PHY 5346
Homework Set 9 Solutions – Kimel
3. 5.19 The system is described by
The effective volume magnetic charge density is zero, since M⃗ is constant within the cylinder. The
effective surface charge density (n̂ ⋅ M⃗ from Eq. (5.99)) is M0, on the top surface and −M0 on the
bottom surface. From the bottom surface the potential is (for z > 0
Φb =
1
4π
−M0 2π ∫
0
a
ρdρ
ρ2 + z2 1/2
= − M0
2
a2 + z2 − z
By symmetry, the potential from the top surface is (on the inside)
Φ t =
M0
2
a2 + L − z2 − L − z
The total magnetic potential is
Φ = Φb + Φ t = −
M0
2
a2 + z2 − z + M0
2
a2 + L − z2 − L − z
So, on the inside of the cylinder,
Hz = − ∂∂z
− M0
2
a2 + z2 − z + M0
2
a2 + L − z2 − L − z
Hz = −
M0
2
2 −
z
a2 + z2
−
L − z
a2 + L − z2
while above the cylinder,
Hz = −
M0
2
−
z
a2 + z2
+
z − L
a2 + L − z2
with a similar expression below the cylinder.
B⃗ = μ0 H⃗ + M⃗
Thus inside the cylinder,
Bz = μ0 −
M0
2
2 −
z
a2 + z2
−
L − z
a2 + L − z2
+ M0
Bz =
μ0M0
2
z
a2 + z2
+
L − z
a2 + L − z2
while above the cylinder,
Bz =
μ0M0
2
z
a2 + z2
−
z − L
a2 + L − z2
First we plot Bz in units of a for L = 5a
gz =
1
2
z
1+z2
+
5−z
1+5−z2
if z < 5
1
2
z
1+z2
−
z−5
1+5−z2
if 5 < z
gz
0
0.2
0.4
0.6
0.8
1
2
4
6
8
10
z
And similarly, Hz in units of a for L = 5a.
fz =
− 12 2 −
z
1+z2
−
5−z
1+5−z2
if z < 5
− 12 −
z
1+z2
+
z−5
1+5−z2
if 5 < z
fz
-0.4
-0.2
0
0.2
0.4
2
4
6
8
10
z