Q 12-04-180JEE MainJEE Main 2020 (9 Jan, Shift 1)Easy
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 $\dfrac{a}{3}$ and $2a$, respectively from the axis of the wire is:
Answer: (A) $\dfrac{2}{3}$
Inside: $B = \dfrac{\mu_0 I r}{2\pi a^2}$, so at $r = a/3$, $B_1 = \dfrac{\mu_0 I}{6\pi a}$.
Outside: $B = \dfrac{\mu_0 I}{2\pi r}$, so at $r = 2a$, $B_2 = \dfrac{\mu_0 I}{4\pi a}$.
$$\frac{B_1}{B_2} = \frac{4}{6} = \frac{2}{3}$$
Solution by Sreeraj P, M.Sc Physics