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Magnetic Dipoles

Key idea: Model magnetic dipoles as current loops, compute μ=IA, and use torque and potential energy in a uniform magnetic field, with worked examples.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • show an understanding that ideal conductors form an equipotential volume, and that the electric field within an ideal conductor is zero
  • show an understanding that electric charge accumulates on the surfaces of a conductor, and that the electric field at the surface of a conductor is normal to the surface
  • recall and apply Gauss’s law 6 for electric and magnetic fields (knowledge of the differential form of Gauss’s law is not required), and
  • recall and apply Ampère’s law 7 relating the line integral of the magnetic field (in a vacuum) around a closed loop with the electric current enclosed by the loop to solve problems involving symmetric field configurations (knowledge of the differential form of Ampère’s law is not required) [Note further that candidates are not required to know Maxwell’s generalisation of Ampère’s law including the term related to the rate of change of electric flux, nor the Biot-Savart law.]
  • define the magnitude of the electric dipole moment as the product of the charge and the separation
  • show an understanding of and use the torque on an electric dipole and the potential energy of an electric dipole to solve related problems
  • define the magnitude of the magnetic dipole moment for a current loop as the product of the current and the area of the loop
  • show an understanding of and use the torque on a magnetic dipole and the potential energy of a magnetic dipole to solve related problems
  • solve problems involving symmetric charge distributions by relating the electric flux (in a vacuum) through a closed surface with the charge enclosed by that surface (ii) show an understanding that the magnetic flux through a closed surface is always zero, suggesting the non-existence of magnetic monopoles

A magnetic dipole can be modelled at this level as a small current loop. It behaves analogously to an electric dipole in torque and energy, while the syllabus framework does not admit magnetic monopoles.

1. Definitions (Must Know)

  • Magnetic dipole moment (magnitude) for a current loop: μ = IA where I is current and A is the area of the loop.
  • For a coil of N identical turns, the moments add: μ = NIA.
  • Magnetic dipole moment vector, vector μ: direction is given by the right-hand rule (curl fingers with current; thumb points along vector μ).
  • Torque on a magnetic dipole in a uniform magnetic field (magnitude): τ = μ B sin θ where θ is the angle between vector μ and vector B.
  • Potential energy of a magnetic dipole in a uniform field: U = - vector μ · vector B = -μ B cos θ
  • Symbols used in this lesson: I (A), A (m²), μ (A m²), vector B (T), τ (N m), U (J), θ (rad).

2. Key Ideas (What Earns Marks)

  • A current loop in a uniform vector B experiences a torque that tends to align vector μ with vector B.
  • Stable equilibrium is aligned (θ = 0), unstable is anti-aligned (θ = π).
  • The analogy with electric dipoles is strong for torque and energy, but the model has no isolated magnetic charges: magnetic field lines form closed loops.

Quick comparison (uniform fields):

Dipole typeDipole momentTorque magnitudePotential energy
Electricp = qdτ = pE sin θU = -pE cos θ
Magneticμ = IAτ = μ B sin θU = -μ B cos θ
Tilted current loop in a uniform magnetic field with magnetic dipole moment normal to the loop according to the right-hand rule
Curl the right-hand fingers with the conventional current; the thumb gives the loop's magnetic dipole moment. For multiple identical turns, use μ = NIA.

3. Detailed Explanations

A. Why a current loop has a dipole moment

Currents in a loop produce a magnetic field pattern similar (at distances large compared to loop size) to the field of a magnetic dipole. The loop is therefore characterised by a single vector vector μ with magnitude μ = IA.

B. Torque and potential energy in uniform fields

In a uniform field, the dipole experiences: τ = μ B sin θ and potential energy: U(θ) = -μ B cos θ

Energy picture:

  • θ = 0 ⇒ Uₘᵢₙ = -μ B (stable),
  • θ = π ⇒ Uₘₐₓ = +μ B (unstable).

C. The key limitation of the analogy (no monopoles)

Electric dipoles are made of + q and -q, so isolated charges exist. For magnetism at this level, there is no evidence of isolated “north” or “south” poles acting as monopoles; instead: ∮ vector B · d vector A = 0 and magnetic field lines do not begin or end.

4. Common Mistakes

  • Mixing up symbols: μ (dipole moment) vs μ₀ (vacuum permeability).
  • Using μ = I/A (wrong); it is μ = IA.
  • Forgetting the right-hand rule direction for vector μ.
  • Dropping the sin θ in torque or the minus sign in energy.

5. Exam Tips

  • Sketch the loop and use the right-hand rule to label vector μ.
  • Use torque for “turning” questions, energy for “stable/unstable” and “work done to rotate”.
  • When asked about analogy limits, mention “no magnetic monopoles / field lines form closed loops”.

6. Worked Examples

Modelled example 1

Magnetic dipole moment of a loop

Core

Problem

A circular loop carries current I = 0.80 A and has radius r = 0.050 m. Find its magnetic dipole moment.
Study the worked solution

  1. Find loop area

    Method

    A = 7.9 × 10⁻³ m².

    Reason

    The dipole relation uses area, not radius.

    Working

    A = π r² = π(0.050)² = 7.9 × 10⁻³ m²

  2. Apply the dipole relation

    Method

    μ = 6.3 × 10⁻³ A m².

    Reason

    For a single-turn loop, μ = IA.

    Working

    μ = (0.80)(7.9 × 10⁻³) = 6.3 × 10⁻³ A m²

Guided practice 2

Torque in a uniform magnetic field

About 5 min

Problem

A dipole has μ = 3.0 × 10⁻² A m² in a uniform field B = 0.40 T. The angle between vector μ and vector B is 60°. Find the torque magnitude.

Try this before viewing the solution

Hints

Hint 1: identify the angle
Use the angle between the dipole-moment vector and the field.
Hint 2: choose the relation
For torque magnitude, use τ = μ B sin θ.
View solution step by step

  1. Choose the torque relation

    Method

    τ = μ B sin θ.

    Reason

    Torque is the turning effect in a uniform magnetic field.

    Working

    τ = μ B sin 60°

  2. Evaluate

    Method

    τ = 1.0 × 10⁻² N m.

    Reason

    All quantities are already in SI units.

    Working

    τ = (3.0 × 10⁻²)(0.40) sin 60° = 1.0 × 10⁻² N m

Common misconception 3

Potential energy difference

Find and correct the mistake

Learner claim

A dipole with μ B = 0.12 J rotates from 0° to 90°. A learner obtains Δ U = -0.12 J by subtracting in the wrong order. Explain the sign error and find the correct energy change.

Try this before viewing the solution

Correct change in potential energy

View solution step by step

  1. Find initial energy

    Method

    Uᵢ = -0.12 J.

    Reason

    U = -μ B cos θ and cos 0° = 1.

    Working

    Uᵢ = -0.12 cos 0° = -0.12 J

  2. Find final energy

    Method

    U_f = 0.

    Reason

    cos 90° = 0.

    Working

    U_f = -0.12 cos 90° = 0

  3. Use final minus initial

    Method

    Δ U = +0.12 J.

    Reason

    Energy change is defined as U_f-Uᵢ.

    Working

    Δ U = 0-(-0.12) = +0.12 J

Examiner practice 4

Required current for a target dipole moment

3 marks

Examination question

A single-turn rectangular loop has area A = 2.5 × 10⁻³ m². The required magnetic dipole moment is μ = 1.0 × 10⁻² A m². Find the current. [3 marks]

Try this before viewing the solution

View solution step by step

  1. State the relation

    1 mark

    Method

    μ = IA for a single-turn loop.

    Reason

    The loop has one turn, so no factor of N is needed.

    Working

    μ = IA

  2. Rearrange and substitute

    1 mark

    Method

    I = μ/A.

    Reason

    Current is the required unknown.

    Working

    I = (1.0 × 10⁻²)/(2.5 × 10⁻³)

  3. Report the current

    1 mark

    Method

    I = 4.0 A.

    Reason

    The ratio has units of amperes.

    Working

    I = 4.0 A

Challenge 5

Earth’s field torque (order-of-magnitude)

Minimal support

Independent transfer

Take Earth’s magnetic field as B = 50 μT = 5.0 × 10⁻⁵ T. A current loop has μ = 0.020 A m² and is oriented at 90° to vector B. Find the torque magnitude and identify why this orientation sets the maximum torque.

Try this before viewing the solution

Hints

Hint 1: use the angular factor
Evaluate sin 90° before multiplying.
View solution step by step

  1. Use the perpendicular orientation

    Method

    sin 90° = 1.

    Reason

    This is the largest possible value of the angular factor.

    Working

    τₘₐₓ = μ B

  2. Calculate

    Method

    τ = 1.0 × 10⁻⁶ N m.

    Reason

    The microtesla field has already been converted to tesla.

    Working

    τ = (0.020)(5.0 × 10⁻⁵) = 1.0 × 10⁻⁶ N m

  3. Interpret

    Method

    The loop experiences a small but maximum-magnitude aligning torque at this angle.

    Reason

    Torque is greatest when vector μ is perpendicular to vector B.

    Working

    τ = μ B sin θ

7. Mind Stretchers

Mind stretcher 1: Why “north” and “south” can’t be separated (classical picture)Extension

If you cut a bar magnet in half, do you get an isolated north pole and an isolated south pole? What do you actually get, and how does this connect to Gauss’s law for magnetism?

Answer

You get two smaller magnets, each with both a north and south pole. This is consistent with ∮ vector B · d vector A = 0: magnetic field lines form closed loops and there are no monopoles in this model.

Mind stretcher 2: Why a dipole can be pulled into a strong-field regionExtension

A small bar magnet is attracted to a region of stronger magnetic field (e.g. near a magnet’s pole). Explain qualitatively why this is possible even if the magnet is already aligned (so the torque is near zero).

Answer

Even when aligned (small torque), a dipole in a non-uniform field can experience a net force because its potential energy U ≈ -μ B depends on position through B. The system tends to move toward lower potential energy, i.e. toward regions of larger B when aligned.

8. Optional/Enrichment: Force on a Dipole in a Non-uniform Field

In a non-uniform field, a dipole can experience a net force (basis of attraction to strong-field regions). This is beyond the uniform-field torque/energy learning outcomes unless the question provides the necessary field variation details.

Next step

Return to the Electric and Magnetic Fields hub for mixed revision, then continue to RLC Circuits.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027