Question

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of \(30\degree\) with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer 1

Length of a side of the square coil, \(l\) = 10 cm = 0.1 m
Current flowing in the coil, \(I\) = 12 A 
Number of turns on the coil, \(n\) = 20 
Angle made by the plane of the coil with magnetic field, \( θ\) = 30° 
Strength of magnetic field, \(B\) = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation, 
                             \( τ = nBIAsinθ\)
Where, 
A = Area of the square coil
= \(l × l = 0.1 × 0.1 = 0.01 m^2\)
So, \(τ = 20 × 0.8 × 12 × 0.01 × sin30\degree \)
= \(0.96 N m\)
Hence, the magnitude of the torque experienced by the coil is \(0.96 N m\).