A rectangular coil of wire of 100 turns and 10×15cm2 size carrying a current of 2 A is in a magnetic field of induction 2×10–3Wb/m2. If the normal to the plane of the coil makes an angle 30° with the field, then the torque on the coil is.

n=100,A=10×15cm2=150×10−4m2B=2×10−3Wb/m2i=2A,θ=30∘τ=iBAnsin⁡θτ=2×2×10−3×150×10−4×100×sin⁡30∘τ=3×10−3Nm

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A rectangular coil of wire of 100 turns and 10×15cm² size carrying a current of 2 A is in a magnetic field of induction 2×10^–3Wb/m². If the normal to the plane of the coil makes an angle 30° with the field, then the torque on the coil is.

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Detailed Solution

$\begin{array}{l} n = 100, A = 10 \times 15 {cm}^{2} \\ = 150 \times 10^{- 4} m^{2} \\ B = 2 \times 10^{- 3} Wb / m^{2} \\ i = 2 A, θ = 30^{\circ} \\ τ = iBAnsin θ \\ τ = 2 \times 2 \times 10^{- 3} \times 150 \times 10^{- 4} \times 100 \times \sin 30^{\circ} \\ τ = 3 \times 10^{- 3} Nm \end{array}$