Question
A rectangular coil of length 0.12 m and width 0.1 m having turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Wb/m2.The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30o with the direction of the field, the torque required to keep the coil in stable equilibrium will be
-
0.15 Nm
-
0.20 Nm
-
0.24 Nm
-
0.12 Nm
Solution
B.
0.20 Nm
Given N = 50
B = 0.2 Wb/m2, I = 2A
θ = 60o, A = 0.12 x 0.1 = 0.012 m2
Thus, torque required to keep the coil in stable equilibrium i.e.,
= NIAB sin θ = 50 x 2 x 0.012 x0.2 sin 60o
= NIAB sin θ = 50 x 2 x 0.012 x0.2 sin 60o
