W9_parallel_resonance.eps

Week 7: Sources of the Magnetic Field 239

dl

x

y

z

z r

dB

a

φ

φ

dBz

θ

dθ

dB

ω

Q

Figure 83: The geometry and coordinates used to compute the magnetic field on the axis of symmetry
of a circular ring of chargeQrevolving at angular velocityω.

figure above) is in thexy-plane concentric with thez-axis. Our job is to find the field on thez-axis
once again.

If this figure reminds you of the one in the last section, it should – they are thesame. In fact,
the solution is going to be the same, except that we have to figure out thecurrentin the case where
we have a revolving ring of charge instead of an actual current in a wire. To do so, we note that
all of the charge in the ring moves past an arbitrary point on the circle of its motion – say, where
it crosses thex-axis – in one period of its revolution. The total charge per unit time passing that
point is thus:

I=

Q

T

=

Qω

2 π

(529)

The field is thus obtained by doing the exact same integrals as before:

B~=km I^2 πa

2

(a^2 +z^2 )^3 /^2

ˆz=km

Qωa^2

(a^2 +z^2 )^3 /^2

ˆz (530)

The main reason to do this here (since it no doubt seems trivial) is thatit is a key step along the
way to finding the magnetic field of a rotatingdiskof chargeQand radiusRon your homework.
On your homework problem you will want to draw the disk of charge and select out a thin ring of
that charge of radiusrand thicknessdr. The field of this rotating ring will depend onr(which is
the same asain this example, the radius of the ring, not the distance from the ringto the point
z). With a bit of care, you can integrateBzoverrto find the total field on thezaxis. Then you
can investigate thez≫Rlimit and (with luck and the use of expressions you derived formof a
rotating disk in the last chapter) show that it is stillkm 2 ~m/z^3 in this much more complicated case.

We aren’t quite “done” with Biot-Savart. There are a few problems Icould reasonably give you
and expect you to be able to at least formulate them as integrals and– with a bit of skill – integrate
to find the total field. Some of them are on your homework, but youcan imagine others – the field
on the axis of a rotating rod of charge. The field on the axis of a solenoid. The field of a rotating
sphere, or spherical shell of charge. Some of these may seem daunting, but in all cases with a bit of
work you could get them.

However, learning to do these increasingly difficultintegralswon’t teach you thephysicsany
better. To get a better grip on the physics, we have to leave the Biot-Savart Law behind, or better
yet, convert it into a more general equation the same way that we converted the field of a point
charge into Gauss’s Law for Electrostatics. In the next section wewill do just that – we will turn
the Biot-Savart Law into our next Maxwell equation,Ampere’s Law.