A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic field of 0.050 T oriented at 30° to the plane of the loop. What torque acts on the loop?
1 answer:
Answer:
0.52 Nm
Explanation:
A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T
Angle between the plane of loop and magnetic field = 30 Degree
Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree
θ = 60°
Torque = N i A B Sinθ
Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60
Torque = 0.52 Nm
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