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?
Answer & explanation
Correct answer: option 3
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°
$Z=(300)(12)(0.024)(6)(\sin 60°)$
$=300×12×0.024×6×\frac{\sqrt{3}}{2}$
$=\frac{518.4×1.71}{2}≃447.8Nm$