A wire X of length 50 cm carrying a current of 2 A is placed parallel to a long wire Y of length 5 m. The wire Y carries a current of 3 A. The distance between two wires is 5 cm and currents flow in the same direction. The force acting on the wire Y is (a) 1.2 × 10⁻⁵ N directed towards wire X. (b) 1.2 × 10⁻⁴ N directed away from wire X. (c) 1.2 × 10⁻⁴ N directed towards wire X. (d) 2.4 × 10⁻⁵ N directed towards wire X.

Answer: (a). By Newton's third law the force on Y equals the force on X. X (50 cm) lies in the field of the long wire Y. F = (μ₀ IX IY lX)/(2π d) = 2×10⁻⁷×2×3×0.50.05 F = 1.2×10⁻⁵ N; currents are in the same direction, so the force is attractive. So the answer is (a) 1.2×10⁻⁵ N towards X.

A wire X of length 50 cm carrying a current of 2 A is placed parallel… | JEE Main 2022 (29 Jul, Shift 2)

Q42JEE Main 2022 · 29 JUL · SHIFT 2Medium

Force on Current-Carrying Conductors

A wire X of length $50\text{ cm}$ carrying a current of $2\text{ A}$ is placed parallel to a long wire $Y$ of length $5\text{ m}$ . The wire $Y$ carries a current of $3\text{ A}$ . The distance between two wires is $5\text{ cm}$ and currents flow in the same direction. The force acting on the wire $Y$ is

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Answer: (a)

By Newton's third law the force on Y equals the force on X. X (50 cm) lies in the field of the long wire Y.
$F = \frac{\mu_0 I_X I_Y l_X}{2\pi d} = \frac{2\times10^{-7}\times2\times3\times0.5}{0.05}$
$F = 1.2\times10^{-5}\text{ N}$ ; currents are in the same direction, so the force is attractive.
So the answer is (a) $1.2\times10^{-5}$ N towards X.

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