Answer to Question #213291 in Electricity and Magnetism for Sneha

Question

A long straight wire of radius R carries a steady current I. The current is uniformly distributed over its cross section. The magnetic field at a distance 2a/3 from its centre is

Answer 1

We know that

$\oint B.dl=\mu_0i$

$B\oint dl=\mu_0i=B.(2\pi a)=\mu_0i$

$B=\frac{\mu_0i}{2\pi a}$

Magnetic field at centre of wire

$B=\frac{\mu_0i}{2\pi a}$

When

$r=\frac{2a}{3}$

$\oint B'.dl=\mu_0i_{en}$

$i_{en}=i\frac{\pi \times( \frac{2a}{3})^2}{\pi a^2}=\frac{4}{9}i$

$B'(2\pi\times\frac{2a}{3})=\mu_0\frac{4i}{9}$

$B'=\frac{\mu_0 I}{3 \pi a}$

$\frac{B'}{B}=\frac{2}{3}$

$B'=\frac{2}{3}B$