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JEE Main 2007 (AIEEE) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2007 (AIEEE)

Q56JEE Main 2007 · AIEEEEasy

Ampere's Law, Cylinders, Solenoid & Toroid

A long straight wire of radius $a$ carries a steady current $i.$ The current is uniformly distributed across its cross section. The ratio of the magnetic field at $a/2$ and $2a$ is

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Answer: (d)

At $a/2$ : $B_1 = \dfrac{\mu_0 i (a/2)}{2\pi a^2} = \dfrac{\mu_0 i}{4\pi a}$ .
At $2a$ : $B_2 = \dfrac{\mu_0 i}{2\pi (2a)} = \dfrac{\mu_0 i}{4\pi a}$ .
So the answer is (d) $1$ .

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