Conductor force, flux density, current balance and parallel currents
Key idea: H2 Physics lessons on current-produced fields, magnetic forces and crossed-field velocity selection.
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The core idea
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Big question: How does a magnetic field push a current-carrying conductor?
A conductor of length l carrying current I in flux density B experiences F = BIl sinθ. Fleming's left-hand rule gives direction. A current balance can measure B, while parallel currents attract if they flow in the same direction and repel if opposite because each wire lies in the other's field.
Apply the force equation to the field-exposed length
A conductor carrying current I through an external flux density B experiences F = BIl sinθ, where θ is between current and field and l is only the length within the field. For a perpendicular conductor, B = F/(Il): magnetic flux density is force per unit current per unit length.
Force is maximum when current is perpendicular to B and zero when parallel. Use Fleming's left-hand rule for force; use the separate right-hand grip rule only to find a current's own field.
Check your understanding: A wire lies at 30° to B. What fraction of maximum force acts?
sin 30° = 1/2.
Measure B and explain paired-wire forces
In a current balance, convert the change in balance reading to magnetic force with Δmg, then use B = F/(Il) for a perpendicular conductor. Reverse current and use the difference between opposite readings to reduce zero error.
Two parallel currents exert equal and opposite forces. Same-direction currents attract and opposite-direction currents repel; this follows from each wire's field acting on the other, not from two unrelated rules.
Check your understanding: Why reverse current in a balance experiment?
The magnetic force reverses while many offsets do not, so half the difference improves the force estimate.
Key ideas to keep
- θ is between current direction and magnetic field.
- No magnetic force acts when the conductor is parallel to the field.
- The forces between parallel wires form an equal and opposite pair.
See the reasoning
Worked example
Explain and calculate the force between parallel currents
Question: Two long parallel wires 5.0 cm apart carry 6.0 A in the same direction. Find force per unit length and explain its direction.
Step 1: Find one wire's field at the other
Why: Each wire experiences the magnetic field produced by its neighbour.
Working: B = μ₀I/(2πd) = 2.4×10⁻⁵ T.
Step 2: Use conductor force per length
Why: The second current is perpendicular to the circular field there.
Working: F/l = BI = 1.44×10⁻⁴ N m⁻¹.
Step 3: Determine direction
Why: The grip rule plus conductor-force rule gives the interaction.
Working: Currents in the same direction attract; the forces are equal and opposite.
Answer: Field from one at the other is μ₀I/(2πd) = 2.4 × 10⁻⁵ T. F/l = BI = 1.44 × 10⁻⁴ N m⁻¹. The right-hand grip rule plus conductor-force direction gives attraction for currents in the same direction.
Check: The result is force per unit length because no active length was specified.
Use a hint if needed
Practise with support
Try this
A wire turns from 90° to 30° relative to B at fixed B, I and l. State the force factor.
Hint: θ is the angle between current direction and field.
Check your answer
F ∝ sin θ, so the factor is sin30°/sin90° = 1/2.
Now work without the hint
Practise independently
Your turn
Define magnetic flux density operationally and explain how field superposition produces forces between parallel currents.
Check your answer
For a conductor perpendicular to B, flux density is force per unit current per unit length: B = F/(Il). Each wire produces a circular field at the other; applying F = BIl and the direction rule gives attraction for same-direction currents and repulsion for opposite currents.
Avoid these traps
Common mistakes
Common mistake
Magnetic force is parallel to current.
What is wrong with this reasoning?
Show better thinking
F = BIl sin θ acts perpendicular to both conventional current and B; use Fleming's left-hand rule for direction.
Common mistake
Parallel currents in the same direction repel.
What is wrong with this reasoning?
Show better thinking
Same-direction currents attract; opposite-direction currents repel.
Write for the examiner
Exam guidance
State field, current and force directions in three dimensions before using a hand rule.
Exam-style practice [6 marks]
A current balance uses l = 0.080 m and I = 2.5 A; rebalancing needs 3.06 g. Find B using g = 9.81 m s⁻².
Plan before you answer
- Convert the balance mass to force.
- Use only the field-exposed length.
- Solve B = F/(Il).
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
F = Δmg = 0.00306(9.81) = 0.0300 N. B = F/(Il) = 0.0300/[2.5(0.080)] = 0.150 T.
Come back in three days
Check what stayed with you
Recall question
A perpendicular wire's current and active length both double. State the force factor.
Check the answer
F = BIl, so the force increases by 2 × 2 = 4.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 17 states no explicit exclusions. Straight-wire, flat-coil and long-solenoid field equations use their named ideal geometries; a ferrous core changes the air-core model. In F = BIl sin θ, θ is between conventional current and B; in F = BQv sin θ, direction is determined for positive charge then reversed for negative charge. Magnetic force does no work. Circular-path formulae require velocity perpendicular to a uniform B. Velocity selection requires uniform mutually perpendicular E, B and beam velocity with opposing forces.
- GCE A-Level H2 PhysicsTopic 17(e) / Topic 17(f) / Topic 17(g) / Topic 17(h) / Topic 17(i) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027