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 circlke concentric with the corss secion. For this loop.
`L=2pir`
`l_(e)=` current enclosed by the loop `=I`
This result is the familiar expression for a logn straight wire `B(2pir)=mu_(0)I`
`B=(mu_(0)I)/(2pir)`
`B prop 1/r" "(r gt a)`
b. Consider the case `r lt a`. THe amperian loop is circle labelled 1. For this loop taking the radius of the circle to be r
`L=2pir`
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((pir^(2))/(pia^(2)))=(Ir^(2))/(a^(2))`
Using ampere.s law
`B(2pir)=mu_(0)(Ir^(2))/(a^(2))`
`B=((mu_(0)I)/(2pia^(2)))r`
`B prop r (r lt a)`
Figure show a plot of the magnetic 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 ca be applied readily.