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

(a)Consider the case r > a. The amperian loop, labelled 2, is a circle cocentric with the cross-section. For this loop,
L=2r
`I_(e)` = current enclosed by the loop = I
This result is the familiar exprssion for a long straight wire `B(2pi r)=mu_(0)I`
`B=mu_(0)I/2pi r`
`B alpha 1/r`(r > a)
(b) Consider the case r < a. The amperian loop is circle labelled 1. For this loop taking the radius of the circle to be r
`L= 2pi r`
Now the current enclosed `I_(e)` is not I, but is less than this value. Since the current distribution is uniform, the current enclosed is
`I_(e)=I(pi r^(2)/pi a^(2))=Ir^(2)/a^(2)`
Using Ampere.s law, `B(2 pi r)=mu_(0)Ir^(2)/a^(2)`
`B=(mu_(0)I/2 pi a^(2))r`<>br>`B alpha r`(r < a)
Figure show a plot of the magnitude of B with distance r from thạentre of the wure The direction of the field is tangential tothe respective circular loop (1 or 2) and given by the right hand rule described earlier in this section