Jackson 5.19 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
A magnetically “hard” material is in the shape of a right circular cylinder of length L and radius a. The
cylinder has a permanent magnetization M0, uniform through-out its volume and parallel to its axis.
(a) Determine the magnetic field H and magnetic induction B at all points on the axis of the cylinder,
both inside and outside.
(b) Plot the ratios B/μ0M0 and H/M0 on the axis as functions of z for L/a = 5
SOLUTION:
(a) Let us place the center of the cylinder at the origin and its axis aligned with the z axis. Then it is
natural to use cylindrical coordinates. Let us state everything we know about this problem outright.
There are no free currents present, so there cannot be any curl to the H field.
∇×H=0
This is valid inside and out. The curl of the gradient of any function is always zero, so we can define a
magnetic scalar potential
H =−∇  M
Outside the object, there is no material, so that the magnetization outside must be zero and the total
field B depends only on the H field.
B= 0 H and M = 0 outside
We should also note that there are no applied external fields, only the fields created by the shape. This
means that all fields should die down to zero infinitely away from the shape.
Inside, the magnetization is a known:
M=M 0 z inside
Generally speaking, the equation that says there are no magnetic monopoles can be cast as a
relationship between M and H:
∇⋅B=0

∇⋅ 0 H0 M =0

∇⋅H=−∇⋅M
This is a totally general results, applicable everywhere, no matter what material is present. It means that
whatever divergence that may exist in the magnetization must be perfectly canceled by the divergence
in the H field in order to keep the actual, real total field B divergence-less. We now substitute in the
definition of H in terms of the magnetic scalar potential and recognize an entity that acts like a
magnetic charge density:
∇ 2  M =−M where  M =−∇⋅M
Because the magnetization is constant inside the cylinder, its divergence is zero there. The magnetic
charge density must reside on the surface. Drawing a closed surface around the surface charge density
and shrinking it down leads to M 2−M1 ⋅n=− M . Outside there is no magnetization M2 = 0 and
inside we know M 1=M 0 z so that we have:
 M =M 0 z⋅n
 M =M 0 on the top of the cylinder,  M =−M 0 on the bottom of the cylinder, and  M =0 on the sides
This is best summed up as:
 M =M 0  z−L /2− z L /2 for ρ < a, and ρM = 0 for ρ > a.
The Poisson equation for the magnetic scalar potential has the general solution:
M =
M
1
d x'
∫
4  ∣x−x '∣
M =
2 ∞
M
1 ∞
 ' d ' d ' d z '
∫
∫
∫
4  −∞ 0 0  2 '2−2   ' cos−' z− z '2
M =
2 a
 M 0  z '− L/2−  z 'L /2
1 ∞
 ' d ' d ' d z '
∫
∫
∫
4  −∞ 0 0  2 '2−2   ' cos−' z− z '2
M
M = 0
4
[
2 a
2 a
 ' d  ' d '
'd 'd  '
−∫ ∫ 2
∫∫ 2 2
2
2
2
0 0   ' −2   'cos −' z− L/2
0 0   ' −2  ' cos −' z L/2
The problem only asks for the solution on the axis of the cylinder, so we can safely set ρ = 0.
M =
M0
4
M =
M0
2
[
[∫ 
M0
2
[[  '  z− L/2 ] −[ '  z L/2 ] ]
M =
M =
2 a
2 a
∫ ∫ 2' d  ' d ' 2 −∫ ∫ '2 d  ' d ' 2
0 0  '  z −L/ 2
0 0  '  z L /2
a
0
a
'd '
'd '
−∫
2
2
2
 '  z− L/2 0  '  z L/ 22
2
2
a
0
2
2
]
]
a
0
M0
[a 2  z −L /22−a 2 z L/ 22 −∣z −L /2∣∣z L /2∣]
2
]
H=−∇  M
H=−
∂ M
z
∂z
H=−
M0
2
[
]
z −L/ 2
z L/ 2
z− L/2 z L/2 
− 2
−

z
2
2 ∣z− L/2∣ ∣z L/2∣
a  z −L /2 a  z L /2
2
A careful evaluation of the absolute value in the different regions leads to:
H out =−
H in =−
M0
2
M0
2
[
]
z −L/ 2
z L/ 2
z on axis
− 2
2
a  z −L /2 a  z L /22
2
[
z −L /2
2
a  z−L /2
2
−
z L /2
a  zL /22
2
]
2 z on axis
Using B= 0 H  0 M , M = 0 outside, and M=M 0 z inside
B=−
0 M 0
2
[
]
z−L / 2
zL / 2
z on axis
− 2
2
a  z− L/2 a  z L/22
2
The total field inside and outside end up having the same form
(b) Plot the ratios B/μ0M0 and H/M0 on the axis as functions of z for L/a = 5
For L/a = 5 we have
H out / M 0 =−
H in /M 0=−
1
2
B/ 0 M 0=−
[
[
[
1
2
1
2
]
z / L−1/ 2
z / L1/ 2
−
z on axis
2
1/25 z / L−1/2 1/25 z / L1/22
]
z / L−1 /2
z / L1/2
−
2 z on axis
2
1/25 z / L−1/ 2 1/25 z / L1 /22
]
z / L−1/2
z / L1/2
z on axis
−
2
1/25 z / L−1 /2 1/25 z / L1/ 22
0.6
0.4
H scaled
0.2
0.0
-0.2
-0.4
-0.6
-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0
z/L
1.0
B scaled
0.8
0.6
0.4
0.2
0.0
-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0
z/L