**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, ^{→}F_{1} = ^{→}I(QR) × ^{→}B**

Since the angle between ^{→}I(QR) and ^{→}B is (90^{0} – θ),

Magnitude of the force **F _{1} = BIb sin (90^{0} – θ)**

**F _{1} = BIb cos θ**

Force on the arm SP, ^{→}F_{2} = ^{→}I(SP) × ^{→}B

Since the angle between ^{→}I(SP) and ^{→}B is (90^{0} + θ),

Magnitude of the force **F _{2} = BIb cos θ**

The forces F_{1} and F_{2} 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 P**Q, ^{→}F_{3} = ^{→}I(PQ) × ^{→}B**

Since the angle between ^{→}I(PQ) and ^{→}B is 90^{0},

Magnitude of the force** F _{3} = BIl sin 90^{0} = BIl**

F_{3} acts perpendicular to the plane of the paper and outwards.

Force on the arm **RS, ^{→}F_{4} = ^{→}I(RS) × ^{→}B**

Since the angle between ^{→}I(RS) and ^{→}B is 90^{0},

Magnitude of the force **F _{4} = BIl sin 90^{0} = BIl**

F_{4} acts perpendicular to the plane of the paper and inwards.

The forces F_{3} and F_{4} are equal in magnitude, opposite in direction and have different lines of action. So, they constitute a couple.

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

**= BI l × 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.

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