Figure shows a long straight wire of a circular cross-section (radius a) carrying steady current l. The current l 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 gt a`. The amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop
`L = 2pi r`
`l_(e)`= current enclosed by the loop = l
This result is the familiar expression for a long straight wire `B(2pi r) = mu_(0)l`
`B = (mu_(0) l)/(2pi r)`
`B prop (1)/(r) " " (r gt a)`
(b) Consider the case `r gt 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 `l_(e)` is not l, but is less than this value. Since the current distribution is uniform, the current enclosed is
`l_(e) = l ((pi r^(2))/(pi a^(2))) = (lr^(2))/(a^(2))`
Using ampere's law, `B (2pi r) = mu_(0) (lr^(2))/(a^(2))`
`B = ((mu_(0)l)/(2pi a^(2))) r`
`B prop r " " (r gt a)`

Figure show a plot of the magnitude of B with distance r from the centre of the wire. Th edirection of the field is tangential to the respective circular loop (1 or 2) and given by the right hand rule described earlier in their secion
This example possesses the reqired symmetry so that Ampere's law can be applied readily