Q 12-04-177JEE MainJEE Main 2021 (27 Aug, Shift 2)Medium
A coaxial cable consists of an inner wire of radius $a$ surrounded by an outer shell of inner and outer radii $b$ and $c$ respectively. The inner wire carries an electric current $i_0$ which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance $x$ from the axis when (i) $x < a$ and (ii) $a < x < b$?
Answer: (D) $\frac{x^2}{a^2}$
By Ampère's law:
(i) $x < a$: enclosed current $i_0\dfrac{x^2}{a^2}$, so $B_1 = \dfrac{\mu_0i_0x}{2\pi a^2}$.
(ii) $a < x < b$: enclosed current $i_0$, so $B_2 = \dfrac{\mu_0i_0}{2\pi x}$.
$$\frac{B_1}{B_2} = \frac{x^2}{a^2}$$
Solution by Sreeraj P, M.Sc Physics