# Physics 70006 Answers to Final Exam May 8, 2006 1. A charge Q is

```Physics 70006
May 8, 2006
1. A charge Q is distributed uniformly along the z axis from z = −a to z = a.
Write out the ﬁrst few terms in the Legendre polynomials expansion of
the potential for the case r &gt; a.
Answer: The potential at a point z &gt; a on the axis may be written
z+a
1 Q a dz 1 Q
Φ(z) =
ln
=
.
4π0 2a −a z − z 4π0 2a
z−a
Expanding in powers of a/z, we ﬁnd
a 1 a 2 1 a 3
1 Q
−
+
+ &middot;&middot;&middot;
Φ(x) =
4π0 2a
z
2 z
3 z
2
a 1 a
1 a 3
− − −
−
+ &middot;&middot;&middot;
z
2 z
3 z
1 Q a 1 a 3 1 a 5
=
+
+
+ &middot;&middot;&middot;
4π0 a z
3 z
5 z
Replacing 1/z (l+1) by 1/r(l+1) Pl (cos θ), we obtain
1 1 a2
Q
1 a4
Φ(x) =
+
P2 (cos θ) +
P4 (cos θ) + &middot; &middot; &middot;
4π0 r
3 r3
5 r5
Answer: For r a, the above potential reduces to the potential of a point
charge at the origin.
Q 1
Φ(x) →
.
4π0 r
2. A point dipole p is imbedded at the center of a dielectric sphere (radius
R and dielectric constant r ). Find the potential inside and outside of the
sphere.
Answer: Set up a boundary value problem: Assume that the potential has
the form
1 p
P1 (cos θ) + ArP1 (cos θ)
Φin =
4π0 r2
B
Φout = 2 P1 (cos θ),
r
where A and B are expansion constants. Boundary conditions at r = R
B
p 1
= AR +
2
R
4π0 R2
2B
2p 1
= r −A +
.
R3
4π0 R3
1
Solving, we ﬁnd
2(r − 1) 1 p
(r + 2) R3 4π0
p
3r
B=
(r + 2) 4π0
A=
By what factor is the dipole moment p enhanced by the presence of the
dielectric sphere?
pout =
3r
p
(r + 2)
3. A long cylinder of radius R has magnetization vector M = kρ3 ẑ, where
k is a constant and ρ is the radius in cylindrical coordinates.
(a) Ignoring end eﬀects, determine the bound current densities Jb =
∇ &times; M and Kb = M &times; n.
Jb = ∇ &times; M = −
dM
φ̂ = −3kρ2 φ̂
dρ
Kb = M &times; n = kR3 φ̂
(b) From the bound currents determine B inside and outside the cylinder
using Ampère’s law.
Answer: For ρ &lt; R choose a rectangular loop; one side of the loop
of length l coincides with the axis, an opposite side is parallel to the
axis at a distance ρ away. These two sides are connected by sides
perpendicular to the axis. The total bound current through this loop
is
ρ
Ib = l 3kρ2 dρ = lkρ3
0
This current ﬂows in the −φ̂ direction. From Ampère’s law, taking
into account the sense of the current, one ﬁnds that B is in the +z
direction and that
Bz (ρ) = μ0 kρ3 = μ0 M (ρ),
ρ &lt; a.
If the above loop encloses the the surface of the cylinder, then the
total enclosed current will be
R
3kρ2 dρ − lkR3 = 0.
Ib = l
0
and we will have
Bz (ρ) = 0,
2
ρ &lt; a.
(c) Determine H inside and outside the cylinder.
Answer: Outside H = B/μ0 = 0. Inside H = B/μ0 − M = 0.
Therefore,
H = 0, everywhere.
4. An inﬁnite straight wire carries a current
0 t≤0
I(t) =
I0 t &gt; 0
(a) Show that the (retarded) vector potential A is in the z direction and
that
Az (ρ, t) = 0
=
μ0 I0
ln
2π
ct +
c2 t2 − ρ2
ρ
ct &lt; ρ
ct ≥ ρ.
Answer: For ct &lt; ρ no signal
reaches ρ. For ct &gt; ρ, only that segment
of the wire with |z| ≤ c2 t2 − ρ2 contributes to the potential at ρ.
For ct &gt; ρ, we have
√c2 t2 −ρ2
dz
μ0 I0
Az =
4π −√c2 t2 −ρ2 z 2 + ρ2
c2 t2 − ρ2 + ct
μ0 I0
ln
=
4π
− c2 t2 − ρ2 + ct
μ0 I0
ct + c2 t2 − ρ2
=
ln
2π
ρ
(b) Find B(ρ, t) and E(ρ, t). Show that your ﬁelds have the correct static
limit as t → ∞.
Answer: E has only a z component since
E=−
We ﬁnd
μ0 I0
Ez =
2π
1
∂A
.
∂t
ct + c2 t2 − ρ2
μ0 I0
c
=
.
2
2π
c t2 − ρ2
As t → ∞, Ez → 0, as expected.
3
c+ c2 t
c2 t2 − ρ2
B has only a φ component and
∂Az
∂φ
1
ρ
μ0 I0 1
+
=
2π ρ
ct + c2 t2 − ρ2
c2 t2 − ρ2
Bφ = −
μ0 I0 1
ct
2π ρ c2 t2 − ρ2
μ0 I0 1
as t → ∞.
→
2π ρ
=
5. A transmission line consists of two long conducting ribbons of width w,
parallel to one another and separated by distance h. A uniformly distributed current I runs up one conductor and back down the other as
illustrated in the ﬁgure.
(a) Determine the direction and magnitude of the B ﬁeld between the
ribbons.
Answer: First note that the magnitude of the surface current density
is K = I/w. Using the coordinates shown in the ﬁgure, the B ﬁeld
between the ribbons is in the −z direction:
B = −μ0 K ẑ = −μ0
I
ẑ.
w
(b) Determine the magnetic energy stored in a section of length l.
1
μ0 hI 2
Wm =
l
d3 rB &middot; H =
2
2w
(c) Find the self inductance per unit length of the transmission line.
Answer: Use Wm = (1/2)LI 2 to ﬁnd
L/l = μ0
h
.
w
(d) Assuming that one end of the transmission line is connected to a
power supply with terminal voltage V and the other end is terminated
by a resistor R, determine the direction and magnitude of Poynting
vector. Express the magnitude in terms of V and R.
Answer: Assuming the upper ribbon is at potential V , the electric
ﬁeld is downward and has value
E=−
V
ŷ
h
The Poynting vector is
S =E&times;H =
4
V I
V2
x̂ =
x̂,
hw
RA
where A = wh is the area between the ribbons normal to the direction
of energy ﬂow (x̂).
(e) Determine the direction and magnitude force/length exerted on the
upper ribbon by the E ﬁeld and by the B ﬁeld.
Answer: The contribution to the force from the E ﬁeld is
1 2
0 V 2
2
da,
dFy = 0 Ey − E (−1) da = −
2
2 h2
thus
0 V 2 w
2h2
The contribution from the B ﬁeld is
1
1
μ0 I 2
da,
dFy =
0 − B 2 (−1) da = +
μ0
2
2 w2
Fye /l = −
and
Fym /l =
5
μ0 I 2
.
2w
```