Massachusetts Institute of Technology
Department of Physics
Physics 8.07
Fall 2005
Problem Set 8 Solutions
Problem 1: Griffiths Problem 7.20 (p. 315)
a) The field on the axis above the big loop is given by Griffiths Example 5.6:
B = (µ0 Ib2 /2)(b2 + z 2 )−3/2 . The flux through the little loop is πa2 B or
Φ=
µ0 Iπa2 b2
.
2(b2 + z 2 )3/2
b) The field produced by current I flowing around the little loop is given by eq.
(5.87) of Griffiths with dipole moment Iπa2 . Using cylindrical coordinates,
~r = z~ez + s~es so
2
i
h
µ0 Ia2
2
2
~ = µ0 Ia [3(~ez · ~er )~er − ~ez ] =
.
B
(2z
−
s
)~
e
+
3zs~
e
z
s
4r3
4(z 2 + s2 )5/2
The flux through the big loop is now
Z
0
b
Rb
0
Bz 2πs ds. Using the quadrature
s(2z 2 − s2 ) ds
b2
=
,
(z 2 + s2 )5/2
(z 2 + b2 )3/2
we get
Φ=
µ0 Iπa2 b2
.
2(b2 + z 2 )3/2
c) The fact that the two fluxes in parts (a) and (b) are the same means that
the mutual inductances are equal, with
M=
µ0 πa2 b2
.
2(b2 + z 2 )3/2
Problem 2: Griffiths Problem 7.32 (p. 324)
The model of the capacitor given here requires that there be a radial surface current on
the capacitor plates so that charges flow out from the z-axis (the axis of the wire) to
uniformly cover the plate. The exact form of this current is not needed for this problem.
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a) The charge on the capacitor plates is ±It giving a surface charge density
σ = It/(πa2 ) and therefore an electric field between the plates
~ =
E
It
~ez .
0 πa2
~ = B~eφ . From Ampere’s law,
b) The magnetic field is circumferential, B
I
Z
2
~ · d~l = B(2πs) = µ0 0 d E
~ · d~a = µ0 Is
B
dt
a2
⇒ B=
µ0 Is
.
2πa2
c) The geometry here is tricky. It is natural to orient the surface so that d~a
points outward from the cylinder, in which case the loop for Stokes’s Theorem
is oriented along −~eφ by the right-hand rule. So,
I
~ · d~
B
= − B(2πs) = µ0 Ienc ≡ µ0
Z
J~ · d~a .
The
enclosed current has a contribution from the left end of the cylinder,
R
~
J · d~a = −I (with the negative sign because J~ k −d~a). There is also
a contribution from the side of the cylinder where it crosses the capacitor
plate. To calculate this, consider a ring of inner radius s and outer radius
ds on the capacitor plate. Let us denote by IC (s, t) the current flowing on
the capacitor plate through radius s at time t. In time dt, the change in the
amount of charge in this ring is
dQ = [IC (s, t) − IC (s + ds, t)]dt = −
∂IC
dσ
2Is
ds dt =
(2πsds)(dt) = 2 ds dt .
∂s
dt
a
(The first equation is charge conservation, accounting for the flow of current
into and out of the ring; then we equate dQ to the charge that is added
for a uniform surface charge density.) Integrating this and requiring that no
charge flow beyond s = a, so IC = 0 at s = a, we get
s2
IC (s, t) = I 1 − 2
a
!
.
Adding the two contributions to Ienc now gives
−B(2πs) = µ0
"
s2
−I + I 1 − 2
a
which gives exactly the same result as part (b).
2
!#
Problem 3: Griffiths Problem 7.45 (p. 336)
The EMF arises from magnetic forces on charges:
E=
Z
f~ · d~l =
Z
~ ) · d~l .
(~v × B
The velocity arises from the spin of the sphere: ~v = ωa sin θ ~eφ where θ is the polar
~ = B0~ez and the path used for computing the EMF is a
angle. The magnetic field is B
meridian of constant longitude from θ = 0 to θ = π/2, with d~l = adθ d~eθ . Combining
the ingredients,
~ ) · d~l = B0 ωa2 sin θ dθ (~eφ × ~ez ) · ~eθ .
(~v × B
Geometry gives (~eφ × ~ez ) · ~eθ = ~es · ~eθ = cos θ. Thus,
E=
Z
0
π/2
1
B0 ωa2 sin θ cos θ = B0 ωa2 .
2
The EMF tends to push positive charges toward the equator and negative charges toward
~
the pole if ω
~ k B.
Problem 4: Griffiths Problem 7.55 (p. 339)
a) With J~ = J(~x ) independent of time, charge conservation
∂ρ
~ · J~ = f (~x )
= −∇
∂t
where f (~x ) is independent of time. At each spatial position ~x, ρ̇ = f , the
general solution of which is
ρ(~x, t) = ρ(~x, 0) + tf (~x ) = ρ(~x, 0) + tρ̇(~x, 0) .
~
b) Using ∇(1/r)
= −r̂/r2 where ~r ≡ ~x − ~x 0 , r = |~r |, and r̂ = ~r/r, we can write
the Biot-Savart law as
~ x 0)
µ0 ~ Z J(~
~
~ ×A
~ c (~x ) .
B(~x ) =
d3 x0 ≡ ∇
∇×
4π
r
~ c (~x ) is, of course, the Coulomb gauge vector potential defined by
Here, A
Griffiths Eq. (5.63). Now we take the curl and use vector identity (11) from
the inside cover of Griffiths:
~ ×B
~ = −∇2 A
~ c + ∇(
~ ∇
~ ·A
~ c) .
∇
But
~c =
∇2 A
Z
µ0 Z ~ 0 2 1
~ x 0 )δ 3 (~x − ~x 0 ) d3 x0 = −µ0 J(~
~ x) .
J(~x )∇
d3 x0 = −µ0 J(~
4π
r
3
~ ·A
~ c ≡ f (~x ). We rewrite ∇f
~ using
Now let ∇
~ · (∇f
~ ) = ∇2 (∇
~ ·A
~ c) = ∇
~ · (∇2 Ac ) = −µ0 ∇
~ · J~ = µ0 ∂ρ
∇
∂t
where in the last step we used charge conservation. Now we use Gauss’s law,
~ · E = ρ, to find
0 ∇


~
~ · (∇f
~ )=∇
~ · µ0 0 ∂ E  .
∇
∂t
Using the Helmholtz theorem (the decomposition of a vector field, here
~
∂ E/∂t,
into longitudinal and transverse parts) this implies
~
~ = µ 0 0 ∂ E + ∇
~ ×C
~
∇f
∂t
~ is some transverse vector field obeying
where C
~ ×E
~).
~ ×∇
~ ×C
~ = −µ0 0 ∂ (∇
∇
∂t
~ ×E
~ = −∂ B/∂t.
~
~ =
From Faraday’s law, ∇
In the current application, B
~ ×A
~ c is independent of time because the current is independent of time.
∇
~ ×E
~ = 0 implying ∇
~ ×C
~ = 0. Combining the partial results
Therefore ∇
from above gives
~
~ ×B
~ = µ0 J~ + ∇f
~ = µ0 J~ + µ0 0 ∂ E ,
∇
∂t
which is Ampère’s law with displacement current.
Problem 5: Griffiths Problem 8.2 (p. 349)
a) Taking the cylinder to lie along the z-axis, using E = σ/0 for a capacitor
~ we get
with surface charge density σ, and using Ampère’s law for B,
~ =
E
It
~ = µ0 Is ~eφ .
~ez , B
2
0 πa
2πa2
b) From the results of part (a),
uEM
1
=
20
It
πa2
2
µ0
+
2
Is
2πa2
2
~=−
, S
I 2 st
~es .
20 (πa2 )2
These imply
I 2t
I 2t
∂uEM ~ ~
+∇·S =
−
=0.
∂t
0 (πa2 )2 0 (πa2 )2
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c) The total energy in the gap is
UEM = w
Z
a
uEM (s) 2πsds =
0
I 2 wt2
µ0 I 2 w
+
.
20 πa2
16π
The total power flowing into the gap is
P =
Z
2
~ · d~a = −2πaw~es · S(s
~ = a) = I wt .
S
0 πa2
As expected, dUEM /dt = P .
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