First slide
Torque on a current carrying loop placed in uniform magnetic field

A rectangular coil of length 0.12 m and width 0.1m 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 required torque is 

τ = NIAB sin θ 
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 θ is the angle between the direction of the magnetic field and normal to the plane of the coil. Here, N = 50, I = 2 A, A = 0.12 m ×  0.1 m= 0.012 m2
B = 0.2 Wb /m2 and  θ = 90° - 30° = 60° 
  τ=(50)(2 A)0.012 m20.2 Wb/m2sin60°

=0.20 Nm

Get Instant Solutions
When in doubt download our app. Now available Google Play Store- Doubts App
Download Now
Doubts App