Coaxial cable mag field?

Question

The coaxial cable shown in the figure http://session.masteringphysics.com/problemAsset/1036042/6/RW-26-68.jpg

consists of a solid inner conductor of radius a and a hollow outer conductor of inner radius b and thickness c. The two carry equal but opposite currents I, uniformly distributed.

Find an expression for the magnetic field strength as a function of radial position r within the inner conductor.

Express your answer in terms of the variables I, a, b, c, r, and constant mu_0.

Find an expression for the magnetic field strength as a function of radial position r between the inner and outer conductors.

Express your answer in terms of the variables I, a, b, c, r, and constant \mu_0

Find an expression for the magnetic field strength as a function of radial position r beyond the outer conductor.

Express your answer in terms of the variables I, a, b, c, r, and constant \mu_0

Answer 1

This is all about Ampere's Law which is difficult to type here but it's

integral of B . dL = mu_0I

For the inner conductor, imagine an Amperian loop inside the wire (just a circle inside the wire), at a distance r from the centre.

Integral of B.dL is just B(2pi r) since an infinitesimal amount of this loop is always in the same direction as the B field and by symmetry the B field is constant.

Now I inside this first wire depends on the amount of wire you have included - if your amperian loop encompasses half the wire you have half the current. The amount of current you have included depends on the area of the loop.

ie. your "mu_0 x I" will be mu_0 x I (pir^2)/(pia^2) = mu_0 I (r/a)^2

So rearranging your Ampere's Law expression for B you get

B = (mu_0 I r^2) / (2pi r a^2)

B = mu_0 I r / (2pi a^2)

Now outside the inner conductor your enclosed current is always I, and the left hand side of Ampere's law is just the same as before so this is easy

B(2pir) = mu_0 I

B = mu_0I / (2pir)

For the outer conductor the left hand side is still the same! But now the current enclosed decreases as r increases because the current is flowing in the opposite direction and it is the net current enclosed in your Amperian loop that matters.

The current enclosed is I at r = b, and 0 at r = b+c

So the right hand side is mu_0 I (1 - fraction of outer wire included)

The fraction of the outer wire included is the area of your amperian loop at a distance r, minus the area of the amperian loop up to the point b (so you're just including an anulus) all divided by the area of the cross section up to the edge of c minus the area of the loop up to b

So that's a pretty horrible expression but it's right, so working through this fraction of the area becomes (r^2 - b^2) / ((b+c)^2 - b^2)

So rearranging your expression you get

B = {mu_0 I (1 - [(r^2 - b^2) / ((b+c)^2 - b^2)] } / 2pir

Yup, hideous, but it's nicer if you write it on a piece of paper. And you can investigate it and see that it does match the required boundary conditions.