Figure shows a long straight wire of a circular cross - section (radius a) carrying steady current I. The current I is uniformly distributed across this cross - section . Calculate the magnetic field in the region `r lt a ` and `r gt a `

Explanation

I When `r lt a ` (i.e inside ), `B=((mu_0i)/(2pia^2))r`
[For part I shown in fig, the current enclosed , i.e , `=(i xx pi r^2)/(pi a^2)`. Applying Ampere.s theorem, `B.2pi r = mu_0 (ir^2)/(a^2) therefore B = ((mu_0i)/(2pia^2))r]` when `r gt a` , `B xx 2 pi r = mu_0 i therefore B = (mu_0 i)/(2 pi r)`