# Magnetic flux outside cylindrical conductor

1. Jan 25, 2006

### mmh37

Hey everyone,

I have failed to show that the magnetic flux outside a cylindrical conductor is zero.

The problem goes like this:

a coaxial cable consists of a solid inner cylindrical conductor of radius a and an outer cylindrical conductor of inner and outer radius b and c. Distributed currents of equal magnitude I flow in opposite directions in the two conductors. Derive expressions for the magnetic flux density B(r) for each of these regions

1) 0<r <a

by amperes law I showed: $$B = \frac {K *l*r} {2*pi*a^2}$$

where K is the permeability of free space, don't know how to write that with latex

2) a<r<b

again by amperes law: $$B = \frac {K *l} {2*pi*r}$$

3) b<r<c

again by amperes law: $$B = \frac {K *l*(r^2-b^2)} {2*pi*r*(c^2-b^2)}$$

4) c<r

I got stuck here, surely it must be zero....but how can that be shown?

thanks for your help - it's very much appreciated

2. Jan 25, 2006

### Staff: Mentor

With the coaxial cable coming out of the paper at you, draw a circular closed path around the outside of it.

INTEG[B dot dL] around that closed path = mu * I(enclosed)

The total I enclosed by the circle is zero, since the current up the inner conductor is the same as the current down the outer conductor, so B (or H) has to be identically zero everywhere outside the coax.

Edit -- BTW, check your answer for (3). When r=c, you should get zero for H.

Last edited: Jan 25, 2006
3. Jan 26, 2006

### mmh37

Thanks Berkeman - that's explained it and it's very much appreciated!! :-) As for the third one it surely has to be (c^2-r^2)/(c^2-b^2)