Torque on a Current-loop: Moving-Coil Galvanometer — Flashcards

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Formula: τ = IAB sinθ. Here I is current in ampere, A is area in m², B is magnetic field in tesla, and θ is the angle between the normal to the loop and B. The torque tends to rotate the loop.

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Each side of the loop experiences a magnetic force. For the two vertical sides, the forces are equal and opposite but act along different lines of action, so they form a couple. The other two sides pr…

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Given: l = 0.20 m, b = 0.10 m, I = 5 A, B = 0.50 T, θ = 30° Step 1: Area A = l × b = 0.20 m × 0.10 m = 0.020 m² Step 2: Use τ = IAB sinθ Step 3: τ = 5 × 0.020 × 0.50 × sin30° N m Step 4: sin30° = 1/2 …

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Vector form: τ⃗ = m⃗ × B⃗ Magnitude: τ = mB sinθ Magnetic moment: m = NIA Unit of magnetic moment: A m² Direction: along the area vector normal to the plane of the coil.

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Given: N = 25, A = 4.0 × 10^-3 m², I = 2 A, B = 0.20 T Step 1: Magnetic moment m = NIA Step 2: m = 25 × 2 × 4.0 × 10^-3 = 0.20 A m² Step 3: Maximum torque occurs when θ = 90° Step 4: τmax = NIAB Step …

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When the plane of the coil is parallel to B, the normal to the coil is perpendicular to B, so θ = 90°. Since sin90° = 1, torque becomes maximum: τmax = NIAB. This is the strongest turning effect.

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When the plane of the coil is perpendicular to B, the normal to the coil is parallel to B, so θ = 0°. Since sin0° = 0, torque is zero: τ = 0. The loop has no turning effect in this position.

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Given: side = 0.10 m, N = 20, I = 12 A, B = 0.80 T, θ = 30° Step 1: Area A = (0.10 m)² = 0.010 m² Step 2: Use τ = NIAB sinθ Step 3: τ = 20 × 12 × 0.010 × 0.80 × sin30° Step 4: τ = 20 × 12 × 0.010 × 0.