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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 aand r gt a |
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Answer» Solution :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 
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