A rectangular current carrying coil of length 2 m and width 4 m having 100 turns is placed in a magnetic field of 0.05 tesla. Find the torque due to magnetic field on the coil if angle between the coil and the magnetic field is 90° and current in the coil is 3 A?

Concept:

• A coil always act as a magnetic dipole of dipole moment M.

The dipole moment of a coil is given by:

Dipole moment = M = NIA

Where N is number of turns, I is current and A is area of the coil

Area vector (A):

• The vector perpendicular to the plane of the coil is called as area vector.
• The magnitude of area vector is equal to area of the coil.

Torque of magnetic field on any magnetic dipole moment is given by;

Torque (τ) = M × B = MB Sinθ

Where M × B is cross product of Area vector A and magnetic field vector B and θ is the angle between area vector A and magnetic field B.

Explanation:

Given that:

Area = A = 2 × 4 = 8 m²

• As the area is perpendicular to coil so angle between area vector and magnetic field B is 90°.

Angle = θ = 90°

Number of turns = N =100

Current = I = 3 A

Magnetic field = B = 0.05 T

Magnetic moment = M = NIA = 100 × 3 × 8 = 2400