Interactive practical

Force on a current-carrying wire

Use a top-pan balance and Hall probe to test how magnetic force depends on current, conductor length and magnetic flux density.

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A-Level Physics

Force on a current-carrying wire

WJEC 1420QS — matched to your course.

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At a glance

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Practical focuses1
Questions available2
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Available question marks4

Practical focus

Practical method

Aim

Investigate how the force on a current-carrying wire depends on current and determine the magnetic flux density B.

Variables

Independent variable: current I
Dependent variable: force F (from the balance reading change)
Control variables: length of wire in the field; the same magnet; wire perpendicular to the field.

Apparatus

Method

  1. Place the magnet on the balance and zero (tare) the reading.
  2. Clamp the wire horizontally between the poles, perpendicular to the field, without touching the magnet.
  3. Pass currents from 1.0 A to 5.0 A and record the change in balance reading.
  4. Force on the wire is equal and opposite to the force on the magnet: F = Δm × g.
  5. Plot F against I: gradient = BL, where L is the length of wire in the field.

Sample results

Length of wire in field L = 0.050 m.

I / ABalance change / gF / mN
1.00.2552.50
2.00.5105.00
3.00.7657.50
4.01.01910.0
5.01.27412.5

Analysis

F is proportional to I with gradient 2.50 × 10⁻³ N A⁻¹ = BL, so B = \(\dfrac{2.50 \times 10^{-3}}{0.050}\) = 0.050 T.

Uncertainty

The balance reads to 0.001 g; at 1.0 A this is 0.4%. The length of wire in the field is uncertain because the field spreads beyond the poles (about ±2 mm in 50 mm, 4%) — the largest uncertainty.

Errors and improvements

ProblemEffectImprovement
Field not uniform beyond the pole edgesEffective L uncertainMeasure L as the pole width and estimate the uncertainty
Wire heatingResistance changes; supply adjustmentsSwitch off between readings
Wire not perpendicular to fieldSmaller forceAlign carefully

Risk assessment

HazardRiskControl measure
Hot wire at high currentBurnsShort readings and switch off between them
Strong magnetsPinched fingersHandle carefully

Common mistakes

Worked analysis

At 4.0 A, Δm = 1.019 g, so F = 1.019 × 10⁻³ × 9.81 = 1.00 × 10⁻² N. B = \(\dfrac{F}{IL}\) = \(\dfrac{0.0100}{4.0 \times 0.050}\) = 0.050 T.

Common misconception

The balance reading changes because of Newton's third law: the magnet experiences an equal and opposite force to the wire.

Assessment coverage

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2 questions available, worth 4 marks in total.

Practical 2
Foundation 1Standard 1
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Retrieval cards

Force on a current in a field?

F = BIL (perpendicular).

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