Question

A square coil whose each side is 10 cm has 20 turns and 12 A current is flowing. The coil is hanging vertically and a perpendicular drawn from the plane of this makes an angle 30° with the uniform magnetic field of 0. 80 T. How much is the magnitude of torque acting on the coil?

Answer 1

Side of square coil = 10 cm
∴ Area of coil A = (10)² = 100cm² = 10^-2 m²
Current flowing I = 12 A
Number of turns in coil N = 20
Magnetic field B = 0.80 T
Angle θ = 30°
Torque acting on coil τ = NIABsinθ
τ = 20 × 10^-2 × 12 × 0.80 × sin 30°
τ = 20 × 12 × 0.80 × \(\frac{1}{2}\) × 10^-2
τ = 0.96 N-m