Figure shows a long straight wire of circular crosssection (radius a) carrying steady current I. The current I is uniformly distributed across this crosssection. 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 = 2 pir`
`I_(e) = `Current enclosed by the loop = I The result is the familiar expression for a long straight wire
`B (2pir) = mu_(0)I`
`B = (mu_(0)I)/(2pir)`
`B prop (1)/(r) [r gt alpha]`
(b) Consider the case `r lt alpha`, The Amperian loop is a circle labelled 1. For this loop, taking the radius of the circle to be r,
`L = 2pi r`
Now the current enclosed Ie 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 (2pi r) = mu_(0) (Ir^(2))/(a^(2))`
`B = ((mu_(0)I)/(2pi a^(2)))r`
`B prop r (r lt a)`
`B prop r (r lt a)`

Figure shows a plot of the magnitude of B with distance r from the centre of the wire. The direction of the field is tangential to the respective circular loop (1 or 2) and given by the right-hand rule described earlier in this section. This example possesses the required symmetry so that Ampere’s law can be applied readily.