W9_parallel_resonance.eps

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.