Figure shows the circular cross-section of a long straight wire of 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` (dashed inner circle) and `r gt a` (dashed outer circle).

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`
`I_(e)=` Current enclosed by the loop = I
The result is the familiar expression for a long straight wire `B(2pi r)=mu_(0)I`
`B=(mu_(0)I)/(2pi r)`
b) Consider the case `r lt a`. The Amperian loop is a circle labelled 1. For this loop, taking the radius of the circle to be r,
`L=2 pi r`
Now the current enclosed `I_(e)` is not I (because `r lt a`), but is less than this value. Since the current distribution is uniform, the current enclosed is,
`I_(e)=I({:(pir^(2)),(pia^(2)):})=(Ir^(2))/(a^(2))`
Using Ampere's law, `B(2pi r)= mu_(0) (Ir^(2))/(a^(2))`
`B=({:(mu_(0)I),(2pia^(2)):})r" "-(2)`
`B prop r (r lt a)`

Figure shows a plot of the magnitude of B with distance r from the centre (axis) 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 dsecribed earlier in this section.
This example possesses the required symmetry so that Ampere's law can be applied readily.