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JEE Main 2023 (24 Jan, Shift 1) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2023 (24 Jan, Shift 1)

Q29JEE Main 2023 · 24 JAN · SHIFT 1Easy

Force on Current-Carrying Conductors

Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F ${}_1$ . If distance between wires is halved and currents on them are doubled, force F ${}_2$ on 10 cm length of wire P will be:

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Answer: (d)

$F = \frac{\mu_0 I^2 l}{2\pi d} \Rightarrow F \propto \frac{I^2}{d}$ for the same length $l$
$\frac{F_2}{F_1} = \frac{(2I)^2}{I^2}\times\frac{d}{d/2} = 4\times2 = 8$
So the answer is (d) $8F_1$ .

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