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(C) For a long straight wire of radius $a$ carrying current $I$ uniformly distributed:$1$. The magnetic field inside the wire at a distance $r < a$ is

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$\frac{2}{3}$

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(C) For a long straight wire of radius $a$ carrying current $I$ uniformly distributed:$1$. The magnetic field inside the wire at a distance $r $2$. The magnetic field outside the wire at a distance $r > a$ is given by $B_{	ext{out}} = \frac{\mu_0 I}{2 \pi r}$.$3$. At $r_1 = \frac{a}{3}$ (inside),$B_1 = \frac{\mu_0 I (a/3)}{2 \pi a^2} = \frac{\mu_0 I}{6 \pi a}$.$4$. At $r_2 = 2a$ (outside),$B_2 = \frac{\mu_0 I}{2 \pi (2a)} = \frac{\mu_0 I}{4 \pi a}$.$5$. The ratio is $\frac{B_1}{B_2} = \frac{\mu_0 I / 6 \pi a}{\mu_0 I / 4 \pi a} = \frac{4 \pi a}{6 \pi a} = \frac{2}{3}$.

$A$ long straight wire,of radius $a$,carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire,at distance $\frac{a}{3}$ and $2a$ respectively from the axis of the wire,is

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