Torque experienced by a current loop in a uniform magnetic field

Let consider a rectangular loop PQRS of length l and breadth b (Figure). It carries a current of I along PQRS. The loop is placed in a uniform magnetic field of induction B. Let θ be the angle between the normal to the plane of the loop and the direction of the magnetic field.

Fig: Torque on a current loop placed in a magnetic field

Force on the arm QR, F1 = I(QR) × B

Since the angle between I(QR) and B is (900 – θ),

Magnitude of the force F1 = BIb sin (900 – θ)

F1 = BIb cos θ

Force on the arm SP, F2 = I(SP) × B

Since the angle between I(SP) and B is (900 + θ),

Magnitude of the force F2 = BIb cos θ

The forces F1 and F2 are equal in magnitude, opposite in direction and have the same line of action. Hence their resultant effect on the loop is zero.

Force on the arm PQ, F3 = I(PQ) × B

Since the angle between I(PQ) and B is 900,

Magnitude of the force F3 = BIl sin 900 = BIl

F3 acts perpendicular to the plane of the paper and outwards.

Force on the arm RS, F4 = I(RS) × B

Since the angle between I(RS) and B is 900,

Magnitude of the force F4 = BIl sin 900 = BIl

F4 acts perpendicular to the plane of the paper and inwards.

The forces F3 and F4 are equal in magnitude, opposite in direction and have different lines of action. So, they constitute a couple.

Hence, Torque = BIl × PN = BIl × PS × sin θ (Figure)

= BIl × b sin θ = BIA sin θ

Fig: Torque

If the coil contains n turns, τ = nBIA sin θ

So, the torque is maximum when the coil is parallel to the magnetic field and zero when the coil is perpendicular to the magnetic field.