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Question

A square loop of side \( 0.15 \, \text{m} \) with 50 turns carries \( 1 \, \text{A} \) in a magnetic
field of \( 0.6 \, \text{T} \). The plane of the loop is at \( 30^\circ \) to the field. What is the
torque?

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Explanation

**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. Torque tau = N I A B sin θ , where A = 0.15 × 0.15 = 0.0225 m² . tau = 50 × 1 × 0.0225 × 0.6 × sin 30° = 0.675 × 0.5 =

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