Question

A straight long wire of radius a carries a current I uniformly distributed along its cross section. The magnetic field B in the region r < a will be proportional to​

Answer 1

Answer:

Field at outside points : The Amperean loop is a circle labelled 2 haveing radius r>a.

Length of the loop, L=2πr

Net current enclosed by the loop =I

By Ampere's circuital law,

BL=μ0I

or B×2πr=μ0I

or B=μ0I2πr

i.e., B∝1r [For outside points]

Field at inside points : The Amperean loop is a circle labelled 1 with r<a.

Length of the loop, L=2πr

So, the current enclosed by loop 1 is less than I. As the current distribution is uniform, the fraction of I enclosed is

I'=1πa2×πr2=Ir2a2

Applying Amphere's law,

BL=μ0I'

or B×2πr=μ0lr2a2

or B=(μ0I2πa2)r[Forr<a]

i.e., B∝r [For inside points ]

Thus, the field B is proportional to r as we move from the axis of the cylinder towards its surface and then it decreases as 1r. The variation of B with distance r from the centre of the wire is shown in figure.

Step-by-step explanation: