216 Week 6: Moving Charges and Magnetic Force
R
M,Q
ω
Figure 74: A rotating ring of charge with massM, radiusR, and chargeQhas a magnetic moment
of~m=Q/ 2 M(M R^2 ~ω) =μ~L.
The “current” in such a ring can easily be evaluated. The total charge in the ring goes around
exactly one time in one period of its revolution. Thus:
I=
Q
T
=
Qω
2 π
(466)
The magnetic moment of the ring in the (right handed)z-direction is thus just:
mz=IA=
Qω
2 π
πR^2 =
Qω
2
R^2 (467)
If we multiple the expression on the right byMM(one!) and rearrange the terms, we get:
mz=
Q
2 M
(
M R^2 ω
)
=μLz (468)
usingLz=M R^2 ωfor a ring of massMrotating symmetrically about thez-axis.
That was almost too easy! Next consider a rotating disk of total chargeQ, total massM, radius
R. The charge of a differential ring of charge of radiusrand thicknessdris justdq= (2πrdr)σ
whereσ=Q/πR^2 is the surface charge density of the uniformly distributed charge.dI=dq 2 ωπas
before. The area inside the ring isA=πr^2. Thus:
dmz=dIA= 2πrdr
Qπr^2
πR^2
ω
2 π
=
Q
R^2
ωr^3 dr (469)
and integrating from 0 toRwe find:
mz=
∫R
0
Q
R^2
ωr^3 dr=
Q
4 R^2
R^4 ω=
Q
4
R^2 ω (470)
One again we multiply byMM, do some rearrangement, and:
mz=
Q
4 M
M R^2 ω=
Q
2 M
(
1
2
M R^2 ω
)
=μLz (471)
Part of your homework for this week will be to re-prove these two cases and several others to show
that this result is quite general. If you are an advanced student (e.g. a physics major) you may
be asked to prove that this is ageneralresult for any figure with enough symmetry about the axis
of rotation so that~ω||L~. This is simple enough if you require that both the mass and the charge
have densities that areidentical functions of coordinatesand writedmzcorrectly in terms of those
densities.