Question

A hundred-turn closely wound circular coil, of main radius 10 cm, carries a current of 3.2A. The coil is placed in a vertical plane and is free to rotate about a horizontal direction which coincides with its diameter. uniform magnetic field of 2T, in the horizontal direction, exists such that initially the axis of the coil is in the direction of the field the call rotates through an angle of 90\(^{\circ}\) under the influence of the magnetic field. calculate the torque in the initial and final position of the coil.

Answer 1

Given:
Number of turns, N = 100
Magnetic field strength, B = 2 T
Mean radius, r = 10 cm = 0.1 m
Current, I = 3.2 A
The angle of rotation, θ = 90°

Initial Position:
In the initial position, when the coil's axis is aligned with the magnetic field (θ = 0°), the torque is zero.

Final Position:
In the final position, when the coil's axis is perpendicular to the magnetic field (θ = 90°), the torque can be calculated using the formula:

Torque (τ) = N \(\times\) B\(\times\)A\(\times\)I\(\times\)sin(θ)

Where A is the area of the coil, A = π \(\times\) r².

Substitute the values:
τ = 100 \(\times\) 2\(\times\) π \(\times\) (0.1)^2 \(\times\) 3.2\(\times\)sin90°
τ = 100 \(\times\) 2 \(\times\) π \(\times\)0.01 \(\times\) 3.2 \(\times\)1
τ = 20 Nm

Therefore, the torque in the final position of the coil is 20 Nm.