A long,straight wire of radius a carries a cu | vedclass

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Question

(A) Let the total current be $I$ and current density be $J = \frac{I}{\pi a^2}$.For a point inside the wire at distance $r < a$,using Ampere's law: $\

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) Let the total current be $I$ and current density be $J = \frac{I}{\pi a^2}$.For a point inside the wire at distance $r $B(2\pi r) = \mu_0 (J \cdot \pi r^2) \Rightarrow B = \frac{\mu_0 J r}{2}$.At $r = \frac{a}{3}$,$B_A = \frac{\mu_0 J (a/3)}{2} = \frac{\mu_0 J a}{6}$.For a point outside the wire at distance $r > a$,the wire acts as a long straight wire carrying current $I$ at its axis.$B(2\pi r) = \mu_0 I = \mu_0 (J \pi a^2) \Rightarrow B = \frac{\mu_0 J a^2}{2r}$.At $r = 2a$,$B_B = \frac{\mu_0 J a^2}{2(2a)} = \frac{\mu_0 J a}{4}$.The ratio is $\frac{B_A}{B_B} = \frac{\mu_0 J a / 6}{\mu_0 J a / 4} = \frac{4}{6} = \frac{2}{3}$.

$A$ 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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