Question

A circular coil of 30 turns and radius 8 cm carrying a current of 6 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 30° with the normal of the coil. The magnitude of the counter torque that must be applied to prevent the coil from turning is

• A. 5.4 Nm
• B. 7.2 Nm
• C. 3.6 Nm
• D. 1.8 Nm

Hint:

For a current-carrying coil in a magnetic field, the torque depends on the number of turns, the current, the area of the coil, and the angle between the magnetic field and the coil.

Answer 1

The Correct Option is D

The torque acting on a current-carrying coil in a magnetic field is given by: \[ \tau = nIBA \sin \theta \] Where: - \( n = 30 \) is the number of turns, - \( I = 6 \, \text{A} \) is the current, - \( B = 1.0 \, \text{T} \) is the magnetic field strength, - \( A = \pi r^2 = \pi (0.08)^2 \, \text{m}^2 \) is the area of the coil, - \( \theta = 30^\circ \) is the angle between the magnetic field and the normal to the coil. Substituting the values: \[ \tau = 30 \times 6 \times \pi \times (0.08)^2 \times \sin 30^\circ = 1.8 \, \text{Nm} \] Thus, the counter torque required is \( \boxed{1.8 \, \text{Nm}} \).