Torque on a current loop; the loop as a magnetic dipole
A loop in a uniform field feels no net force, only a torque τ = m × B, where m = NIA is its magnetic moment.
Derivation: torque on a rectangular coil
A rectangular coil PQRS of N turns carries current I in a uniform field B. Sides PQ and RS have length a and stay perpendicular to B. Sides QR and SP have length b. The normal to the coil makes an angle θ with B.
- 1Forces on QR and SP are equal and opposite, and act along the same line (the axis of rotation). They cancel and give no torque.
- 2Forces on PQ and RS: each side is perpendicular to B, so each force has size F = IaB for one turn. They are equal and opposite but act along different lines, forming a couple.
- 3The perpendicular distance between the lines of these two forces is b sin θ.
- 4Torque on one turn: τ = F × b sin θ = IaB × b sin θ = IAB sin θ, where A = ab is the area.
- 5★For N turns: τ = NIAB sin θ.
- 6★Define m = NIA, along the normal. Then τ = mB sin θ, or in vector form τ = m × B.
- 7So, the net force is zero, but the torque turns the coil until m lines up with B.
★ key step
The current loop as a magnetic dipole
A current loop has a magnetic dipole moment m = NIA. Its direction is along the normal to the loop, given by the right-hand rule: curl the fingers along the current and the thumb points along m. The SI unit of m is A m², also written J T⁻¹. The face where the current looks anticlockwise is the north pole, and m points out of it. Far from the loop on its axis, B = (μ₀/4π)(2m/x³). This has the same form as the field of an electric dipole, E = (1/4πε₀)(2p/x³). In a field the loop feels a torque m × B, just as an electric dipole feels p × E. So a current loop behaves like a magnetic dipole, a tiny bar magnet.
Electric dipole and current loop side by side
| Quantity | Electric dipole | Current loop (magnetic dipole) |
|---|---|---|
| Dipole moment | p = q × 2a, from −q to +q | m = NIA, along the normal |
| SI unit | C m | A m² |
| Torque in a uniform field | τ = p × E | τ = m × B |
| Net force in a uniform field | Zero | Zero |
| Far field on the axis | (1/4πε₀)(2p/r³) | (μ₀/4π)(2m/r³) |
| Stable equilibrium | p along E | m along B |
Key positions of the coil
- θ = 0°: m along B, plane of the coil perpendicular to B. τ = 0, stable equilibrium.
- θ = 90°: plane of the coil parallel to B. τ = NIAB, the largest value.
- θ = 180°: m opposite to B. τ = 0, unstable equilibrium: a small push makes the coil flip over.
- Trap: θ is measured from the normal, not from the plane. If the plane makes an angle α with B, then θ = 90° − α.
- In a non-uniform field, the loop also feels a net force.
Worked example: torque on a coil
A coil of 100 turns and area 2.0 × 10⁻³ m² carries 0.50 A in a uniform field of 0.20 T. The plane of the coil makes 60° with the field. Find (a) its magnetic moment, (b) the torque on it and (c) the largest torque possible.
- (a) m = NIA = 100 × 0.50 × 2.0 × 10⁻³ A m² = 0.10 A m².
- (b) The plane makes 60° with B, so the normal makes θ = 90° − 60° = 30° with B.
- τ = mB sin θ = 0.10 × 0.20 × sin 30° N m = 0.10 × 0.20 × 0.5 N m = 1.0 × 10⁻² N m.
- (c) τ(max) = mB = 0.10 × 0.20 N m = 2.0 × 10⁻² N m, when the plane is parallel to B.
(a) 0.10 A m²; (b) 1.0 × 10⁻² N m; (c) 2.0 × 10⁻² N m
How to draw the coil in the field (view from above)
- Draw the field B as parallel arrows from left to right.
- Seen from above, the coil is a straight line of length b, tilted in the field. Mark its two vertical sides as ⊙ at one end and ⊗ at the other.
- Draw the forces F on these two sides: equal, opposite and perpendicular to B.
- Draw the normal m perpendicular to the coil and mark θ between m and B.
- Mark the perpendicular distance b sin θ between the two lines of force.
Every electric motor, from a mixer-grinder to the motor of an e-rickshaw, spins because a magnetic field keeps turning its current-carrying coils by this torque.
3 marks: “Derive an expression for the torque on a rectangular current loop placed in a uniform magnetic field.” Forces on the two sides of length b cancel (equal, opposite, same line); forces IaB on the sides of length a form a couple; arm b sin θ; τ = NIAB sin θ = mB sin θ; vector form τ = m × B.
2 marks: “Write the magnetic dipole moment of a circular loop of radius R carrying current I. What is its direction?” m = IπR² (NIπR² for N turns), along the normal to the loop: curl the right-hand fingers along the current and the thumb gives m.
1 marks: “In which orientation is a current loop in stable equilibrium in a uniform magnetic field?” When m is parallel to B (plane of the loop perpendicular to B), θ = 0°.
The torque wants m to line up with B. τ = NIAB sin θ, with θ measured from the normal, not the plane.