A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 weber/m2. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30 with the direction of the field, the torque required to keep the coil in stable equilibrium will be
- 0.24 Nm
- 0.12 Nm
- 0.15 Nm
- 0.20 Nm
Answer
(4) 0.20 Nm
Solution
Plane of the coil makes 30 with B its normal makes = 60 with B.
= NIAB
= 50 2 (0.12 0.1) 0.2 60
= 0.24 0.866
0.20 N m
An equal and opposite torque is needed to keep the coil in equilibrium.
Answer: (4)