Question

A circular coil of 50 turns, each of radius 0.1 m, carries a current of 2 A. If the coil is placed in a uniform magnetic field of 0.5 T perpendicular to its plane, what is the magnitude of the torque acting on the coil?

• A. \( 0.157 \, \text{N·m} \)
• B. \( 0.785 \, \text{N·m} \)
• C. \( 1.57 \, \text{N·m} \)
• D. \( 3.14 \, \text{N·m} \)

Hint:

For torque on a current-carrying coil, ensure the angle \( \theta \) is between the magnetic field and the normal to the coil’s plane. When the field is perpendicular to the coil’s plane, \( \sin\theta = 1 \), maximizing the torque.

Answer 1

The Correct Option is B

The torque on a current-carrying coil in a magnetic field is given by: \[ \tau = N I A B \sin\theta \] Where: - \( N = 50 \) (number of turns), - \( I = 2 \, \text{A} \) (current), - \( A \) is the area of the coil, - \( B = 0.5 \, \text{T} \) (magnetic field strength), - \( \theta \) is the angle between the magnetic field and the normal to the coil’s plane. Since the magnetic field is perpendicular to the plane of the coil, the normal to the coil is parallel to the field, so \( \theta = 90^\circ \), and \( \sin 90^\circ = 1 \). The area of the circular coil is: \[ A = \pi r^2 \] Where \( r = 0.1 \, \text{m} \): \[ A = \pi \times (0.1)^2 = \pi \times 0.01 = 0.0314 \, \text{m}^2 \] Using \( \pi \approx 3.14 \): \[ A \approx 3.14 \times 0.01 = 0.0314 \, \text{m}^2 \] Now calculate the torque: \[ \tau = N I A B = 50 \times 2 \times 0.0314 \times 0.5 \] \[ \tau = 50 \times 2 \times 0.0314 \times 0.5 = 100 \times 0.0157 = 1.57 \, \text{N·m} \] Recalculating for accuracy: \[ \tau = 50 \times 2 \times \pi \times (0.1)^2 \times 0.5 \] \[ \tau = 100 \times 3.14 \times 0.01 \times 0.5 = 100 \times 0.0157 = 0.785 \, \text{N·m} \] \[ \tau = 50 \times 2 \times 0.0314 \times 0.5 = 1.57 \times 0.5 = 0.785 \, \text{N·m} \] Thus, the magnitude of the torque acting on the coil is \( 0.785 \, \text{N·m} \).