Current Balance (Measuring Magnetic Flux Density)

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 = BIlsinθ
  • 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 Rightarrow 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 = (mdown-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 = BIlsinθ so: B = F/Ilsinθ

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 sinθ 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 sinθ = 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

Example 1: Find B from a single mass changeCore

A wire of length l = 0.12 m in a uniform field carries current I = 2.0 A. The balance reading increases by 3.0 g. Find B.

Show Answer

Δ m = 3.0 g = 3.0 × 10⁻³ kg; F = Δ m g = (3.0 × 10⁻³)(9.81) = 2.94 × 10⁻² N; B = F/Il = 2.94 × 10⁻²/(2.0)(0.12); = 1.23 × 10⁻¹ T ≈ 0.12 T

Example 2: Using two readings (current reversed)Core

A balance reads mdown = 0.523 kg for one current direction and mᵤp = 0.517 kg when the current is reversed. The wire length in the field is 0.15 m and the current is 4.0 A. Find B.

Show Answer

F = (mdown-mᵤp)g/2 = (0.523-0.517)(9.81)/2; = 2.94 × 10⁻² N; B = F/Il = 2.94 × 10⁻²/(4.0)(0.15) = 4.90 × 10⁻² T

Example 3: Wire at an angleCore

A wire of length 0.050 m carries 3.0 A in a uniform field B = 0.80 T. The wire makes an angle θ = 30^° to the field. Find the force magnitude.

Show Answer

F = BIlsinθ; = (0.80)(3.0)(0.050)sin 30^°; = (0.80)(3.0)(0.050)(0.50) = 6.0 × 10⁻² N

Example 4: Predict the reading change for a given BCore

In a current balance, a wire segment of length l = 0.10 m carries I = 3.0 A perpendicular to a uniform field B = 0.25 T. Find the apparent mass change Δ m for one current direction. Take g = 9.81 m s⁻².

Show Answer

F = BIl = (0.25)(3.0)(0.10) = 7.5 × 10⁻² N; Δ m = F/g = 7.5 × 10⁻²/9.81 = 7.64 × 10⁻³ kg = 7.6 g

Example 5: Find the current from two readingsCore

A balance reads mdown = 0.502 kg and mᵤp = 0.498 kg when the current is reversed. The wire length in the field is l = 0.080 m and the magnetic flux density is B = 0.30 T. Find the current magnitude. Take g = 9.81 m s⁻².

Show Answer

F = (mdown-mᵤp)g/2 = (0.502-0.498)(9.81)/2; = 1.96 × 10⁻² N; I = F/Bl = 1.96 × 10⁻²/(0.30)(0.080) = 8.18 × 10⁻¹ A ≈ 0.82 A

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 Rightarrow 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 = (mdown-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: mdown = m₀ + Δ m. With current reversed, the magnetic force subtracts: mᵤp = m₀-Δ m.

Subtracting eliminates m₀: (mdown-mᵤp) = 2Δ m Rightarrow F = Δ m g = (mdown-mᵤp)g/2

8. Practice, Quiz and Next Step

Close your notes and use Current Balance (Measuring Magnetic Flux Density) in the supplied context below. This requires a constructed explanation or working, not recognition of an option.

Fresh context: An unfamiliar data set or physical system requires you to apply Current Balance (Measuring Magnetic Flux Density) while stating the model, regime and assumptions.

1. Retrieve: define current balance (measuring magnetic flux density) in your own words, including units, sign or conditions where relevant.
2. Represent: Choose and label an appropriate diagram, graph, table or symbolic model; derive or justify the relationship used.
3. Apply: Reach a conclusion, then evaluate it using units, uncertainty, a limiting case and one practical or modelling limitation.

Check the response before looking back

• The model, regime, coordinates and assumptions are explicit.
• The derivation or multi-step reasoning is visible rather than implied.
• The conclusion is tested against units, data quality and a limiting case.
• A practical control, uncertainty or model limitation is evaluated where applicable.

If one check fails, name that exact gap, revisit the matching explanation or worked example, and redo the task with different values or a different situation. Then use theA-Level Physics course hub orpractice browser for an independent re-test.