A 50.5-turn circular coil of radius 4.75 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.455 T. If the coil carries a current of 22.5 mA, find the magnitude of the maximum possible torque exerted on the coil.

The maximum torque, τ = 3.67 × 10⁻³ Nm

Explanation:

The torque τ = NiABsinθ where N = number of turns of circular coil = 50.5, i = current in circular coil = 22.5 mA = 0.0225 A, A = area of circular coil = πr² where r = radius of circular coil = 4.75 cm = 0.0475 m, B = magnetic field strength = 0.455 T and θ = 90° for maximum torque.

So, τ = NiABsinθ

τ = Niπr²Bsinθ

τ = 50.5 × 0.0225 A × π × (0.0475 m)² × 0.455 T × sin90°

τ = 0.003665 Nm

τ = 3.665 × 10⁻³ Nm

τ ≅ 3.67 × 10⁻³ Nm

So the maximum torque, τ = 3.67 × 10⁻³ Nm

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33: answer is c

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