A square coil ABCD of side ' L ' is carrying a current in clockwise direction. A straight conductor carrying current I₂ (upward direction) is kept parallel to side AB at a distance L/3 in the plane of ABCD. The net force on the coil ABCD is ( μ₀ = magnetic permeability) (a) (μ₀ I₁ I₂)/(3π) (b) (3μ₀ I₁ I₂)/(4π) (c) (3μ₀ I₁ I₂)/(8π) (d) (9μ₀ I₁ I₂)/(8π)

Answer: (d). Forces on the two sides perpendicular to the wire cancel. Near side AB is at L/3, far side CD at 4L/3, with opposite currents. FAB = (μ₀ I₁ I₂ L)/(2π (L/3)) = (3μ₀ I₁ I₂)/(2π), FCD = (μ₀ I₁ I₂ L)/(2π (4L/3)) = (3μ₀ I₁ I₂)/(8π), in opposite directions. Fnet = (3μ₀ I₁ I₂)/(2π) - (3μ₀ I₁ I₂)/(8π) = (9μ₀ I₁ I₂)/(8π) So the answer is (d) (9μ₀ I₁ I₂)/(8π).

A square coil ABCD of side ' L ' is carrying a current in clockwise… | MHT CET 2025 (23 Apr, Shift 1)

Q14MHT CET 2025 · 23 APR · SHIFT 1Medium

Force on Current-Carrying Conductors

A square coil ABCD of side ' L ' is carrying a current in clockwise direction. A straight conductor carrying current $I_2$ (upward direction) is kept parallel to side AB at a distance $\frac{L}{3}$ in the plane of $ABCD$ . The net force on the coil ABCD is ( $\mu_0 =$ magnetic permeability)

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Answer: (d)

Forces on the two sides perpendicular to the wire cancel. Near side AB is at $L/3$ , far side CD at $4L/3$ , with opposite currents.
$F_{AB} = \frac{\mu_0 I_1 I_2 L}{2\pi (L/3)} = \frac{3\mu_0 I_1 I_2}{2\pi}$ , $F_{CD} = \frac{\mu_0 I_1 I_2 L}{2\pi (4L/3)} = \frac{3\mu_0 I_1 I_2}{8\pi}$ , in opposite directions.
$F_{net} = \frac{3\mu_0 I_1 I_2}{2\pi} - \frac{3\mu_0 I_1 I_2}{8\pi} = \frac{9\mu_0 I_1 I_2}{8\pi}$
So the answer is (d) $\frac{9\mu_0 I_1 I_2}{8\pi}$ .

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