W9_parallel_resonance.eps

238 Week 7: Sources of the Magnetic Field

However, one day you might need toshowthat the⊥field vanishes thehard wayby actually
integrating it. Fortunately, this really isn’t that hard. If you look carefully at the picture, you can
see that:

dBz = km

Idl a

r^3

(520)

dBx = km

Idl z

r^3 cos(θ) (521)

dBy = km

Idl z

r^3

sin(θ) (522)

The only thing remaining is a variable we can integrate over. Hopefully itis obvious that integrating
overxandyis a really bad choice, while integrating overθis a good one. We note thatdl=adθ,
substitute this in, and we are ready to go:

Bz = km

∫ 2 π

0

Ia^2 dθ

r^3

= km

I 2 πa^2

r^3

(523)

Bx = km

∫ 2 π

0

I az

r^3

cos(θ)dθ= 0 (524)

By = km

∫ 2 π

0

I az

r^3

sin(θ)dθ= 0 (525)

We conclude that:

B~=km I^2 πa

2

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

ˆz (526)

It is instructive to write this in terms of themagnetic momentof the loop,~m=Iπa^2 zˆ:

B~=^2 km~m

(a^2 +z^2 )^3 /^2 (527)

which isexactly the same formas that of the electric field on the axis of an electric dipole,E~=
2 ke~p/(a^2 +z^2 )^3 /^2 , that we derived several weeks ago, with the substitution ofkmforkeand~pfor
~m. This (hopefully) continues to motivate the idea that electric and magnetic fields have certain
characteristic shapes– those of monopoles, dipoles, quadrupoles, and so on – and that if we ever
learn to evaluate theirmultipolar moments for arbitrary charge-current distributions, we will be
able to easily reconstruct at least a good approximation to the total electromagnetic field of those
distributions.

In that spirit, we can easily find the form of the field whenz≫a:

B~=^2 km~m

z^3

(528)

where we used the binomial expansion, sort of – we only had to keep the leading term after factoring
out thezso it was pretty easy.

Evaluating the magnetic field using the Biot-Savart Law becomes increasingly difficult from here
on. At the very least, it becomes an exercise in increasingly difficultcalculus, even though the
physicalconceptis the same and you can always write down an integral that – if you could do it –
would lead you to the answer. There is one more worth at least laying out to help get you set up
for your homework.

Example 7.4.3: Field of a Revolving Ring of Charge on its Axis

In the figure above, a circular ring of charge with chargeQ, radiusa, and angular velocityω(which
really points in the vectorz-direction, recall – I’m just indicating the direction of rotation in the