Finding H inside and outside a current-carrying copper rod

[Saraharris38]

Homework Statement

A long copper rod of radius R has uniformly distributed free current I. Find the value of H inside and outside the rod

Homework Equations

∫Hdl=I(free enclosed)

The Attempt at a Solution

Copper is diamagnetic so the magnetization will be circumferential and opposite of B, producing a downwards bound current inside and an upwards bound current on the surface. We can use ampere's law to calculate H:

∫Hdl=I(free enclosed). The path dl is the amperian loop inside, with s<R, so

H(2πs)= I(free enclosed)

How do I find I(free enclosed)? Isn't it just I?

Thanks! If the explanation could be as explicit as possible, that would be great.

 

[nickjer]

You would be right if the loop you made was outside of the rod. Then the total current enclosed is I. But, if the loop is inside the rod then only a fraction of the current exists in your loop. Knowing that the current is uniform is a hint on how to calculate the fraction of current inside of your loop.