# Flux density inside and outside a coaxial cable.

1. Jan 15, 2014

### lam58

For question 4b in the problem sheet attached below, is my working correct for B outside and inside the coaxial cable?

Ans:

Outside coaxial:∫B.dl = μI (the integral is between 0 and 2∏r)
=> B = (μI)/(2∏r)

Inside the coaxial: $$\int_{2\pi a}^{2\pi r} B.dI$$
= $$μI.\frac{a^2}{\pi r^{2}b^{2}}$$

=> $$B[2\pi r - 2\pi a] = μI.\frac{a^2}{\pi r^{2}b^{2}}$$
=>$$B = μI.\frac{a^2}{\pi r^{2}b^{2}} * \frac{1}{2\pi r - 2\pi a}$$

#### Attached Files:

• ###### coax.jpg
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2. Jan 19, 2014

### lightgrav

It's not obvious what you mean by "outside the coaxial" ... do you mean "between the conductors"?
Notice that the inner current Ia is leftward, but the outer current Ib is rightward.
So, for r>b (outside the outer mesh), how much total I is piercing the Area?

It looks like your "inside the coaxial" is trying to be for r<a, that is, inside the inner conducting wire, assuming uniform current density. This region was not asked for in the problem (perhaps because the current density is not really uniform, in real situations). The region between a and b is filled with insulator, having zero current flow in it (so the right-hand side is constant non-zero).

Last edited: Jan 19, 2014