Torque on a Rectangular Current Loop in a Uniform Magnetic Field

When a current-carrying loop is placed in a uniform magnetic field, it does not experience a net force, but it does experience a torque that tends to rotate it. This principle forms the basis of devices like the electric motor and the moving-coil galvanometer. The behaviour of such a loop is identical to that of a magnetic dipole in a field.

A vector quantity that measures the strength and orientation of a current loop as a magnetic source is called the magnetic dipole moment.

The rotational effect experienced by a current-carrying loop placed in a uniform magnetic field is called torque.

\[\tau=NIAB\sin\theta\]

Also written as:

\[\vec{\tau}=\vec{m}\times\vec{B}\]

\[m=NIA\]

Setup: A rectangular loop ABCD (length l, breadth b, current I, turns N) placed in a uniform magnetic field B.

Step 1 - Force on each side:

Step 2 - Case 1: Plane of loop parallel to B (θ = 90°)

Torque = Force × perpendicular distance between the two forces:

τ = F₁ × l = BIb × l = BIA

For N turns:

τ_max ⁡= NIAB

Step 3 - Case 2: Loop tilted at angle θ (general case)

The perpendicular distance between the forces reduces to b sin θ:

τ = BIl × b sin ⁡θ = BIA sin⁡ θ

For N turns: τ = NIAB sin⁡ θ

Step 4 - Vector form:

\[\vec τ\] = \[\vec m\] × \[\vec B\] where \[\vec m\] = NI\[\vec A\]

Given:

• N = 100 turns, circular coil
• Radius r = 10 cm = 0.1 m → Area A = πr² = π × 10⁻² m²
• Current I = 3.2 A
• External field B = 2 T
• Moment of inertia of coil = 0.1 kg·m²
• Initial position: axis of coil along \[\vec B\] (θ = 0°)
• Final position: coil turned 90° (θ = 90°)

(a) Magnetic field at the centre of the coil

\[B_\mathrm{centre}=\frac{\mu_0NI}{2R}=\frac{4\pi\times10^{-7}\times100\times3.2}{2\times0.1}\]

B_centre = 2 × 10⁻³ T

(b) Magnetic moment of the coil

m = NIA = 100 × 3.2 × π × (0.1)²

𝑚 ≈ 10⁢ A·m²

(c) Torque in initial and final positions

• Initial (θ = 0°):
  τ_i = mB sin⁡0° = 0

• Final (θ = 90°):
  𝜏_𝑓 = 𝑚⁢𝐵⁢ sin⁡ 90° = 10 × 2 = 20 N·m

(d) Angular speed after rotating through 90°

Energy conservation: Loss in PE = Gain in KE

ΔU = mB(cos⁡0° − cos⁡90°) = mB = 10 × 2 = 20 J

\[\frac {1}{2}\] I_moi ω² = 20

\[\frac {1}{2}\] (0.1) ω² = 20 ⟹ ω² = 400

ω = 20 rad s⁻¹.

A current-carrying loop is placed on a smooth horizontal surface in a uniform magnetic field.

(a) Can a uniform \[\vec B\] make the loop spin about its own vertical axis?

• For a horizontal loop, the area vector \[\vec A\] is vertical.
• Torque = \[\vec m\] × \[\vec B\] — this torque acts in the horizontal plane, not about the vertical axis.
• A uniform field cannot produce the required torque to spin the loop about its own vertical axis.

Answer: No.

(b) In what orientation is the loop in stable equilibrium? What happens to flux?

• Stable equilibrium occurs when \[\vec m\]  ∥ \[\vec B\] → θ = 0°, torque = 0.

• In this position, \[\vec A\] is aligned with \[\vec B\].

• The magnetic field of the loop and the external field point in the same direction → total flux is maximum.

Answer: Stable equilibrium when \[\vec m\] || \[\vec B\]; total magnetic flux is maximum.

(c) Why does a flexible current loop become circular when placed in a magnetic field?

• A flexible loop adjusts its shape to maximise magnetic flux.
• For a fixed perimeter, a circle encloses the maximum area.
• More area → larger magnetic moment → larger flux.
• Therefore, the loop deforms into a circle.

Answer: It becomes circular because a circle has the maximum area for a given perimeter, maximising the magnetic flux through the loop.