Question

A coil of \(50\) turns carrying a current of \(2\) A in a magnetic field of \(0.5\) T. The torque acting on the coil is

• A. \(0.4\) Nm clockwise
• B. \(0.2\) Nm anticlockwise
• C. \(0.4\) Nm anticlockwise
• D. \(0.2\) Nm clockwise
• E. \(0.8\) Nm anticlockwise

Hint:

Torque on a current loop: \[ \tau = NIAB\sin\theta \] Use the right-hand rule to decide clockwise or anticlockwise direction.

Answer 1

The Correct Option is C

From the figure on page 13: - number of turns \(N=50\) - current \(I=2\) A - magnetic field \(B=0.5\) T - coil dimensions \(10\text{ cm}\times 8\text{ cm}\) Area: \[ A=0.10\times 0.08=0.008\text{ m}^2 \] Torque on the coil: \[ \tau = N I A B \sin\theta \] Here the plane of the coil is parallel to the field direction shown, so \(\sin\theta=1\). \[ \tau = 50\times 2\times 0.008\times 0.5 \] \[ \tau = 0.4\text{ Nm} \] From the field direction \(N\to S\) and current shown in the figure, the torque is anticlockwise. Hence, \[ \boxed{(C)\ 0.4\text{ Nm anticlockwise}} \]