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JEE Main 2022 (26 Jun, Shift 2) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2022 (26 Jun, Shift 2)

Q39JEE Main 2022 · 26 JUN · SHIFT 2Easy

Force on Current-Carrying Conductors

Two 10 cm long, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10 $-$ 5 N, then separation between the wires is ____________ cm.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

$F = \frac{\mu_0 I^2 l}{2\pi d} \Rightarrow d = \frac{\mu_0 I^2 l}{2\pi F}$
$d = \frac{2\times10^{-7}\times25\times0.1}{10^{-5}} = 0.05\text{ m}$
Separation $= 5\text{ cm}$

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