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JEE Main 2020 (9 Jan, Shift 1) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2020 (9 Jan, Shift 1)

Q47JEE Main 2020 · 9 JAN · SHIFT 1Easy

Ampere's Law, Cylinders, Solenoid & Toroid

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 2 $a$ , respectively from the axis of the wire is:

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Answer: (d)

At $a/3$ (inside): $B_1 = \dfrac{\mu_0 I (a/3)}{2\pi a^2} = \dfrac{\mu_0 I}{6\pi a}$ .
At $2a$ (outside): $B_2 = \dfrac{\mu_0 I}{2\pi (2a)} = \dfrac{\mu_0 I}{4\pi a}$ .
$\dfrac{B_1}{B_2} = \dfrac{4}{6} = \dfrac{2}{3}$ .
So the answer is (d) $\dfrac{2}{3}$ .

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