Question

A long straight wire of a circular cross-section of radius 'a' carries a steady current 'I'. The current is uniformly distributed across the cross-section. Apply Ampere's circuital law to calculate the magnetic field at a point 'r' in the region for (i) r < a and (ii) r > a.

Answer 1

Dear Student ,
(i) Consider the case r<a . The Amperian loop is a circle .
for this loop , taking the radius of the circle to be r
so L = 2$\mathrm{\pi }$r
Now the current enclosed I_e is not I , but less than this value .
Since the distribution is uniform ,
the current enclosed is
$$\begin{gathered} I_e=I\left(\frac{{\mathrm{\pi r}}^2}{{\mathrm{\pi a}}^2}\right) \\ {I}_e=\frac{I{r}^2}{a^2} \end{gathered}$$
further you can solve it .
Regards