A coaxial cable consists of an inner wire of radius 'a' surrounded by… | JEE Main 2021 (27 Aug, Shift 2)

Q45JEE Main 2021 · 27 AUG · SHIFT 2Difficult

Ampere's Law, Cylinders, Solenoid & Toroid

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 i0, 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?

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Answer: (a)

Ampere's law, $x < a$ : enclosed current $i_0 \frac{x^2}{a^2}$ , so $B_1 = \frac{\mu_0 i_0 x}{2\pi a^2}$ .
$a < x < b$ : enclosed current $i_0$ , so $B_2 = \frac{\mu_0 i_0}{2\pi x}$ .
$\frac{B_1}{B_2} = \frac{x^2}{a^2}$
So the answer is (a) $\frac{x^2}{a^2}$ .

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