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Question

Physics Question on Moving charges and magnetism

A long straight wire of radius a carries a steady current I. The current uniformly distributed over its cross-section. The ratio of the magnetic fields B and B', at radial distance a2\frac{a}{2} and 2a respectively, from the axis of the wire is :

A

12\frac{1}{2}

B

1

C

4

D

14\frac{1}{4}

Answer

1

Explanation

Solution

For points inside the wire
B=μ0Ir2πR2(rR)B = \frac{\mu_0 I r}{2 \pi R^2} (r \le R)
For points outside the wire
B=μ0I2πr(rR)B = \frac{\mu_0 I}{2 \pi r} (r \ge R)
according to the question
BB=μ0I(a/2)2πa2μ0I2π(2a)=1:1\frac{B}{B'} = \frac{\frac{\mu_0 I (a /2)}{2 \pi a^2}}{\frac{\mu_0 I }{2 \pi(2a)}} = 1 : 1