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/m². The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be?

The torque on a loop in uniform magnetic field is given by
τ = NIABsinθ
where N is the number of turns, I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the plane of the loop and the magnetic field.
Given:
N = 50 turns
I = 2 A
A = length × width = 0.12 m × 0.1 m = 0.012 m²
B = 0.2 Weber/m²
θ = 30°
Substituting the values into the equation:
τ = 50 × 2 A × 0.012 m² × 0.2 Weber/m² × sin(30°)
τ = 1.2 Nm × 0.5
τ = 0.6 Nm
However, this is not one of the given options. There must be an error in the question or options provided. Let's check the calculation again.
τ = NIABsinθ
τ = 50 × 2 × 0.012 × 0.2 × sin(30°)
τ = 50 × 2 × 0.012 × 0.2 × 0.5
τ = 0.12 Nm
Therefore, the correct answer is 0.12 Nm.