1
Strong Little Magnets
Appendix (to “Strong Little Magnets” by Michael J. Moloney, September 2007)
Magnetic field near the surface of a cylindrical disc magnet.
Sheet current densities. One may calculate the magnetic
field of a uniformly magnetized cylindrical disc
magnet based on a surface sheet current density1
K = M x n̂ , where n̂ is the surface normal unit vector.
M
K
K is the sheet current density circumferential to the disc,
in A/m, so the cylindrical disc magnet is equivalent to
a solenoid of current I and n turns per unit length
with K = nI = M. The magnetic field on the axis of a
solenoid of length L and radius R is [Ref. 2, Sect. 8-4]
Bp =
μon I
(cos 2 - cos 1).
2
(A1)

p
1
2
z
L
xxxxxxxxxxxxxxxxxxxxxxxxxxxx
(If the solenoid were to grow in both directions until point p was inside an infinite solenoid, then
cos 1  -1, cos 2  +1, and Bp  o nI = o M as it should.)
The expression for Bp in terms of z, L and the solenoid radius R is
Bp =
μoM
zL
[
2
R 2  ( z  L) 2
z
R2  z2
],
which was used to calculate the disc magnet values in Table 1, with z = 0.50 mm and L = 3 mm.
(A2)
2
Magnetic surface pole densities. Since there are no free currents present in the cylindrical discs, one may
employ a scalar magnetic potential2,3 to calculate the magnetic field anywhere outside the discs, instead
of using the sheet current density K. (Magnetic surface pole densities give incorrect magnetic fields
inside the discs, however.) The surface magnetic pole density m for a solid of uniform magnetization M
is given by
m = M n̂ ,
(A3)
where n̂ is the outward surface normal unit vector. This corresponds to the electrical surface charge
density e in C/m2. The cylindrical disc magnets with magnetization M parallel to the axis (as shown
above) may be regarded as two flat, uniformly charged discs, one positive and one negative.
For a single disc of radius R and uniform electric charge density +e as shown in the sketch below, the
electric field on the disc axis, a distance z above the disc is readily found to be
p
z
++++++++++++++
z
Ep = 2  ke e [1 -
Ep = ½ e/o [1 -
],
z  R2
2
z
z2  R 2
(A4)
].
(A5)
(When R  infinity, E  e/(2o), the familiar E field on either side of an infinite sheet of charge.)
A magnetized cylindrical disc is then regarded as having two uniformly 'charged' circular discs of
opposite polarities. To convert Eq. (A5) to that of a magnetic field, we substitute km = o/(4) for ke =
1/(4o) and m for e. This gives the magnetic field B on the axis of a disc of radius R and length L as
z
B = 2  km m { [1 -
z R
2
] - [1 -
2
zL
(z  L) 2  R 2
] },
(A6)
and then rearranging and employing m = M we have
B = ½ o M {
zL
(z  L)  R
a result identical to Eq. (A2).
2
2
-
z
z  R2
2
},
(A7)
3
The force between a cylindrical magnetized disc and a flat, highly magnetically permeable sheet.
Electric image charges are often employed where the boundary of an electrical conductor has sufficient
symmetry. In a corresponding manner, image magnetic current surface densities K and image magnetic
pole surface densities m may be used where there is high symmetry and a high magnetic permeability4.
M
Method 1 - magnetic pole surface densities.
Using magnetic pole surface densities m =  M, and
image pole densities inside the steel, one must calculate
the total force between the disc and its image. The
sketch at the right shows a side view of the disc and
its image inside the steel.
1 ++++++++++++++++++
disc magnet
2 -----------------------------steel
3 +++++++++++++++++
image disc
4 ------------------------------
The magnet itself is represented by circular sheets 1 and 2,
and the image by sheets 3 and 4. Sheets 2 and 3 are separated by 0.84 mm, due to the 0.42 mm thickness
of the plastic, while discs 1 and 2 are separated by the 3 mm disc thickness, as are discs 3 and 4.
The net force between the magnet and its image is the sum of attractive force between 2 and 3 and
between 1 and 4, minus the repulsive forces between 1 and 3, and between 2 and 4.
y
The figure at the right shows a top view of
any two discs, and the right-handed coordinate
system employed, with the z axis coming out of
the page.

r

w
x
p
Point p lies on the 'upper' disc while the
integration is carried out over the 'lower'
disc to obtain the z-component of the field at p.
Point p is a distance w from the disc axis. The source
point is a distance  from the disc axis, and makes an
angle  to the x-axis.
r is the vector from the source point on the lower disc to the field point p on the upper disc. The distance
between disc centers is z, so r can be written
r = î (w -  cos ) - ĵ ( sin ) + k̂ z .
(A8)
4
The integral is done exactly as it would be for the z-component of the electric field with electric charge
density e the disc which creates the field:
R
Ez(p) = ke e
 ρ dρ
0
2

dθ
0
z
.
(z  ρ  w  2 w ρ cosθ ) 3/2
2
2
2
(A9)
It is desirable to go to dimensionless variables  = /R,  = w/R and  = z/R. Then one writes the
integral in magnetic notation as
μ Mγ
Bz(p) = o
4π
1
2
 α dα 
0
0
dθ
1
.
(γ  α  β  2αβ cos θ) 3/2
2
2
2
(A10)
Maple 10 will do the theta integral in terms of an elliptic E function, felliptic(, , ), but one must be
sure to declare all variables to be real ( assume(theta>=0), assume(alpha>=0), etc.). This leaves an 
integral which, if values of  and  are supplied, Maple can numerically integrate and evaluate.
Then to find the force between discs when =z/R is given, one must integrate Bz(p) over :
1
Fz = m  Bz(p) 2 w dw = M R2
B
z
(p) 2 π β dβ
(A11)
0
Fz =
1
μoM2R 2 1
γ  βdβ  αdα f elliptic(α, β , γ) .
2
0
0
(A12)
For a given , the outer integral in  was done with a 40-step trapezoidal method, at each step in the loop
having Maple numerically execute the  integral. Three different  values were needed because of the
distances between 2 and 3, 1 and 4, and between 1 and 3 (the same as between 2 and 4). Liftoff forces
for discs of 6-mm and 9-mm radius were within 0.2 % of the results to be discussed in the next section.
Method 2 - Attraction between shallow solenoids.
The second way to calculate the force between the disc magnet and its image is to treat each as a
circumferential current surface density K = M x n̂ . One employs the same coordinate system as shown
above, with the x-direction lying in the plane of the steel sheet, and integrates over the image sheet
current (lower disc in the sketch, bottom of next page) to find the x-component of the B field at an
arbitrary point p on the disc sheet current (upper disc), a distance z' above the the strip of current we are
integrating.
5
y
dl
In the Biot-Savart law dB = o/(4) I dl x r/r3
K
r
dl = R d ( - î sin  + ĵ cos ) .

p
The field point p is a distance s above the bottom
of the image magnet, (diagram, bottom of this page) and
a distance z' from the strip of image magnet being
integrated.
x
K
K
r is the distance between the source and field points, given by
r = î R (1 - cos ) - ĵ R sin  + k̂ z'.
We wish to calculate the x-component of the magnetic field at p, because this will exert an attractive
force on the (CCW) sheet current density K in the disc magnet. The element of Bx at p due to
dl = Rd of the incremental current I is
dδBx (p) 
μo  I
z' cosθ
R d
2
4π (z' 2R 2 (1  cos θ )) 3/2
(A13)
disc magnet
A side view of the calculation is shown
at the right. The lower disc is the image,
inside the steel plate, and the upper disc
is the magnet itself, above the plate.
We begin by calculating Bx at point
p from a strip of the image disc a
vertical distance z' from p.
p
h
z' = s-z
s
L
z
image magnet
The thickness of the strip is dz, which carries a current I = I/z dz = M dz. We must integrate z from
0 to L to obtain Bx at p, a distance s from the bottom of the image disc.
6
L
μ o R 2π
z' Mdz
(A14)
cosθ
dθ
2
2
3/2


4π 0
0 (z' 2R (1  cos θ))
Now we change the integration variable to z' (z' = s – z, and d z' = -dz), integrating z' from z1' = s-L to z2'
= s. (Later, we will integrate s over the top disc magnet from h +L to h + 2L to get the force.)
Bx(s) =
μ MR
d Bx(s) = o
4π
2π
s-L
 cosθ dθ  (z'
0
2
s
- z' dz'
2R (1  cos θ)) 3/2
(A15)
2
Carrying out the z' integral results in
μ M
Bx(s) = o
4π
2π
 cosθ dθ {
0
R
(s - L)  2R (1  cos θ )
2
2
-
R
s  2R 2 (1  cos θ )
2
}.
(A16)
This is Bx at p due to the image disc. The integral each term of Eq. (A16) over theta in Maple gives a
closed form in elliptic functions, Gelliptic(s) . Then Eq. (A16) becomes
Bx(s) =
μoM
{ Gelliptic(s-L) - Gelliptic(s) }
4π
(A17)
A strip of height ds on the upper disc magnet has a current dI = Mds. The downward force on this strip,
of circumference 2R is
dFdown = M ds Bx(s) (2R)
(A18)
Finally, we integrate numerically to obtain the total force between the disc and its image, or the force
between one magnet and the steel plate.
Fdown =
μoM2R 2
2
h  2L
 ds/R ( G
elliptic
(s  L)  G elliptic(s))
(A19)
hL
The liftoff force from Eq. (A19) is the essentially the same as that from the pole-density method of the
previous section. (For a 20-step pole-density integration of Eq. (A12), the agreement was around 0.5%,
and for a 40-step integration, the agreement was better than 0.2 %.)
The calculated liftoff force was 10.9 N for the 6-mm diameter disc magnets, and 18.2 N for the larger 9mm diameter discs.
Note. To do the theta integral in Eq (A16), it is best to ask Maple to integrate the general form
2π

o
cosθ dθ
ζ 2  2(1  cosθ )
(A20)
7
which it will readily do in terms of elliptic integrals. Then in Eq. (A19), one enters specific numerical
values for the limits:  = s/R goes from h/R to (h+L)/R for the first term, and from (h+L)/R to (h+2L)/R
for the second.
References
1. D.J. Griffiths, Introduction to Electrodynamics, 3rd Edition, (Prentice-Hall, N. J., 1999), Section
6.2.1.
2. J.R Reitz and F.J. Milford, Foundations of Electromagnetic Theory, 2nd Ed. (Addison-Wesley, New
York, 1967), Section 10-3.
3. A. Sommerfeld, Electrodynamics, (Academic Press, New York, 1964), Section 12
4. Nathan Newbury et al, Princeton Problems in Physics with Solutions (Princeton University Press,
Princeton, NJ, 1991), Problem 2.6.