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

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 We have `B(2 pi r ) = mu_(0) `
`B= (mu_0 I)/(2 pi r)`
`B prop 1/r ( r gt a )`
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, =`2ir`
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_0I)/(2pia^2))r`
`B prop r " " (r lta)`
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