Question

A long, hollow wire with inner radius a and outer radius b carries a uniform current density J. What is the magnetic field as a function of r, the distance from the center of the wire within the wire's material (i.e. what is B(r) in the region a < r < b)?

Answer 1

Answer:

The magnetic field in the region a < r < b is $B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$

Explanation:

If we have the a < r < b. The formula of current is:

$J=\frac{I_{total} }{A}$

Where:

A = area enclosed by the loop.

Itotal = total current in loop.

$J=\frac{I}{\pi b^{2}-\pi a^{2} }$

$I_{enclosed} =JA_{enclosed}$

$I_{enclosed} =\frac{I(\pi r^{2}- \pi a^{2})}{\pi b^{2}-\pi a^{2} }$

If we have the Ampere`s law:

$\int\limits^a_b {B} \, ds =u_{0} I_{enclosed} \\2B\pi r=u_{0} (\frac{I(\pi r^{2}-\pi ^{2} }{\pi ^{2}-\pi a^{2} } )\\B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$