Question

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/$m^2$. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30$^{\circ}$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be

• A. 0.24 Nm
• B. 0.12Nm
• C. 0.15Nm
• D. 0.20 Nm

Answer 1

The Correct Option is D

The required torque is $t = NIAB \sin \theta;$
where N is the number of turns in the coil, I is the current through the coil, B is the uniform magnetic field, A is the area of the coil and $?$ is the angle between the direction of the magnetic field and normal to the plane of the coil.
Here, $N = 50, I =2A , A = 0.12 \times 0.1 m = 0.012 m^2$
$B = 0.2 Wb / M^2 and \, ? = 90^\circ - 30^\circ = 60^\circ$
$\therefore t = (50) (2A) (0.012 M^2) (0.2 W b / m^2) sin60^\circ $
= 0.20 Nm

Answer 2

Whenever a coil (carried by current) comes in contact with a magnetic field, it experiences a torque that makes the plane of the coil perpendicular to the direction of the magnetic field. The same magnitude of opposing torque is similarly applied to keep the coil in a stable state.

The magnitude of torque (𝜏) experienced by a coil of N turns to carry a current I and having cross-sectional area A when kept in a magnetic field B is given by

𝜏 = N AIBsinθ------(1)

Where θ is the angle made by the perpendicular to the plane with the direction of the magnetic field.

This torque has a direction as it is trying to make the plane of the coil perpendicular to the magnetic field. Let's understand the question.

Number of turns of the coil N=50

Length of the coil=0.12m

The breadth of the coil=0.1m

Therefore, the cross-sectional area of the rectangular coil

A=length×width=0.12×0.1=0.012m²

Current in the coil I=2A

Strength of magnetic field B=0.2Wb/m²

Since the plane of the coil is inclined at an angle of 30° with the field, the perpendicular to the plane will be inclined at 90°- 30° = 60° with the field.

Hence, θ = 60°

Using (1), the torque(𝜏) experienced by the coil will be

𝜏 = 50×0.012×2×0.2×sin 60° = 0.12×2×\(\frac{\sqrt{3}}{2}\) ≈ 0.20Nm(sin 60° =\(\frac{\sqrt{3}}{2}\))

Therefore, the magnitude of the torque exerted by the magnetic field on the coil (and the magnitude of the torque required to keep it in equilibrium) is 0.20Nm.

Hence, the correct answer is (D) 0.20Nm.