A 300 turns rectangular coil of length 20 cm and breadth 12 cm is carrying a current of 12 A, when placed in a magnetic field of 6 T. The plane of the coil is making an angle of 60° with the magnetic field. What is the torque acting on the coil?

447.8 N m

The correct answer is Option (3) → 447.8 N m

The torque (Z) acting on a current carrying conductor coil is given by -

$Z=nIAB\sin θ$

where,

n = number of turns in coil = 300

I = current through the coil = 12 A

A = Area of coil = $0.2 m × 0.12 m = 0.024m^2$

θ = Angle between the coil and magnetic field = 60°

$Z=(300)(12)(0.024)(6)(\sin 60°)$

$=300×12×0.024×6×\frac{\sqrt{3}}{2}$

$=\frac{518.4×1.71}{2}≃447.8Nm$