Question

A circular coil of wire 50 turns and a cross sectional area of 10.0 cm^2 carries a current of 4.0 A.It is placed in a magnetic field of 2.0 T with the coil making an angle of 60 degrees with the magnetic field. What is the torque exerted on the coil by the field?

A) 0.70 N*m

B) 0.20 N*m

C) 0.80 N*m

D) 0.40 N*m

E) 0.35 N*m

(I got 0.80 N*m,but I feel like I did something wrong or did not complete the whole process.)

Answer 1

The expression for the torque, T is given as -

T = BINAcos(θ)

B = Magnitude of the magnetic field = 2.0 T
I = Current in the coil = 4.0 A
N = Number of turns in the coil = 50
A = Cross-sectional area of the conductor = 10.0 cm^2 = 10 x 10^-4 m^2
θ = Angle between the coil and the magnetic field = 60 deg.

Put the above values in the expression for torque -

T = 2.0 x 4.0 x 50 x 10 x 10^-4 x cos60 = 0.20 N*m

Hence, option (B) is the correct answer.