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Magnetic fields and force on a conductor

Key idea: Currents and permanent magnets produce magnetic fields. A conductor carrying current across a field can experience a force, allowing magnetic flux density to be defined and measured.

  • GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 2 of 3

Check your understanding

By the end of this lesson, you should be able to

  • Sketch fields around a straight wire, flat coil and long solenoid.
  • Use F = BIℓ sin θ and Fleming’s left-hand rule.
  • Define magnetic flux density and explain a current-balance measurement.

Learn the idea

Big question: How do current, magnetic field and force fit together without mixing up the two hand rules?

Construct fields made by currents

Magnetic fields are force fields produced by permanent magnets as well as by current-carrying conductors. Field-line arrows outside a permanent magnet run from its north pole towards its south pole.

Use the right-hand grip rule for a current's own magnetic field: thumb follows conventional current and curled fingers show the field. A straight wire has concentric circles; a flat coil has a bar-magnet-like pattern; a long solenoid has nearly parallel, equally spaced internal lines.

Closer field-line spacing represents a stronger field qualitatively. Reversing current reverses every field arrow but does not change the basic pattern.

Check your understanding: What happens to a solenoid's north and south ends when current reverses?

They swap because the magnetic field direction reverses.

Find force in an external field

A length ℓ of current-carrying conductor in an external field experiences F = BIℓ sinθ. The force is maximum when current and field are perpendicular and zero when parallel.

Use Fleming's left-hand rule for force: first finger field, second finger conventional current, thumb force. This is a different task from the right-hand grip rule. For a perpendicular conductor, B = F/(Iℓ) defines magnetic flux density; 1 T = 1 N A⁻¹ m⁻¹.

Check your understanding: Which length belongs in F = BIℓ?

Only the conductor length inside the magnetic field.

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.
Measuring magnetic flux density with a current balanceA horizontal wire segment of known length lies perpendicular to a magnetic field between pole pieces. Current through the wire produces a vertical magnetic force, recorded as a change in balance reading.NScurrent out of pageFbalance: reading changes by Δmknown field length ℓ
Scroll diagram horizontally to read all labels.
Convert the balance-reading change Δm to force Δmg. With the wire perpendicular to the field, B = F/(Iℓ), where ℓ is only the length inside the field.

Key ideas

  • Do not mix the right-hand grip rule with Fleming’s left-hand rule.
  • ℓ is the conductor length inside the magnetic field.
  • Magnetic flux density has unit tesla, T = N A⁻¹ m⁻¹.

Relationships to know

  • F = BIℓ sin θ
  • for θ = 90°, B = F/(Iℓ)

Follow the reasoning

Worked example

Determine flux density with a current balance

Question: A 0.060 m wire segment is perpendicular to a magnetic field. Switching on 3.0 A changes a balance reading by 0.45 g. Find B and explain why reversing current reverses the reading change. Use g = 9.81 m s⁻².

  1. Step 1: Convert balance change to force

    Why: The balance measures an equivalent weight change, not magnetic force directly.

    Working: F = Δmg = 0.00045(9.81) = 4.41 × 10⁻³ N.

  2. Step 2: Use the perpendicular definition

    Why: sin90° = 1 for the field-exposed length.

    Working: B = F/(Iℓ) = 4.41 × 10⁻³/[3.0(0.060)] = 2.45 × 10⁻² T.

  3. Step 3: Apply the direction rule

    Why: Fleming's rule contains current as one of the perpendicular directions.

    Working: Reversing I with B unchanged reverses F, so the balance change changes sign.

Answer: B = 2.45 × 10⁻² T; reversing current reverses the magnetic force and hence the balance deflection.

Check: The unit N/(A m) is tesla, and the measured force is only a few millinewtons, consistent with the gram-scale change.

Now try it with support

Practise with support

A current balance shows a mass-reading change of 0.60 g when 2.5 A flows through 0.080 m of wire perpendicular to the field. Find B.

Hints

  1. Convert 0.60 g to kg and then to force using g.
  2. Use B = F/(Iℓ).
View the guided answer

F = 0.00060(9.81) = 5.89 × 10⁻³ N. B = 5.89 × 10⁻³/[2.5(0.080)] = 2.94 × 10⁻² T.

Your turn

Practise independently

A 0.15 m wire carries 3.0 A perpendicular to a 0.40 T field. Find the force and determine its direction for current east and field north.

Check your answer

F = 0.40(3.0)(0.15) = 0.18 N. With current east and field north, Fleming’s left-hand rule gives force vertically upward.

Common mistakes and exam guidance

Watch out for

  • Using the current direction as the magnetic-field direction.
  • Forgetting the sin θ factor or using the full wire length when only part lies in the field.

In an exam

  • Draw the three perpendicular arrows before using a hand rule.
  • In a current-balance question, convert the balance-reading change to force before calculating B.

Put the ideas together

Exam-style practice [8 marks]

A 0.12 m conductor carrying 4.0 A lies in a 0.35 T field at 30° to the field. Find the force. Describe a current-balance experiment to measure B and state two ways to improve its reliability.

Plan before you answer

  • Use the angle between current and field.
  • Turn balance-reading change into force.
  • Include reversal and repetition in the method.
View the marking points and model answer

Marking points

  1. Uses F = BIℓsinθ.
  2. Obtains F = 0.084 N.
  3. Places a known conductor length in a uniform field/perpendicular for measurement.
  4. Measures current and balance change.
  5. Converts mass change using Δmg.
  6. Uses B = F/(Iℓ).
  7. Reverses current and uses half the difference/averages opposite readings.
  8. Repeats at several currents or zeros the balance and plots force against I.

Model answer

F = 0.35(4.0)(0.12)sin30° = 0.084 N. For a current balance, place a measured wire length perpendicular to the field on the balance assembly, zero the balance, pass a measured current and record the mass-reading change. Convert it using F = Δmg and calculate B = F/(Iℓ). Reverse current and use half the difference between readings to reduce zero effects; repeat for several currents and use the gradient of F against I.

Finish from memory

Three-question recap

  1. Which rule gives the field around a current?

    Check

    The right-hand grip rule.

  2. Which rule gives force on a current in a field?

    Check

    Fleming's left-hand rule.

  3. When is magnetic force on a conductor zero?

    Check

    When current is parallel or antiparallel to the field.

Try this next

Replace the current element by one moving charge and track how charge sign affects force direction.

Continue to Force and deflection of moving chargesReturn to ElectromagnetismReview Motion in a uniform electric field

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H1 Physics
Edition
GCE A-Level H1 Physics 2027