NEET | 2016Magnetic field due to currentHard

Question

A long straight wire of radius a carries a steady current I. the current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B′, at radial distances a/2 and 2a respectively, from the axis of the wire is:

Options

A.

1/4

B.

1/2

C.

1

D.

4

Solution

Inside the wireBy ampere's law
∫B→. dl→=μ0(ienclosed)
∫B . dI cos 0 = μ01πa2. πa22
B ∫dI =μ0I4
B2πa2=μ0I4
B=μ0I4πa
Outside the wire,
B=μ0I2πr=μ0I2π(2a)=μ0I4πa
So, BB'=1

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