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Current Balance (Measuring Magnetic Flux Density)

Key idea: Use a current balance to measure magnetic flux density B from mass changes, including reversed-current readings, correct directions and units (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Analyse forces on current-carrying conductors, current balances and interactions between parallel currents.

1. Definitions (Must Know)

  • Magnetic flux density, B (T): for a wire perpendicular to the field, B = F/Il where F is the magnetic force on the wire, I is the current and l is the length of wire in the field.
  • Force on a current-carrying conductor in a uniform magnetic field:
    • F = BIl sin θ
    • θ is the angle between the current direction and the magnetic field direction.
  • Current balance: an apparatus that measures B by measuring the force on a current-carrying conductor placed in a magnetic field.

2. Key Ideas (What Earns Marks)

  • Use Fleming’s left-hand rule to get the direction of the magnetic force.
  • Identify the effective length, l: only the portion of wire actually inside the (approximately uniform) field region.
  • Most current-balance questions use the perpendicular case (θ = 90°):
    • F = BIl ⇒ B = F/Il
  • The balance measures an apparent mass change, Δ m:
    • F = Δ m g
  • If you are given two balance readings with the current reversed:
    • F = ((m_down-mᵤₚ)g)/2

3. Detailed Explanations

A. What a current balance measures

In a current balance, a straight wire segment is placed between the poles of a magnet so that it experiences a magnetic force when current flows.

  • If the magnetic force is downward, the balance reading increases.
  • If the current is reversed, the magnetic force reverses and the balance reading decreases.

The key physics is that the balance converts a force into a mass-equivalent reading: F = Δ m g

B. Measuring B from the balance reading

For a wire perpendicular to the magnetic field: F = BIl

Combine with F = Δ m g: B = (Δ m g)/Il

C. If the wire is not perpendicular

Use the general form: F = BIl sin θ so: B = F/(Il sin θ)

4. Common Mistakes

  • Using Δ m directly as a force (forgetting to multiply by g).
  • Using the total length of wire instead of the length inside the field.
  • Forgetting the sin θ factor when the wire is not perpendicular to the field.
  • Mixing up units: g vs kg, cm vs m.

5. Exam Tips

  • State the perpendicular case explicitly: “wire is perpendicular, so sin θ = 1”.
  • If directions are asked, sketch the three perpendicular directions: current, field, force.
  • If readings are given for both current directions, use the half-difference to eliminate zero offset.

6. Worked Examples

Modelled example 1

Find B from a single mass change

Core

Problem

A perpendicular 0.12 m wire carries 2.0 A in a uniform field. The balance reading increases by 3.0 g. Find B.
Study the worked solution

  1. Convert the balance change

    Method

    Use Δ m = 3.0 × 10⁻³ kg.

    Reason

    The measured gram change is a mass-equivalent reading, not yet a force.

    Working

    3.0 g = 3.0 × 10⁻³ kg

  2. Convert mass change to force

    Method

    F = 2.94 × 10⁻² N.

    Reason

    The magnetic force produces the apparent weight change.

    Working

    F = Δ mg = (3.0 × 10⁻³)(9.81) = 2.94 × 10⁻² N

  3. Infer flux density

    Method

    B = 1.23 × 10⁻¹ T ≈ 0.12 T.

    Reason

    The wire is perpendicular, so F = BIl.

    Working

    B = F/Il = (2.94 × 10⁻²)/(2.0)(0.12) = 0.123 T

Guided practice 2

Using two readings (current reversed)

About 6 min

Problem

A balance reads 0.523 kg for downward magnetic force and 0.517 kg after current reversal. The active length is 0.15 m and current is 4.0 A. Find B.

Try this before viewing the solution

Hints

Hint 1: remove the common baseline
The reading difference represents twice the magnetic mass-equivalent change.
View solution step by step

  1. Use the half-difference

    Method

    F = 2.94 × 10⁻² N.

    Reason

    Reversal changes + F to -F, so subtracting readings produces a 2F effect and cancels the common offset.

    Working

    F = ((0.523-0.517)(9.81))/2 = 2.94 × 10⁻² N

  2. Calculate flux density

    Method

    B = 4.90 × 10⁻² T.

    Reason

    The active wire is perpendicular to the field.

    Working

    B = (2.94 × 10⁻²)/(4.0)(0.15) = 4.90 × 10⁻² T

Common misconception 3

Wire at an angle

Find and correct the mistake

Learner claim

A 0.050 m wire carries 3.0 A at 30° to a 0.80 T field. A learner uses F = BIl because the wire is inside the field. Diagnose and calculate.

Try this before viewing the solution

Unit: N

View solution step by step

  1. Restore the angle factor

    Method

    Use F = BIl sin 30°.

    Reason

    BIl alone applies only when current and field are perpendicular.

    Working

    F = BIl sin θ

  2. Evaluate

    Method

    F = 6.0 × 10⁻² N.

    Reason

    sin 30° = 0.50, so the force is half its perpendicular value.

    Working

    F = (0.80)(3.0)(0.050)(0.50) = 6.0 × 10⁻² N

Examiner practice 4

Predict the reading change for a given B

3 marks

Examination question

A perpendicular 0.10 m wire carries 3.0 A in a 0.25 T field. Find the magnetic force and apparent mass change for one current direction. Take g = 9.81 m s⁻². [3 marks]

Try this before viewing the solution

View solution step by step

  1. Calculate force

    1 mark

    Method

    F = 7.5 × 10⁻² N.

    Reason

    The wire is perpendicular, so F = BIl.

    Working

    F = (0.25)(3.0)(0.10) = 7.5 × 10⁻² N

  2. Convert to mass-equivalent

    1 mark

    Method

    Δ m = 7.64 × 10⁻³ kg.

    Reason

    The balance reading satisfies F = Δ mg.

    Working

    Δ m = F/g = 7.5 × 10⁻²/9.81 = 7.64 × 10⁻³ kg

  3. Express in grams

    1 mark

    Method

    Δ m ≈ 7.6 g.

    Reason

    Balance changes are commonly displayed in grams.

    Working

    7.64 × 10⁻³ kg = 7.64 g.

Challenge 5

Find the current from two readings

Minimal support

Independent transfer

A balance reads 0.502 kg and 0.498 kg for reversed current directions. The perpendicular active length is 0.080 m and B = 0.30 T. Find the current and explain why any constant balance offset cancels.

Try this before viewing the solution

Hints

Hint 1: extract force before current
Use half the reading difference to find F, then rearrange F = BIl.
View solution step by step

  1. Extract magnetic force

    Method

    F = 1.96 × 10⁻² N.

    Reason

    The magnetic contribution reverses sign, so the full reading difference is twice its mass-equivalent.

    Working

    F = ((0.502-0.498)(9.81))/2 = 1.96 × 10⁻² N

  2. Infer current

    Method

    I = 0.818 A ≈ 0.82 A.

    Reason

    For the perpendicular active length, I = F/(Bl).

    Working

    I = (1.96 × 10⁻²)/(0.30)(0.080) = 0.818 A

  3. Explain offset cancellation

    Method

    A constant zero offset appears equally in both readings and disappears on subtraction.

    Reason

    Only the magnetic term changes sign with current reversal.

    Working

    (m₀ + Δ m)-(m₀-Δ m) = 2Δ m.

7. Mind Stretchers

Mind stretcher 1: Example: Current needed to balance a weightExtension

A mass m is supported by a wire segment of length l inside a uniform magnetic field B, with the wire perpendicular to the field. Show that the current needed so the magnetic force balances the weight is I = mg/Bl.

Show Answer

Balance condition: magnetic force equals weight. BIl = mg ⇒ I = mg/Bl

Mind stretcher 2: Example: Why reverse the current?Extension

Explain why taking two balance readings with the current reversed and using F = ((m_down-mᵤₚ)g)/2 reduces systematic errors.

Show Answer

If the balance has an offset (zero error) or there is a constant background force, it affects both readings in the same way.

With current one way, the magnetic force adds to the reading: m_down = m₀ + Δ m. With current reversed, the magnetic force subtracts: mᵤₚ = m₀-Δ m.

Subtracting eliminates m₀: (m_down-mᵤₚ) = 2Δ m ⇒ F = Δ m g = ((m_down-mᵤₚ)g)/2

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