
 
τ = µ × B ext
Force on a Magnetic Dipole
   
F = ∇( µ ⋅ B)
Fz = µ z
∂Bz
∂z
Source Equations:
 
 
dq
dq′(r − r′)
E(r ) = ke ∫ 2 rˆ = ke ∫
  3
r
r − r′
source
source

 
µo
Id s × r̂
B(r) =
∫ r2
4π source

 
 
µo
Id s′ × (r − r ′ )
B(r) =
∫
  3
4π source
r − r′
r̂ points from source to field point
Ampere’s Law
 

∫ B ⋅ d s = µ0 ∫∫ J ⋅ n̂ da
closed
path
S
Faraday’s law
∫
closed
fixed path
 

d
E ⋅ ds = −
B ⋅ n̂ da
∫∫
dt open
surface S
ε = I ind R
Gauss’s Law for Magnetism:


∫∫ B ⋅ n̂ da = 0
closed surface
Gauss’s Law:
  qenc
E
⋅ dA =

∫∫
εo
closed surface
Electric Potential Difference
Electrostatics:
b
 
ΔV = Vb − Va ≡ − ∫ E ⋅ d s
a


E = −∇V
Potential Energy:
ΔU = qΔV
Capacitance:
1 Q2 1
Q
UE =
= CΔV 2
C=
2 C 2
ΔV
Inductance:
L=
ε back = − LdI / dt
Φ B,Total
I
=
N ΦB
I
U M = 12 LI 2
Energy Density Stored in Fields:
uE = 12 ε 0 E 2 ; uB = 12 B 2 / µ0
1
Current Density and Current:

I = ∫∫ J ⋅ n̂ da
Power Dissipated in Resistor:
PJoule = I 2 R = ΔV 2 / R
Ohm’s Law: ΔV = I R


J = σ c E where σ c is the conductivity


E = ρ r J where ρ r is the resistivity
Constants:
µ0 ≡ 4π × 10−7 T ⋅ m ⋅ A -1
 
Power: P = F ⋅ v
ke = 1 / 4πε 0  9.0 × 109 N ⋅ m 2 ⋅ C-2
open surface
ε 0 ≡ 1 / µ 0 c 2  8.85 × 10-12 C2 ⋅ N -1 ⋅ m -2
Power from Voltage Source:
Psouce = I ΔV
2
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2014
Exam 3 Practice Problems Solutions
Problem 1 Biot Savart
A current loop, shown in the figure below,
consists of two arc segments as shown, with a
common center at P. One arc segment has an
opening angle of 120 degrees and the other arc
segment has an opening angle of 240 degrees.
Two straight line segments join the arc
segments. One arc segment has radius R and
the other arc segment has radius R / 2 . A
current I1 flows clockwise in the loop in the
direction shown.

a) What is the direction and magnitude of the magnetic field B at the point P ?
Answer: The sides will not contribute because the cross product is zero there. So using
the Biot-Savart Law we have that

µ

B outer arc (r) = o
4π

µ

B inner arc (r) = o
4π
∫

I1d s × r̂
µI
µI 1
µI
= − ẑ o 12 ∫ ds = − ẑ o 12 ( 2π R ) = − ẑ o 1 .
2
6R
r
4π R
4π R 3
∫

I1d s × r̂
µo I1
µo I1
2µ I
2
R
= − ẑ
ds = − ẑ
2π = − ẑ o 1 .
2
2 ∫
2
2
3R
r
4π ( R / 2 )
4π ( R / 2 ) 3
Therefore the total field at P is given by
 
µI
4µ I
5µ I
B(r) = − ẑ o 1 − ẑ o 1 = − ẑ o 1 .
6R
6R
6R
Two fixed conducting rails are arranged as shown in the figure below. A metal bar of
length s is placed at the origin, initially held in place, and a current I 2 runs through the
rails and bar clockwise. The bar is then released. You may assume that the length of the
bar is very short and that the magnetic field you calculated in part a) is uniform over the
length of the bar. You may neglect the magnetic field due to the current through the rails.
Solving 7-3
b) What is the direction and magnitude of the magnetic force acting on bar the
instant it is released?
Answer: The field at point P is inward from our calculations above, and the current in
the bar is upward, so the total force on the bar is in the negative x direction, with
c)

5µ I
F = − x̂ o 1 I 2 s
6R
Solving 7-4
Problem 2: Current Ring
A circular ring of radius R has a current I in the counterclockwise direction as seen
from above.
c) Calculate the magnetic field due to the current at an arbitrary point a distance z
along the z -axis passing through the center of the ring perpendicular to the plane
of the ring.
z = +∞
 
d) Calculate the line integral ∫ B ⋅ d s along the z-axis from z = −∞ to z = +∞ .
z = −∞
e) The path is not a closed and yet it satisfies Ampere’s Law. Why?
Solution:
We shall apply the Biot-Savart law to calculate the magnetic field at the point P a
distance z along the positive z-axis

  
θ = 2π

µ0 I d s × r̂ θ = 2π µ0 I d s × (r - r ′)
B(P) = ∫
= ∫
  3
4π r 2
4π
r - r′
θ =0
θ =0
Choose a coordinate system and unit vectors as shown in the figure below.
Solving 7-5



Then d s = Rdθ θ̂ , r = z k̂ , and r ′ = R r̂ . Then the Biot Savart Law becomes
θ = 2π

µ I (Rdθ θ̂) × (z k̂ - R r̂)
.
B(z) = ∫ 0
3
4
π
θ =0
z k̂ - R r̂
(1.1)
We now explicitly calculate the cross products noting that θ̂ × k̂ = r̂ and θ̂ × (−r̂) = k̂
and Eq. (1.1) becomes

B(z) =
θ = 2π
µ0 IzR
4π (z 2 + R 2 )3/ 2
∫
dθ r̂ +
θ =0
θ = 2π
µ0 IR 2
4π (z 2 + R 2 )3/ 2
∫
dθ k̂
(1.2)
θ =0
The first integral vanishes because as we integrate around the circle the unit vector r̂
always points radially outward and hence sums to zero. Equivalently if we choose
coordinates such that r̂ = cosθ î + sin θ ĵ then
=
θ = 2π
µ0 IzR

B(z) =
4π (z 2 + R 2 )3/ 2
µ0 IzR
4π (z 2 + R 2 )
∫
dθ r̂ =
θ =0
θ = 2π
µ0 IzR
4π (z 2 + R 2 )3/ 2
∫
dθ (cosθ î + sin θ ĵ)
θ =0

θ = 2π
θ = 2π
(sin
θ
î
+
−
cos
θ
ĵ)
=
0
3/ 2
θ =0
θ =0
The second integral in Eq. (1.2) is straightforward and so the magnetic field along the zaxis is given by

B(z) =
µ0 IR 2
4π (z 2 + R 2 )3/ 2
z = +∞
We now calculate the line integral
∫
µ0 IR 2
θ = 2π
θ θ =0 k̂ =
2(z 2 + R 2 )3/ 2
k̂ .
(1.3)
 
B ⋅ d s along the z-axis from z = −∞ to z = +∞ .
z = −∞
z = +∞
∫
z = −∞
  z = +∞
B ⋅ ds = ∫
z = +∞
µ0 IR 2
2
2 3/ 2
z = −∞ 2(z + R )
k̂ ⋅ dzk̂ =
∫
µ0 IR 2
2
2 3/ 2
z = −∞ 2(z + R )
dz
(1.4)
This integral is equal to
z = +∞
∫
z = −∞
  z = +∞
B ⋅ ds = ∫
z = −∞
µ0 IR 2
2(z 2 + R 2 )3/ 2
dz =
µ0 Iz
2(z 2 + R 2 )1/ 2
z = +∞
=
z = −∞
µ0 I
µI
− − 0 = µ0 I
2
2
(1.5)
Solving 7-6
where we have used the fact that
µ0 I(z)
µ0 I
µ0 I(z)
µ0 I
and
.
lim
=
lim
=
−
z→∞ 2((z) 2 + R 2 )1/ 2
z→−∞ 2((z) 2 + R 2 )1/ 2
2
2


∫ B ⋅ d s = µ I
Recall that Ampere’s Law states that
0 enc
closed line integral.
however the integral above is not a

However if we integrate B along a semi-circular path at infinity from z = +∞ to z = −∞
and add our two line integrals together we get a closed line integral
z = +∞
∫
 
B ⋅ ds +
z = −∞
∫
semi-circle
atinfinity
 
 
B ⋅ ds = 
∫ B ⋅ ds
(1.6)

Although we have only calculated B(z) along the z-axis, we see that


µ0 IR 2
. This holds everywhere on the semi-circle at infinity
lim B(z) = lim
k̂
=
0
z→∞
z→∞ 2(z 2 + R 2 )3/ 2
so
  
B
(1.7)
∫ ⋅ ds = 0
semi-circle
atinfinity
Therefore combining Eqs. (1.7), (1.6), and (1.5) we have that
 
B
∫ ⋅ d s = µ0 I
(1.8)
I enc = I
(1.9)
From the figure above we see that
Therefore Ampere’s Law is satisfied.
Solving 7-7
Problem 3: Magnetic fields of Current Loops
(a) Find the magnetic field at point P due to current loop shown in the figure below.
Solution:
(b) There is no magnetic field due to the straightsegments because point P is along
the lines. Using the general expression for dB obtained in (a), for the outer
segment, we have
π

⎛µ I⎞
µ Idθ
B out = ∫ 0
k̂ = ⎜ 0 ⎟ k̂ .
4π b
⎝ 4b ⎠
0
Similarly, the contribution to the magnetic field from the inner segment is
0

⎛µ I⎞
µ Idθ
B in = ∫ 0
k̂ = − ⎜ 0 ⎟ k̂ .
4π a
⎝ 4a ⎠
π
Therefore the net magnetic field at Point P is



µ I ⎛ 1 1⎞
B net = B out + B in = − 0 ⎜ − ⎟ k̂ .
4 ⎝ a b⎠
Solving 7-8
Problem 4: Torque on Circular Current Loop
A wire ring lying in the xy-plane with its
center at the origin carries a
counterclockwise current I. There is an
external uniform magnetic field

B = Bx î + By ĵ such that By > 0 and

Bx < 0 .The magnetic moment vector µ
is perpendicular to the plane of the loop
and has magnitude µ = IA and the
direction is given by right-hand-rule
with respect to the direction of the
current. What is the direction and
magnitude of the torque on the loop?
Solution: The torque on a current loop in a uniform field is given by
  
τ = µ × B,


where µ is the magnetic dipole moment with magnitude µ = IA and points
perpendicular to the plane of the loop and right-handed with respect to the direction of
current flow. The magnetic dipole moment is given by


µ = IA = I π R 2k̂ .
Therefore,
  
τ = µ × B = (I π R 2 )k̂ × (Bx î + By ĵ) = I π R 2 (Bx ĵ − By î) .
Define an angle θ between the negative x - and y -axes as shown in the figure to the
right. Then the direction of the torque is given by
θ = tan(− Bx / By ) > 0
and the magnitude is

τ = I π R 2 Bx2 + By2 .
Solving 7-9
Problem 5: Ampere’s Law
Non-Uniform Current Density A long cylindrical cable consists of a conducting

cylindrical shell of inner radius a and outer radius b. The current density J in the shell is
out of the page (see sketch) and varies with radius as J (r) = α r for a < r < b and is zero
outside of that range, where α is a positive constant with units A ⋅ m -3 . Find the
magnetic field in each of the following regions, indicating both magnitude and direction
(i) r < a , (ii) a < r < b , and (iii) r > b . For each region make a figure clearly showing
your choice of Amperian loop.
Solution:
(a) For r < a , the Amperian loop is shown in the figure below.
The enclosed current is I enc = 0 . Therefore, by Ampere’s law, B = 0
(b) In the region a < r < b , the Amperian loop and integration element is shown in the
figure below.
10
The enclosed current is
I enc =
 
J ⋅ dA =
∫∫
open surface
∫
r
a
r
α r ′ 2π r ′ dr ′ = 2πα ∫ r ′ 2 dr ′ =
a
2πα 3
(r − a 3 )
3
Applying Ampere’s law, we have
 
B
∫ ⋅ d s = µ0 Ienc
2πα 3
(r − a 3 )
3
⇒ B(2π r) = µ0
or

α
1
B = Bϕ̂ = µ0 (r 3 − a 3 ) ϕ̂
3
r
The magnetic field points in the azimuthal direction.
(b) In the region r > b , the Amperian loop is shown in the figure below. We use the same
integration element as in part b).
Then the enclosed current is
I enc =
 
J ⋅ dA =
∫∫
open surface
∫
b
a
α r ′ 2π r ′ dr =
2πα
3
∫
b
a
r ′ 2 dr ′ =
2πα 3
(b − a 3 )
3
Applying Ampere’s law, we have


∫ B ⋅ d s = µ I
0 enc
⇒ B(2π r) =
µ0 2πα 3
(b − a 3 )
3
or

α
1
B = B ϕ̂ = µ0 (b3 − a 3 ) ϕ̂
3
r
11
Collecting our results we have that
⎧
⎪0 ;
r<a
⎪
 ⎪ α
1
B = ⎨ µ0 (r 3 − a 3 ) ϕ̂; a < r < b
r
⎪ 3
⎪ α 3
3 1
⎪⎩ µ0 3 (b − a ) r ϕ̂ ; r > b
A graph of the magnitude of B vs. r is shown below.
12
Problem 6: Ampere’s Law
The figure below shows two slabs of current. Both slabs of current are infinite in the x
and z directions, and have thickness d in the y-direction. The top slab of current is

located in the region 0 < y < d and has a constant current density J out = J k̂ out of the
page. The bottom slab of current is located in the region −d < y < 0 and has a constant

current density J in = − J k̂ into the page.
(a) What is the magnetic field for y > d ? Justify your answer.
The magnitude of the magnetic field of a single slab of current of thickness d outside the
slab is µo Jd / 2 , and the field reverses depending on the sign of y, with y > d . The
field of the two slabs shown above is zero for y > d because when you are above the
sheet or below the sheet by superposition of the field due to the slab carrying current out
of the page just cancels the field due to the slab carrying current into the page.
(b) Use Ampere’s Law to find the magnetic field at y = 0 . Show the Amperian Loop
that you use and give the magnitude and direction of the magnetic field.
We take an Amperean loop whose bottom is at y = 0 and whose top is at y ≥ d , of width
w.
Ampere’s Law becomes
13
∫
closed
path
 

B ⋅ d s = ∫ B(0) ⋅ dx î = wBx (0) = µ0 Jwd ⇒ Bx (0) = µ0 Jd > 0
Our calculation gives a positive value for the magnetic field at y = 0 . Because we are
circulating counterclockwise, we traverse the leg at y = 0 in the positive x-direction and
therefore at y = 0 , the magnetic field points in the positive x-direction. Therefore

B(0) = µ0 Jd î .
(c) Use Ampere’s Law to find the magnetic field for 0 < y < d . Show the Amperian
Loop that you use and give the magnitude and direction of the magnetic field.
We take an Amperean loop whose bottom is at y = 0 and whose top is at 0 ≤ y ≤ d , of
width w .
Ampere’s Law now becomes
∫
closed
path
 


B ⋅ d s = ∫ B(0) ⋅ dx î + ∫ B( y) ⋅(− dx î )
= wBx (0) − wBx ( y) = µ0 Jwd − wBx ( y) = µ0 Jwy

⇒ B( y) = µ0 J (d − y) î
(d) Make a plot of the x-component of the magnetic field as a function of the distance y
from the center of the two slabs.
14
Problem 7: Two Current Sheets Ampere’s Law
Consider two infinitely large sheets lying in the xy-plane separated by a distance d


carrying surface current densities K1 = K ˆi and K 2 = − K ˆi in the opposite directions, as
shown in the figure below (The extent of the sheets in the y direction is infinite.) Note
that K is the current per unit width perpendicular to the flow.

a) Find the magnetic field everywhere due to K1 .

b) Find the magnetic field everywhere due to K 2 .
c) Find the magnetic field everywhere due to both current sheets.
d) How would your answer in (c) change if both currents were running in the same


direction, with K1 = K 2 = K ˆi ?
Solution:

(a) Find the magnetic field everywhere due to K1 .
B

K1
l
B
Consider the Ampere’s loop shown above. The enclosed current is given by
15
 
I enc = ∫ J ⋅ dA = Kl
Applying Ampere’s law, the magnetic field is given by
B(2l) = µ0 Kl
or B =
µ0 K
2
Therefore,
⎧ µ0 K
 ⎪⎪− 2 ĵ, z >
B1 = ⎨
⎪ µ0 K ĵ,
z<
⎪⎩ 2

(b) Find the magnetic field everywhere due to K 2 .
d
2
d
2
The result is the same as part (a) except for the direction of the current:
⎧ µ0 K
ĵ,
⎪⎪

2
B2 = ⎨
⎪− µ0 K ĵ,
⎪⎩ 2
d
2
d
z<−
2
z>−
(c) Find the magnetic field everywhere due to both current sheets.
⎧
d
d
⎪⎪ µ0 K ĵ, − 2 < z < 2
  
B = B1 + B 2 = ⎨
d
⎪0,
| z |>
⎪⎩
2
(d) How would your answer in (c) change if both currents were running in the same


direction, with K1 = K 2 = K ˆi ?

In this case, B1 remains the same but
⎧ µ0 K
d
ĵ, z > −
⎪⎪−

2
B2 = ⎨ 2
d
⎪ µ0 K ĵ,
z<−
⎪⎩ 2
2
Therefore,
16
⎧
d
⎪− µ0 K ĵ, z > 2
⎪
  
d
d
⎪
B = B1 + B 2 = ⎨0,
− <z<
2
2
⎪
d
⎪
z<−
⎪ µ0 K ĵ,
2
⎩
Finally, to make it twice as hard to turn that means twice as much work, which means
that the resistance must be half as much. This is called “loading” the generator – where
an increase in load is actually a decrease in the resistance.
17
Problem 8: Straight Wire and Rectangular Loop of Wire
At t = 0 , there is a current I 0 through a very long wire. (You may assume this is an
infinite wire). No current is flowing in a rectangular loop of wire with dimensions length
l and width w , and has a resistance R . The side of the square loop closest to the wire
is a distance s away from the wire. At t = 0 , the current in the wire is decreased
according to
⎧⎪(1 − α t)I 0 , 0 < t < 1 / α
I(t) = ⎨
⎪⎩0, t > 1 / α
a) In what direction does the induced current in the rectangular loop flow?
b) What total charge passes a given point in the loop during the time this current flows?
Solution:
a) In what direction does the induced current in the rectangular loop flow?
Answer:
For the time interval 0 < t < 1 / α , the current is decreasing, therefore the magnetic field
generated by this current is also decreasing, and hence the magnetic flux in the
rectangular loop is also decreasing into the plane of the figure. By Lenz’z Law, there is
an induced current in the rectangular loop in the clockwise direction to oppose this
change in magnetic flux. We can use Ampere’s Law to calculate the magnetic field due to
the wire, and then Faraday’s Law in the form
I ind R = −

d
B
∫∫ ⋅ n̂da
dt loop
to calculate the induce current.
In the figure below we show our choice of a circle for the Amperian loop.
18
Then Ampere’s Law
 
B
∫ ⋅ d s = µ0 I enc becomes B2π r = µ0 I and so the magnetic field
C
points into the plane of the figure and the magnitude varies with the distance r from the
wire according to
µI
B= 0 .
2π r
In order to apply Faraday’s Law, we choose n̂ = k̂ into the plane of the figure below, we
choose for an integration area element da = ldr , and the magnetic field in that element is
 µ I
given by B = 0 k̂ .
2π r
Therefore the left hand side of Faraday’s Law becomes

d
d
− ∫∫ B ⋅ n̂da = −
dt loop
dt
=−
r = s+ w
∫
r=s
µ0 I
µ0 l dI r = s+ w dr
k̂ ⋅ k̂ldr = −
2π r
2π dt r∫= s r
µ0 l dI ⎛ s + w ⎞
ln
2π dt ⎜⎝ s ⎟⎠
From our function that describes the decreasing current we have that
dI ⎧⎪−α I 0 , 0 < t < 1 / α
=⎨
dt ⎪⎩0, t > 1 / α
Therefore for the interval 0 < t < 1 / α ,
19
−

µ lα I ⎛ s + w ⎞
d
B ⋅ n̂da = 0 0 ln ⎜
.
∫∫
dt loop
2π
⎝ s ⎟⎠
Hence Faraday’s Law becomes
I ind R =
µ0 lα I 0 ⎛ s + w ⎞
.
ln ⎜
2π
⎝ s ⎟⎠
We can now solve for the induced current
I ind =
µ0 lα I 0 ⎛ s + w ⎞
.
ln ⎜
2π R
⎝ s ⎟⎠
Note that I ind > 0 , and because we choose n̂ = k̂ into the plane of the figure, a positive
I ind means that the current is in the clockwise direction consistent with our argument
using Lenz’s Law.
b) What total charge passes a given point in the loop during the time this current flows?
Answer:
During the time interval 0 < t < 1 / α , the total charge passing a given point in the loop is
given by
Q=
t =1/ α
∫
t =0
I ind dt =
t =1/ α
∫
t =0
µ0 lα I 0 ⎛ s + w ⎞
µ lI ⎛ s + w ⎞
ln ⎜
dt = 0 0 ln ⎜
⎟
2π R
2π R ⎝ s ⎟⎠
⎝ s ⎠
20
Problem 9 Ampere’s Law and Faraday’s Law
Consider an infinite cylindrical solid wire that has radius a . The wire has a time varying
current with the current density as a function of time given by the following expression:

⎧0;
t≤0
 ⎪
J = ⎨(J et / T ) k̂; 0 ≤ t ≤ T ,
⎪
T ≤t
⎩ J e k̂;
where J e is positive constant with units of amps per square meter.
a) Find the direction and magnitude of the magnetic field for the interval 0 ≤ t ≤ T
in the regions: (i) 0 ≤ r ≤ a ; (ii) r ≥ a , where r is the distance form the
symmetry axis of the wire.
Clearly show all your work. Answers without justification will receive no credit.
0 ≤ r ≤ a : We take an Amperean loop which is a circle of radius r<a. We have
r
 

 r
⎛ J t⎞
⎛ J t⎞
B
⋅
d
s
=
2
π
rB
=
µ
J
⋅
n̂
da
=
µ
2π r ′ dr ′ ⎜ e ⎟ = µ0π r 2 ⎜ e ⎟ ⇒ B = µ0 (J e t / T )θ̂
θ
0 ∫∫
0∫

∫
2
⎝ T ⎠
⎝ T ⎠
closed
S
0
path
θ̂ is counterclockwise looking from left
r > a : We take an Amperean loop which is a circle of radius r>a. We have
a
 

 a2
⎛ J et ⎞
2 ⎛ J et ⎞
∫ B ⋅ d s = 2π rBθ = µ0 ∫∫S J ⋅ n̂ da =µ0 ∫0 2π r ′ dr ′ ⎜⎝ T ⎟⎠ = µ0π a ⎜⎝ T ⎟⎠ ⇒ B = 2r µ0 (J e t / T )θ̂
closed
path
θ̂ is counterclockwise looking from left
21
b) Suppose a square conducting loop with resistance R , and side s is placed in the
region r > a , such that the nearest side of the loop to the wire is a distance b from
the axis as shown in the end view 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 clockwise as viewed from above. In the time interval
0 ≤ t ≤ T , we have
b+s

dΦ
d
d
a2
=
B ⋅ n̂ da =
dz dr µ0 (J e t / T )
dt
dt open∫∫
dt ∫b
2r
surface S
b+s
⎛ b + s⎞
sa
dr sa 2
=
µ0 (J e t / T ) ∫
=
µ0 (J e t / T )ln ⎜
2
r
2
⎝ b ⎟⎠
b
2
I =
sa 2 J e ⎛ b + s ⎞
1 dΦ
= µ0
ln ⎜
⎟
R dt
2RT ⎝ b ⎠
c) 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 ?
The force on the far size and the near side of the loop are equal and opposite, and cancel.
The force on the left side of the loop is to the right and the force on the right side is to the
left, giving a net force pushing the loop away from the wire. If x̂ is to the right, then

⎛ a2
⎞⎛1
1 ⎞
sa 2 J e ⎛ b + s ⎞
⎡
⎤
where I = µ0
F = x̂sI ⎣ B r=b+s − B r=b ⎦ = x̂sI ⎜ µ0 (J e t / T )⎟ ⎜ −
ln ⎜
⎟
⎝ 2
⎠ ⎝ b b + s ⎟⎠
2 RT ⎝ b ⎠
22
Problem 10: Mutual Inductance of Two Rings. Consider the circuits shown in the
figure below, two coplanar, concentric rings, a small ring C2 of radius R2 and a much
larger ring C1 of radius R1 with current I1 . What is the mutual inductance? You may

assume that R2 << R1 , so that you may neglect the variation of the magnetic field B1 over
the interior of the small ring.

Solution: We begin by calculating the magnetic field B1 at the center of ring C2 using
the Biot-Savart Law
 µ0 Id s × r̂
dB =
.
4π r 2
Choose cylindrical coordinates with unit vectors ( ρ̂, θ̂, k̂) . Note ρ̂ is the unit vector that

points radially outward in the plane. We have for the current element Id s = I1 R1dθθ̂ , the
distance from the current element to the center of ring C2 is r = R1 , and the unit vector
from the current element to the field point at the center of the rings is r̂ = − ρ̂ . Therefore
the Biot-Savart Law becomes


µ0 Id s × r̂ µ0 (IR1dθθ̂ × − ρ̂ ) µ0 Idθ
dB1 =
=
=
k̂
4π r 2
4π
4π R1
R12
Integrating around the circle we get that

µ0 I1 2π
µ0 I1
B1 =
d
θ
k̂
=
k̂ .
4π R1 ∫0
2R1

Because we are assuming that R2 << R1 , the magnetic field B1 is approximately uniform
over the small ring C2 so the magnetic flux Φ 21 through the ring C2 due to the magnetic
field of ring C1
Φ 21 = B1π R22 =
µ0 I1π R22
.
2R1
Therefore the mutual inductance is given by
Φ 21 µ0π R22
.
M 21 =
=
I1
2R1
23
Problem 11: Generator A square loop (side L ) spins with angular frequency ω in
uniform magnetic field of magnitude B . It is hooked to a load with resistance R .
a) What is the current I(t) generated in the loop?
b) What is the work done by the generator per revolution?
c) To make it twice as hard to turn, what do you do to R?
Solution: The flux is changing generating an EMF that drives a current given by
I(t) =
ε (t)
1 dΦ B
1 d ( BAcos(ω t) ) BL2
=−
=−
=
ω sin(ω t)
R
R dt
R
dt
R
The work done by the generator is the integral of the power:
2
2π ω
2π ω
⎛ BL2ω ⎞
B 2 L4ω 2
2
2
P= I R=⎜
⎟ Rsin (ω t) → W = ∫ P(t) dt =
∫ sin (ω t) dt
R
⎝ R ⎠
t =0
t =0
2
Using the fact that the time-averaged average value
2π ω
∫
t=0
sin 2 (ω t) dt =
1 2π π
=
2ω ω
(to see this, think sin 2 (ω t) + cos 2 (ω t) = 1 and they both must have the same average
value or do the integral), we find:
W=
B 2 L4ω 2 ⎛ 1 2π ⎞ π B 2 L4ω
⋅
=
R ⎜⎝ 2 ω ⎟⎠
R
Finally, to make it twice as hard to turn that means twice as much work, which means
that the resistance must be half as much. This is called “loading” the generator – where
an increase in load is actually a decrease in the resistance.
24
25
Problem 12: Stored Magnetic Energy
a) A magnetic field exists in most of interstellar space in our galaxy. There is
evidence that its strength in most regions is between 10 −11 T and 10 −10 T .
Adopting 3 × 10 −11 T as a typical value, find, in order of magnitude, the total
energy stored in the magnetic field of the galaxy. For this purpose you may
assume that the galaxy is a disk roughly 10 21 m in diameter and 1019 m thick.
b) What is the magnetic energy density of the earth’s magnetic field of 5.0 × 10 −5 T ?
c) Assuming the magnetic field is relatively constant over distances small compared
with the earth’s radius and neglecting the variations near the earth’s magnetic
poles, how much energy would be stored in a shell between the earth’s surface
and 15 km above the surface? (Look up any quantities that you may need.)
Solution:
a) Because we are approximating that the magnetic field is uniform, and modeling
the galaxy as a flat disk, the energy stored in the magnetic field is
1
U=
2 µ0
B2
2
∫ B dV  2µ π R s
disk
0
2
where B  3 × 10 −11 T , R  0.5 × 1021 m and s  1019 m . Then
U
(3 × 10−11 T)2
π (0.5 × 1021 m)2 (1019 m) = 0.3 × 1048 J  1047 J
2(4π × 10−7 T 2 ⋅ m 3 ⋅ J −1 )
b) The magnetic energy density of the earth is
B2
(5.0 × 10−5 T)2
uB =
=
= 2.0 × 10−3 J ⋅ m 3
2 µ0 (4π × 10−7 T 2 ⋅ m 3 ⋅ J −1 )
c) The mean radius of the earth is R = 6.4 × 10 6 m and so the magnetic energy
stored in a shell between the earth’s surface and s = 15 × 10 3 m above the
surface is approximately
U
B2
4π R 2 s = (2.0 × 10−3 J ⋅ m 3 )(4π )(6.4 × 106 m)2 (15 × 103 m) = 1.5 × 1016 J
2 µ0
26
Problem 13: The figure below shows two loops of wire having the same axis. The
smaller loop has radius a and resistance R and the larger loop has radius b . The
smaller loop is above the larger one, by a distance z , which is large compared to the
radius b of the larger loop, ( z >> b ). Hence with current I through the larger loop as
indicated, the consequent magnetic field is nearly constant through the plane area
bounded by the smaller loop. Suppose now that z is not constant but is changing at the
positive constant rate vz = dz / dt > 0 ( z increasing).
a) Determine the magnetic flux across the area bounded by the smaller loop as a
function of z .
b) Compute the emf generated in the smaller loop and the induced current at the
instant the loop is located at z and moving with z -component of velocity
vz = dz / dt > 0 . Determine the direction of the induced current in the smaller
loop.
c) What is the induced force on the small ring at the instant the loop is located at z
and moving with z -component of velocity vz = dz / dt > 0 ?
Solution:
a) We begin by calculating the magnetic field at the center of the smaller upper loop
due to the lower loop using the Biot Savart Law
 µ0 Id s × r̂
dB =
.
4π r 2
Choose cylindrical coordinates with unit vectors ( ρ̂, θ̂, k̂) . Note ρ̂ is the unit vector that

points radially outward in the plane. We have for the current element Id s = Ibdθθ̂ , the
distance from the current element to the center of the upper ring is r = (z 2 + b)1/ 2 , and the
27
unit vector from the current element to the field point at the center of the rings is
r̂ = (zk̂ − bρ̂ ) / (z 2 + b2 )1/ 2 . Therefore the Biot-Savart Law becomes


µ0 Id s × r̂ µ0 (IR2 dθθ̂ × (zk̂ − bρ̂ )) µ0 IR2 dθ (x ρ̂ + bk̂)
dB1 =
=
=
4π r 2
4π
4π (z 2 + b2 )3/ 2
(z 2 + b2 )3/ 2
Integrating around the larger ring the non-zero contribution is

B1 =
µ0 Ib2
2(z 2 + b2 )3/ 2
k̂ .
Because the small loop is very far away from the large loop we can approximate the
magnetic flux through the small loop by
Φ
µ0 Ib2
2(z 2 + b2 )3/ 2
π a2 .
b) We can use the chain rule to calculate the change in magnetic flux
3µ0 Ib2π a 2 z
dΦ dΦ dz

=−
v .
dt
dz dt
2(z 2 + b2 )5/ 2 z
Now we use Faraday’s Law to calculate the emf at z = α b
ε=
3µ0 Ib2π a 2 z
2(z 2 + b2 )5/ 2
vz .
The induced current is therefore
I ind
3µ0 Ib2π a 2 z
ε
= =
v .
R 2(z 2 + b2 )5/ 2 R z
We chose a unit normal for the upper loop in the positive k̂ -direction corresponding to a
counterclockwise circulation direction (as seen from above). Because the emf is positive
this means that the induced current is in the counterclockwise direction as seen from
above. We also note that the magnetic field is up, the flux is up and decreasing as the
small loop moves upward, therefore we need and induced magnetic field pointing
upwards which corresponds to a induced current counterclockwise as seen from above.
c) The small ring now acts like a magnetic dipole with magnetic moment
28
3µ0 Ib2π a 2 z

µ=
v π a 2 (k̂) ,
2(z 2 + b2 )5/2 R z
where the z -component of the magnetic dipole moment is given by
3µ0 Ib2π 2 a 4 zvz
µz =
2(z 2 + b2 )5/2 R
The z -component of the force on the upper ring is then
Fz = µ z
∂Bz
.
∂z
From our calculation in part a), the partial derivative of the magnetic field at
∂Bz
3µ0 Ib2 z
=−
∂z
2(z 2 + b2 )5/ 2
Therefore the z -component of the force on the upper ring is then
Fz = µ z
⎛ 9 µ0 2 I 2 b4π 2 a 4 z 2 vz ⎞
∂Bz ⎛ 3µ0 Ib2π 2 a 4 zvz ⎞ ⎛
3µ0 Ib2 z ⎞
=⎜
−
=
−
⎜ 4(z 2 + b2 )5 R ⎟ < 0 ,
∂z ⎝ 2(z 2 + b2 )5/2 R ⎟⎠ ⎜⎝ 2(z 2 + b2 )5/2 ⎟⎠
⎝
⎠
opposing its motion.
29
Problem 14: Power
In the circuits below, all of the batteries and light bulbs are identical.
A
B
C
Let the power consumed by the top bulb in each of the above circuits be PA, PB and PC
respectively. Rank these powers from smallest to largest (write = between powers which
are equal)
Answer: Smallest __ PA = PB_< PC __Largest
Both A and B have one battery voltage dropping across two resistors in series, meaning
that each resistor is going to get half of the voltage drop across it. In C, however, the two
batteries add to double the voltage of a single battery, meaning that each resistor has
double the voltage drop across it. Note that putting batteries in parallel, as in B, does
nothing useful except possibly increase the amount of current that can be sourced (in case
one battery couldn’t handle it).
30
Problem 15: Kirchoff’s Laws
In the multi-loop circuit shown on the right, which of the following set of equations is
correct.
I 2 + I 3 = I1
a) + I 2 R2 − ε1 + I1 R1 − ε 2 = 0
ε 2 − I1 R1 − I 3 R3 = 0
I 2 + I 3 = I1
b) − I 2 R2 − ε1 − I 3 R3 = 0
ε 2 + I1 R1 − I 3 R3 = 0
I1 + I 2 = I 3
c) − I 2 R2 − ε1 + I1 R1 − ε 2 = 0
ε 2 − I1 R1 − I 3 R3 = 0
I 2 + I 3 = I1
d) + I 2 R2 + ε1 − I 3 R3 = 0
ε 2 + I1 R1 − I 3 R3 = 0
I1 + I 2 = I 3
e) − I 2 R2 − ε1 + I1 R1 + ε 2 = 0
ε 2 − I1 R1 + I 3 R3 = 0
Answer: (c)
31
Problem 16: Pressure Transducers
A common way of measuring small pressure changes (for example, in car transmission
systems, airplane cabin pressure monitors, and Parenchymal catheter tips, used, for
example, to measure intercranial pressure) is to use piezoresistive materials, that is,
materials that change resistance when a pressure is applied to them. Often, four identical
piezoresistors are connected as shown below, in such a way that when a pressure is
applied Ra and Rd increase by some fraction and Rb and Rc decrease by about the same
fraction.
a) Determine an expression for Vout / Vin in terms of Ra , Rb , Rc , and Rd , where
Vin = V2 − V1 , and Vout = V3 − V4 .
Answer: Introduce currents in each branch according to the figure below.
Then
Vin = I1 (Ra + Rc ) ⇒ I1 =
Vin
Ra + Rc
Vin = I 2 (Rb + Rd ) ⇒ I 2 =
Vin
.
Rb + Rd
We also know that
V4 − V1 = I1 Rc
V3 − V1 = I 2 Rd
32
Then
Vout = (V3 − V4 ) = (V3 − V1 ) − (V4 − V1 )
= I 2 Rd − I1 Rc
=
.
Vin Rd
V R
− in c
Rb + Rd Ra + Rc
Therefore
Vout
Rd
Rc
=
−
Vin Rb + Rd Ra + Rc
(10)
Note that this is like the Wheatstone bridge, where an ammeter was connected across
the potential difference Vout = (V3 − V4 ) .
b) If a 1 V battery is connected across points 1 and 2 such that Vin = V2 − V1 , what
voltage will be measured at Vout = V3 − V4 if the system is unstrained and all four
resistors have the same resistance R ?
Answer: If they all have the same resistance then by symmetry the output voltage will be
zero.
Vout =
Vin R
V R
− in = 0
R+ R R+ R
c) An unfortunate problem with piezoresistors is that they are also temperature
dependent. If a temperature change causes the resistance of all four resistors to
increase by 5%, what is the new output voltage at Vout = V3 − V4 ?
Answer: They all still have the same resistance so the output voltage will still be zero.
d) Starting from the condition where all four resistors have resistance R , a pressure
is applied that causes Ra and Rd to increase by 1% and Rb and Rc to decrease by
1%. What now is measured at Vout = V3 − V4 ?
Answer: We can substitute the values Rd = Ra = 1.01R and Rb = Rc = 0.99R into Eq.
(10) yielding
⎛ 1.01R 0.99R ⎞
Vout = (1.0 V) ⎜
−
= 0.01 V .
2R ⎟⎠
⎝ 2R
33
Problem 17: Battery Life
AAA, AA, … D batteries have an open circuit voltage (emf) of 1.5 V. The difference
between different sizes is in their lifetime (total energy storage). A typical AAA battery
can deliver a total charge of about 0.5 A ⋅ hr while a typical D battery can deliver a total
charge of about 10 A ⋅ hr . (Note the dimensions are the product of current and time,
which is equal to charge.) Of course these numbers depend on how quickly you
discharge them and on the manufacturer, but these numbers are roughly correct. One
important difference between batteries is their internal resistance – alkaline (now the
standard) D cells are about 0.1 Ω .
Suppose that you have a multi-speed winch that is 50% efficient ( 50% of energy used
does useful work) run off a D cell, and that you are trying to lift a mass of 60 kg . The
winch acts as load with a variable resistance RL that is speed dependent.
a) Suppose the winch is set to super-slow speed. Then the load (winch motor)
resistance is much greater than the battery’s internal resistance and you can
assume that there is no loss of energy to internal resistance. How high can the
winch lift the mass before discharging the battery?
Answer: This is just a question of energy. The energy stored in a D battery is
U tot = Q totV = (10 A ⋅ hr)(1.5 V) = (10 A ⋅ hr)(3600 s ⋅ hr −1 )(1.5 V) = 54 kJ .
When the battery lifts the object, the available chemical potential energy of the battery
U available = U tot / 2 is converted into gravitational potential energy mgh . So the maximum
height the object can reach
h=
U available
54kJ ⋅ 12
=
= 46 m .
mg
(60 kg )(9.8 m s)
The factor of a half is there because the winch is only 50% efficient.
b) To what load resistance RL should the winch be set in order to have the battery lift
the mass at the fastest rate? What is this fastest rate (m/sec)? HINT: You want to
maximize the power delivery to the winch (power dissipated by RL ).
Answer:
First we need to determine how to maximize power delivery. Consider a battery with
potential difference V and internal resistance ri that is connected to the winch that has a
load has resistance RL . The power dissipated in the load is:
34
2
⎛ V ⎞
R
P = I R = ⎜ 0 ⎟ R = V02
.
2
⎝ R + ri ⎠
R
+
r
( )
2
i
We want to maximize this by varying RL :
(
)
−2
−2
−3
dP d
=
V02 R ( R + ri ) = V02 ⎡( R + ri ) − 2R ( R + ri ) ⎤ = 0 .
⎢
⎥⎦
⎣
dR dR
(
)
Multiply both sides by V0−2 ( R + ri ) yielding R + ri − 2R = 0.
3
So, to get the fastest rate of lift (most power dissipation in the winch) we need to set the
winch resistance equal to the battery internal resistance, R = ri = 0.1 Ω . Using this we can
get the lift rate from the power:
50%eff
2
 d
⎛ V0 ⎞
V02
V02
dh
P = I RL = ⎜
⎟ RL = 4r = 2 dt ( mgh ) ⇒ v = dt = 8r mg .
⎝ RL + ri ⎠
i
i
2
Thus we find a list rate of v = 4.8 mm s
c) At this fastest lift rate how high can the winch lift the mass before discharging the
battery?
Answer:
This is just part a over again, except now we waste half the energy in the internal resistor,
so the winch will only rise half as high, to 23 m
d)
Compare the cost of powering a desk light with D cells as opposed to plugging it
into the wall. Does it make sense to use rechargeable batteries? The current price (Jan
2014) for residential electricity in Massachusetts is $0.1638 / kW ⋅ hr .
Answer:
A D cell can deliver a total of 10 A-hr, corresponding to a total energy storage of
U tot = Q totV = (10 A ⋅ hr)(1.5 V) = 15 W ⋅ hr = .015 kW ⋅ hr . We could convert that to about
54 kJ but kW-hour are a useful unit to use because electricity is typically charged by the
kW-hour so this will make comparison easier. A D battery costs about $1 (you can pay
more, but why?) So D batteries cost about $1/ .015 kW ⋅ hr = $67 / kW ⋅ hr $1/0.015 kwh
or $70/kwh. So the battery is nearly three orders of magnitude more expensive. It
definitely makes sense to use rechargeable batteries – even though the upfront cost is
slightly more expensive you will get it back in a couple recharges. As for your desk
35
light, or anything that can run on batteries or wall power, plug it in. If it is 60 Watts, for
every hour you pay about only $0.01 = 1cent with wall power but run through $4 in D
batteries.
36