A 300-turn coil of area 25 cm² is placed in an 80 mT magnetic field. If the coil carries a current of 3.3 A and if the normal to the area of the coil makes an angle of 20° with the direction of the magnetic field, what is the magnitude of the torque on the coil?

Question

Quizplus · Accepted Answer

The torque on a current carrying coil in a magnetic field is given by the formula:

τ = NIABsinθ

where τ is torque, N is the number of turns in the coil, I is the current in the coil, A is the area of the coil, B is the strength of the magnetic field and θ is the angle between the normal to the area of the coil and the direction of the magnetic field.

Plugging in the given values,

N = 300
I = 3.3 A
A = 25 cm² = 0.0025 m²
B = 80 mT = 0.08 T
θ = 20°

τ = (300)(3.3)(0.0025)(0.08)sin(20°) = 0.068 N·m

Therefore, the answer is D.

A 300-Turn Coil of Area 25 Cm2 Is Placed in an 80

A 300-Turn Coil of Area 25 Cm² Is Placed in an 80