A uniform magnetic field exerts no net force on a current loop but it does exert a net torque. Torque (τ) is the twisting force that tends to cause rotation. The axis of rotation is the point where the object rotates.

**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),

The 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),

The 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},

The 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},

The 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.

Torque, τ = F × r

Where, F = force applied and, r = distance between the center of the axis of rotation and to the point where force is applied.

**Mathematical Explanation with Example:**

Motors are the most common function of the magnetic force on current-carrying wires. Motors contain loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft. Electrical energy is converted into mechanical work in the procedure.

[Find the maximum torque on a 100-turn square loop of a wire of 10.0 cm on a side that carries 15.0 A of current in a 2.00-T field.]

**Strategy – **Torque on the loop can be found using τ = NIABsinθτ = NIAB sinθ. Maximum torque occurs when *θ *= 90º and sin *θ* = 1.

**Solution**

For sin *θ* = 1, the maximum torque is

τmax=NIABτmax=NIAB

Entering known values yields

τmax = (100)(15.0 A)(0.100 m2)(2.00 T) = 30.0 N⋅mτmax=(100)(15.0 A)(0.100 m2)(2.00 T) = 30.0 N⋅m.