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(B) For a long straight wire of radius $a$ carrying a steady current $I$ uniformly distributed over its cross-section:$1$. The magnetic field $B$ at a

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(B) For a long straight wire of radius $a$ carrying a steady current $I$ uniformly distributed over its cross-section:$1$. The magnetic field $B$ at a distance $r$ inside the wire $(r For $r = \frac{a}{2}$,we have $B = \frac{\mu_0 I (a/2)}{2 \pi a^2} = \frac{\mu_0 I}{4 \pi a}$.$2$. The magnetic field $B'$ at a distance $r$ outside the wire $(r > a)$ is given by $B' = \frac{\mu_0 I}{2 \pi r}$.For $r = 2a$,we have $B' = \frac{\mu_0 I}{2 \pi (2a)} = \frac{\mu_0 I}{4 \pi a}$.$3$. The ratio of the magnetic fields is $\frac{B}{B'} = \frac{\frac{\mu_0 I}{4 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = 1$.

$A$ 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 $\frac{a}{2}$ and $2a$ respectively,from the axis of the wire is

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