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Magnetic Fields Due to Currents

Key idea: Use standard results for B due to a long straight wire, circular coil centre and long solenoid, and solve B and force-per-length questions (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Calculate and represent magnetic fields produced by currents.
  • Sketch magnetic field lines due to currents in a long straight wire, a flat circular coil and a long solenoid.
  • Analyse forces on current-carrying conductors, current balances and interactions between parallel currents.

1. Definitions (Must Know)

A. Magnetic field (due to currents)

A magnetic field is a field of force produced by current-carrying conductors (and also by permanent magnets).

B. Field patterns you must be able to sketch

  • Long straight wire: concentric circles centred on the wire.
  • Flat circular coil: field lines concentrate through the centre of the coil.
  • Long solenoid: nearly uniform field inside (straight, parallel, equally spaced lines); outside field is weaker and loops back.

C. Magnetic flux density due to a long straight wire

At distance r from a long straight wire carrying current I:

B = (μ₀ I)/(2π r)

D. Magnetic flux density at the centre of a flat circular coil

For a coil of N turns and radius r carrying current I:

B = (μ₀ N I)/2r

E. Magnetic flux density inside a long solenoid

For a long solenoid (inside the solenoid):

B = μ₀ n I

where n is turns per unit length (m⁻¹).

F. Force between two parallel current-carrying wires

Two long parallel wires separated by distance d carrying currents I₁ and I₂ exert forces on each other. The force per unit length on each wire is:

F/l = (μ₀ I₁ I₂)/(2π d)

  • same current direction → attractive
  • opposite current directions → repulsive

2. Key Ideas (What Earns Marks)

  • Use the correct result for the geometry (wire / coil / solenoid).
  • For a straight wire, B ∝ 1/r: doubling r halves B.
  • Direction of field lines around a current-carrying wire: use the right-hand grip rule.
  • A ferrous core inside a solenoid increases B (the core becomes magnetised and reinforces the field).
  • Parallel wires: same current directions attract; opposite directions repel.
Syllabus scope (9478)

This lesson covers A Level 9478 learning outcomes 17a–17d and 17i.

3. Detailed Explanations

A. Right-hand grip rule (direction)

Point your right thumb along conventional current; your curled fingers show the magnetic field direction. Apply the rule separately to each current element in a coil or solenoid.

Magnetic field patterns produced by currentsThree labelled diagrams show concentric magnetic field lines around a straight wire, the axial field through a circular coil, and the nearly uniform field inside a long solenoid.Long straight wirecurrent out of pageconcentric circles; B decreases with rFlat circular coilB at centrefield is strongest through the centreLong solenoidinside: parallel, equally spaced field lines
Scroll diagram horizontally to read all labels.
Use the right-hand grip rule for every current direction. Inside a long solenoid, the closely spaced parallel lines represent an approximately uniform field.

B. Why the field gets weaker with distance from a straight wire

For a straight wire, B ∝ 1/r. As you move further away, the same “circulating” field is spread over a larger circumference, so the field strength decreases.

Long straight wire: B decreases as 1/r (scaled)

A 1/r curve showing how magnetic flux density decreases with distance from a long straight current-carrying wire.

Scroll across the graph to read all labels.

A 1/r curve showing how magnetic flux density decreases with distance from a long straight current-carrying wire.A 1/r curve showing how magnetic flux density decreases with distance from a long straight current-carrying wire.
Doubling the distance halves the flux density. This is a common ratio-method exam question.
Open full-size graph
View figure data
Values for Long straight wire: B decreases as 1/r (scaled)
Distance from wire (r / R)B ∝ 1/r
11
20.5
30.333
40.25
50.2
60.1667

C. Long solenoid field and ferrous cores

Inside a long solenoid, field lines are nearly straight, parallel and evenly spaced, so B is approximately uniform away from the ends.

Adding a ferrous core increases the flux density because the core magnetises and strengthens the field.

D. Why parallel currents attract/repel (direction reasoning)

Each wire produces a magnetic field, and the other wire (a current-carrying conductor) experiences a force in that field.

You only need the rule:

  • same direction currents → attract
  • opposite direction currents → repel

4. Common Mistakes

  • Using cm/mm for r or d without converting to metres.
  • Mixing symbols: r (distance from wire / coil radius) vs d (wire separation).
  • Using the solenoid formula for a short coil (the “long solenoid” approximation).
  • Drawing field lines without direction arrows.

5. Exam Tips

  • Label r, d, N, and n on the diagram before substituting.
  • If asked “how does B change?”, use proportional reasoning first (e.g. B ∝ 1/r).
  • When sketching a solenoid field, make the inside lines denser than the outside lines (stronger inside).

6. Worked Examples

Modelled example 1

Straight wire: find B at a distance

Core

Problem

A long straight wire carries 6.0 A. Find B at 4.0 cm from it. Take μ₀ = 4π × 10⁻⁷ H m⁻¹.
Study the worked solution

  1. Choose the geometry

    Method

    Use B = μ₀I/(2π r).

    Reason

    The source is specified as a long straight wire.

    Working

    B = μ₀I/(2π r)

  2. Convert distance

    Method

    Use r = 0.040 m.

    Reason

    The permeability value is in SI units.

    Working

    4.0 cm = 0.040 m.

  3. Evaluate

    Method

    Obtain 3.0 × 10⁻⁵ T.

    Reason

    The equation gives field magnitude; direction follows the right-hand grip rule.

    Working

    B = ((4π × 10⁻⁷)(6.0))/2π(0.040) = 3.0 × 10⁻⁵ T

Guided practice 2

Circular coil: field at the centre

About 5 min

Problem

A flat circular coil has 20 turns, radius 5.0 cm and current 2.0 A. Find the field at its centre.

Try this before viewing the solution

Unit: T

Hints

Hint 1: select the coil formula
Use B = μ₀NI/(2r), not the long-solenoid expression involving turns per metre.
View solution step by step

  1. Identify the turn quantity

    Method

    Use total turns N = 20.

    Reason

    The centre-field formula for a flat coil sums the field from every turn.

    Working

    B = μ₀NI/2r

  2. Substitute and evaluate

    Method

    Obtain 5.0 × 10⁻⁴ T.

    Reason

    The radius is 0.050 m.

    Working

    B = ((4π × 10⁻⁷)(20)(2.0))/2(0.050) = 5.0 × 10⁻⁴ T

Common misconception 3

Parallel wires: force per unit length

Find and correct the mistake

Learner claim

Two long wires 3.0 mm apart carry 10 A and 6.0 A in the same direction. A learner says like current directions repel. Diagnose the direction and find F/l.

Try this before viewing the solution

Force direction

View solution step by step

  1. Calculate force per length

    Method

    Obtain 4.0 × 10⁻³ N m⁻¹.

    Reason

    The named long-parallel-wire result includes both currents and their separation.

    Working

    F/l = ((4π × 10⁻⁷)(10)(6.0))/(2π(3.0 × 10⁻³)) = 4.0 × 10⁻³ N m⁻¹

  2. Determine direction

    Method

    The wires attract.

    Reason

    Same-direction parallel currents produce forces towards each other.

    Working

    Same directions ⇒ attraction.

Examiner practice 4

Long solenoid: field inside

3 marks

Examination question

A long air-core solenoid has 800 turns over 0.40 m and carries 1.5 A. Calculate its turn density and internal magnetic flux density. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Calculate turn density

    1 mark

    Method

    n = 2.0 × 10³ m⁻¹.

    Reason

    The long-solenoid equation uses turns per unit length, not total turns alone.

    Working

    n = N/L = 800/0.40 = 2.0 × 10³ m⁻¹

  2. Select the solenoid result

    1 mark

    Method

    Use B = μ₀nI.

    Reason

    The stated long air-core geometry supports this model.

    Working

    B = μ₀nI

  3. Evaluate

    1 mark

    Method

    B = 3.77 × 10⁻³ T ≈ 3.8 × 10⁻³ T.

    Reason

    Substitute the calculated turn density and current.

    Working

    B = (4π × 10⁻⁷)(2.0 × 10³)(1.5) = 3.77 × 10⁻³ T

Challenge 5

Force between two wires (find F)

Minimal support

Independent transfer

Two long parallel wires 5.0 mm apart carry 8.0 A in the same direction. Find the magnitude and direction of the force on a 0.20 m length of either wire.

Try this before viewing the solution

Hints

Hint 1: calculate density before total force
First find F/l, then multiply by the specified active length.
View solution step by step

  1. Find force per length

    Method

    F/l = 2.56 × 10⁻³ N m⁻¹.

    Reason

    The long-wire interaction is naturally expressed per unit length.

    Working

    F/l = ((4π × 10⁻⁷)(8.0)(8.0))/(2π(5.0 × 10⁻³)) = 2.56 × 10⁻³ N m⁻¹

  2. Find force on the segment

    Method

    F = 5.12 × 10⁻⁴ N.

    Reason

    Multiply the force per metre by 0.20 m.

    Working

    F = (2.56 × 10⁻³)(0.20) = 5.12 × 10⁻⁴ N

  3. State direction

    Method

    The force on each segment is towards the other wire.

    Reason

    The currents are in the same direction, so the interaction is attractive.

    Working

    Direction: attraction.

7. Mind Stretchers

Mind stretcher 1: Doubling distance from a wireExtension

For a long straight wire with fixed current, what happens to B if you double the distance from the wire?

Show Answer

For a straight wire, B ∝ 1/r. Doubling r halves B.

Mind stretcher 2: Link F/l to B (synthesis)Extension

Show that the force per unit length between two parallel wires can be written as F/l = I₂B₁, where B₁ is the magnetic field produced by wire 1 at the location of wire 2.

Show Answer

At the position of wire 2, the field due to wire 1 is: B₁ = (μ₀ I₁)/(2π d)

Wire 2 experiences force on length l given by F = BIl (perpendicular case), so: F = B₁I₂l ⇒ F/l = I₂B₁

Substituting for B₁ gives: F/l = (μ₀ I₁ I₂)/(2π d)

8. Optional (Enrichment)

A. Beyond these three geometries

More complex current geometries are handled using Biot–Savart or Ampère’s law. In this syllabus, you use the provided results for straight wire, circular coil centre, and long solenoid.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027