A square loop ABCD of side L carrying a current I₁ is placed at a distance (L/3) from a conductor coplaner with a straight conductor XY carrying current I₂ as shown in figure. The net force on the loop will be ( μ₀ = magnetic permeability) (a) (μ₀ I₁ I₂)/(3π) (b) (3μ₀ I₁ I₂)/(8π) (c) (9μ₀ I₁ I₂)/(8π) (d) (3μ₀ I₁ I₂)/(4π)

Answer: (c). Forces on BC and DA cancel. AB (current parallel to I₂) is attracted; CD (opposite current) is repelled. FAB = (μ₀ I₁ I₂ L)/(2π (L/3)) = (3μ₀ I₁ I₂)/(2π) FCD = (μ₀ I₁ I₂ L)/(2π (4L/3)) = (3μ₀ I₁ I₂)/(8π) Fnet = (3μ₀ I₁ I₂)/(2π) - (3μ₀ I₁ I₂)/(8π) = (9μ₀ I₁ I₂)/(8π) (towards the wire) So the answer is (c).

A square loop ABCD of side L carrying a current I₁ is placed at a… | MHT CET 2026 (20 Apr, Shift 1)

Q9MHT CET 2026 · 20 APR · SHIFT 1Medium

Force on Current-Carrying Conductors

A square loop ABCD of side L carrying a current $I_1$ is placed at a distance $(L/3)$ from a conductor coplaner with a straight conductor XY carrying current $I_2$ as shown in figure. The net force on the loop will be ( $\mu_0$ = magnetic permeability)

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Answer: (c)

Forces on BC and DA cancel. AB (current parallel to $I_2$ ) is attracted; CD (opposite current) is repelled.
$F_{AB} = \frac{\mu_0 I_1 I_2 L}{2\pi (L/3)} = \frac{3\mu_0 I_1 I_2}{2\pi}$
$F_{CD} = \frac{\mu_0 I_1 I_2 L}{2\pi (4L/3)} = \frac{3\mu_0 I_1 I_2}{8\pi}$
$F_{net} = \frac{3\mu_0 I_1 I_2}{2\pi} - \frac{3\mu_0 I_1 I_2}{8\pi} = \frac{9\mu_0 I_1 I_2}{8\pi}$ (towards the wire)
So the answer is (c).

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