12 min readConcept sheetSection A objective (2 marks, two parts) on the definition of the ampere; Section B short answer (3 marks): force between parallel wires or torque on a coil; numericals

Force on a current-carrying wire, force between parallel currents, and the current loop as a magnetic dipole

A wire in a field feels F = IlB sin θ, and parallel currents attract. A current loop is a magnetic dipole, M = NIA, which a field turns.

Force on a wireF = IlB sin θ: zero along B, largest (IlB) at right angles to B

Force between parallel wiresF/l = μ₀I₁I₂/2πr; currents in the same direction attract, opposite ones repel

Torque on a coilτ = NBIA sin θ = MB sin θ, where θ is the angle between B and the normal to the coil

Magnetic moment of a coilM = NIA; SI unit A m² (also written J T⁻¹)

Derivation: force on a current-carrying wire

A straight wire of length l and cross-section A lies at right angles to a uniform field B. It carries a current I.

1
The current is carried by free electrons drifting with speed v_d. Each one feels a Lorentz force ev_dB.
2
If n is the number of free electrons per unit volume, the wire holds nAl electrons.
3
Total force: F = (nAl) × ev_dB.
4
The current is I = neAv_d. So F = IlB.
5★
If the wire makes an angle θ with B, F = IlB sin θ. Fleming's left-hand rule gives its direction.
6
So, B = F/Il. One tesla is the field in which a 1 m wire carrying 1 A at right angles to it feels a force of 1 N.

★ key step

Derivation: force between two parallel wires, and the ampere

Two long straight wires lie parallel, a distance r apart, and carry currents I₁ and I₂.

1
Wire 1 makes a field at wire 2: B₁ = μ₀I₁/2πr, at right angles to wire 2.
2
A length l of wire 2 feels a force F = B₁I₂l = μ₀I₁I₂l/2πr.
3★
Force per unit length: F/l = μ₀I₁I₂/2πr. Wire 1 feels an equal and opposite force.
4
Currents in the same direction attract; currents in opposite directions repel.
5
Put I₁ = I₂ = 1 A and r = 1 m: F/l = μ₀/2π = 2 × 10⁻⁷ N m⁻¹.
6★
So, one ampere is defined like this. Two long parallel wires lie 1 m apart in vacuum and carry equal currents. If each wire feels a force of 2 × 10⁻⁷ N per metre of its length, the current in each is one ampere.

★ key step

Derivation: torque on a rectangular coil

A rectangular coil of N turns, length L and breadth b, carries a current I in a uniform field B. The normal to the coil makes an angle θ with B.

1
The two sides of length b feel equal and opposite forces along the same line. They cancel and give no torque.
2
Each side of length L is at right angles to B and feels a force NBIL. These two forces are equal and opposite but act along different lines, so they form a couple.
3
The perpendicular distance between their lines is b sin θ.
4
Torque = force × distance = NBIL × b sin θ = NBIA sin θ, where A = Lb is the area.
5★
So, τ = NBIA sin θ = MB sin θ, with M = NIA. In vector form, τ = M × B.
6
The net force is zero, but the torque turns the coil until its plane is at right angles to B. The torque is largest, NBIA, when the plane of the coil is parallel to B.

★ key step

The current loop as a magnetic dipole; the revolving electron

A current loop behaves like a small bar magnet. One face acts as N and the other as S, and the two faces cannot be separated. Its magnetic dipole moment is M = NIA, directed along the normal to the loop. Far from it, its field on the axis is B = (μ₀/4π)(2M/x³), and on the equatorial line it is (μ₀/4π)(M/x³). These have the same form as the field of an electric dipole, (1/4πε₀)(2p/x³). In a uniform field, the loop feels no net force but a torque, just like an electric dipole. An electron going round a nucleus is also a tiny current loop. If it makes ν rounds per second in a circle of radius r, the current is I = eν. The field at the centre is B = μ₀eν/2r, and the magnetic moment is M = eνπr². This is why atoms can act as tiny magnets. The smallest such moment, the Bohr magneton, is μ_B = eh/4πm ≈ 9.27 × 10⁻²⁴ J T⁻¹. (Your NIOS lesson works it out as 9.34 × 10⁻²⁴ J T⁻¹ because it uses rounded values of e, h and m.)

Worked example: parallel wires, a coil and a revolving electron

Problem

(a) Two long parallel wires 5.0 cm apart carry 10 A and 20 A in the same direction. Find the force on 2.0 m of each wire. (b) A coil of 50 turns and area 4.0 × 10⁻³ m² carries 2.0 A, with its plane parallel to a field of 0.30 T. Find its magnetic moment and the torque on it. (c) An electron goes round a circle of radius 5.3 × 10⁻¹¹ m, 6.6 × 10¹⁵ times a second. Find the field at the centre. (μ₀ = 4π × 10⁻⁷ T m A⁻¹, e = 1.6 × 10⁻¹⁹ C)

Working

(a) F/l = μ₀I₁I₂/2πr = 2 × 10⁻⁷ × 10 × 20/0.050 N m⁻¹ = 8.0 × 10⁻⁴ N m⁻¹.
F = 8.0 × 10⁻⁴ × 2.0 N = 1.6 × 10⁻³ N. The currents are in the same direction, so the force is attractive.
(b) M = NIA = 50 × 2.0 × 4.0 × 10⁻³ A m² = 0.40 A m².
The plane is parallel to B, so the normal is at θ = 90° to B and sin θ = 1.
τ = MB sin θ = 0.40 × 0.30 × 1 N m = 0.12 N m, the largest possible torque.
(c) Current I = eν = 1.6 × 10⁻¹⁹ × 6.6 × 10¹⁵ A ≈ 1.06 × 10⁻³ A.
B = μ₀I/2r = (4π × 10⁻⁷ × 1.06 × 10⁻³)/(2 × 5.3 × 10⁻¹¹) T ≈ (1.33 × 10⁻⁹)/(1.06 × 10⁻¹⁰) T ≈ 12.5 T.

Answer

(a) 1.6 × 10⁻³ N, attractive; (b) 0.40 A m² and 0.12 N m; (c) about 12.5 T

How to draw: parallel wires, and a coil in a field

Parallel wires: draw two long vertical lines, wire 1 on the left and wire 2 on the right, a distance r apart. Put upward arrows on both for currents I₁ and I₂.
At wire 2, show the field B₁ of wire 1 as ⊗ (into the page). Draw the force on wire 2 pointing towards wire 1, and the force on wire 1 pointing towards wire 2. Write 'attract'.
Coil in a field, seen from above: draw the field B as parallel arrows from left to right.
Draw the coil as a tilted straight line of length b, with ⊙ at one end and ⊗ at the other for the current.
Draw equal and opposite forces F on the two ends, at right angles to B. Draw the normal to the coil and mark the angle θ between it and B.
Mark b sin θ, the perpendicular distance between the lines of the two forces.

In real life

Every electric motor, from a ceiling fan to a mixer grinder or an e-rickshaw, turns because a current-carrying coil feels a torque in a magnetic field. The same force on a wire makes a loudspeaker cone move.

Exam angle

2 marks: “Define one ampere in terms of the force between two parallel current-carrying wires.” One ampere is the current which, flowing in each of two long straight parallel wires 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre of length between them.

3 marks: “Derive an expression for the force per unit length between two long parallel wires carrying currents I₁ and I₂. When is the force attractive?” B₁ = μ₀I₁/2πr at wire 2; F = B₁I₂l; F/l = μ₀I₁I₂/2πr; attractive when the currents are in the same direction (Fleming's left-hand rule), repulsive when opposite.

3 marks: “Derive the torque on a rectangular coil of N turns in a uniform magnetic field. When is it largest?” Forces on the sides of length b cancel; the sides of length L feel NBIL, forming a couple with arm b sin θ; τ = NBIA sin θ = MB sin θ; largest when the plane of the coil is parallel to B (θ = 90°).

Remember it

Like currents attract, unlike currents repel: the opposite of the rule for charges. Torque τ = NBIA sin θ.