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Magnetic Flux Density

Key idea: Define magnetic flux density B in tesla, use F = BIl sinθ, and solve force and B calculations for wires in magnetic fields (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)

A. Magnetic flux density, B

Magnetic flux density, B, is defined as the force per unit current per unit length on a straight conductor placed perpendicular to the magnetic field:

B = F/Il

  • F is the magnetic force on the wire (N).
  • I is the current in the wire (A).
  • l is the length of wire inside the field (m).

B is a vector quantity (it has direction).

B. Unit of B

  • unit: tesla (T)
  • 1 T = 1 N A⁻¹m⁻¹

C. Force on a current-carrying conductor

For a straight wire in a uniform field, the force magnitude is:

F = BIl sin θ

where θ is the angle between the current direction and the magnetic field direction.

2. Key Ideas (What Earns Marks)

  • Magnetic flux density is defined using the perpendicular case, so sin θ = 1 and B = F/(Il).
  • Use the general force formula for any angle: F = BIl sin θ.
  • Directions are perpendicular: current direction, field direction, and force direction are mutually perpendicular (use Fleming’s left-hand rule).
  • The effective l is the part of wire actually in the magnetic field.
Fast unit check

If B is in tesla, Il has unit A·m, so BIl has unit (N/A·m)(A·m) = N.

3. Detailed Explanations

A. Why the definition uses a wire

Magnetic fields can exert forces on moving charges. In a metal wire, moving charge carriers make up an electric current, so measuring the force on a current-carrying conductor is a practical way to define and measure B.

B. Perpendicular vs angled wire

  • If the wire is perpendicular to the field, θ = 90° and sin θ = 1: F = BIl
  • If the wire is parallel to the field, θ = 0° and sin θ = 0: F = 0

This matches the idea that the magnetic force depends on the component of the current that is perpendicular to the field.

4. Common Mistakes

  • Using the full wire length instead of the length inside the field region.
  • Forgetting the sin θ factor when the wire is not perpendicular.
  • Mixing centimetres and metres for l.
  • Writing the unit as °T or treating tesla as a “derived symbol” with no unit meaning (use N A⁻¹ m⁻¹ if you want a base-unit check).

5. Exam Tips

  • State the angle: “wire is perpendicular, so sin θ = 1”.
  • If the question asks for direction, sketch the three perpendicular directions and apply Fleming’s left-hand rule.
  • If the question gives mass readings (current balance), convert to force: F = Δ m g.

6. Worked Examples

Modelled example 1

Find the force on a perpendicular wire

Core

Problem

A straight 0.12 m wire carries 2.0 A perpendicular to a uniform 0.30 T field. Find the magnetic force magnitude.
Study the worked solution

  1. Use the perpendicular case

    Method

    Use F = BIl.

    Reason

    The angle between conventional current and field is 90°, so sin θ = 1.

    Working

    F = BIl sin 90° = BIl

  2. Substitute and evaluate

    Method

    Obtain 7.2 × 10⁻² N.

    Reason

    Only the 0.12 m segment in the field contributes.

    Working

    F = (0.30)(2.0)(0.12) = 7.2 × 10⁻² N

Guided practice 2

Find B from a measured force

About 4 min

Problem

A perpendicular wire segment of length 0.080 m carries 3.0 A and experiences 0.060 N. Find B.

Try this before viewing the solution

Unit: T

Hints

Hint 1: make B the subject
For the perpendicular case, rearrange F = BIl.
View solution step by step

  1. Use the definition

    Method

    B = F/(Il).

    Reason

    Flux density is defined by force per unit current per unit perpendicular length.

    Working

    B = F/Il

  2. Evaluate

    Method

    Obtain 0.25 T.

    Reason

    The wire is perpendicular, so no angular correction is needed.

    Working

    B = 0.060/(3.0)(0.080) = 0.25 T

Common misconception 3

Wire at an angle

Find and correct the mistake

Learner claim

A 0.050 m wire carries 4.0 A at 30° to a 0.60 T field. A learner uses cos 30° because the wire is “mostly along the field”. Diagnose and calculate the force.

Try this before viewing the solution

Unit: N

View solution step by step

  1. Select the perpendicular component

    Method

    Use F = BIl sin θ with θ = 30°.

    Reason

    Only the current-length component perpendicular to B produces force.

    Working

    F = BIl sin 30°

  2. Evaluate

    Method

    Obtain 6.0 × 10⁻² N.

    Reason

    The force must lie between zero for a parallel wire and BIl = 0.12 N for a perpendicular wire.

    Working

    F = (0.60)(4.0)(0.050) sin 30° = 6.0 × 10⁻² N

Examiner practice 4

Find the length of wire inside the field

3 marks

Examination question

A wire carries 3.0 A perpendicular to a uniform 0.75 T field and experiences 0.18 N. Find the active wire length and explain what “active” means. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Rearrange

    1 mark

    Method

    Use l = F/(BI).

    Reason

    The wire is perpendicular, so F = BIl.

    Working

    l = F/BI

  2. Evaluate

    1 mark

    Method

    l = 8.0 × 10⁻² m.

    Reason

    Divide the force by flux density times current.

    Working

    l = 0.18/(0.75)(3.0) = 8.0 × 10⁻² m

  3. Interpret active length

    1 mark

    Method

    It is the wire length inside the magnetic-field region.

    Reason

    Segments where B ≈ 0 do not contribute appreciably to the magnetic force.

    Working

    Active length = 0.080 m within the field.

Challenge 5

Find the current needed for a required force

Minimal support

Independent transfer

A 0.20 m wire segment is at 60° to a uniform 0.40 T field. Find the current needed for a 0.10 N force, and state how the required current would change if the wire became perpendicular.

Try this before viewing the solution

Hints

Hint 1: isolate current with the sine factor
Use I = F/(Bl sin θ), then compare sin 60° with sin 90°.
View solution step by step

  1. Calculate the current

    Method

    Obtain 1.44 A ≈ 1.4 A.

    Reason

    The wire is angled, so retain sin 60°.

    Working

    I = 0.10/((0.40)(0.20) sin 60°) = 1.44 A

  2. Compare perpendicular orientation

    Method

    The required current would decrease to 1.25 A.

    Reason

    A perpendicular wire has the maximum sine factor of one for the same force.

    Working

    I₉₀ = 0.10/(0.40)(0.20) = 1.25 A

7. Mind Stretchers

Mind stretcher 1: When is the force maximum?Extension

For a given B, I and l, what wire orientation gives the maximum force? Explain.

Show Answer

From F = BIl sin θ, the force is maximum when sin θ = 1, i.e. when θ = 90° and the wire is perpendicular to the field.

Mind stretcher 2: Why only the wire in the field mattersExtension

In F = BIl (perpendicular case), why is l the length of wire inside the magnetic field region, not the total wire length?

Show Answer

Magnetic force only acts where there is a magnetic field. The force is the result of the local interaction between current and the field, so wire segments outside the field (where B ≈ 0) contribute negligible force.

For a wire passing through a finite field region, only the part inside that region experiences the B field and contributes to the net force.

8. Optional (Enrichment)

A. Measuring B using a current balance

In a current balance, the force on the wire produces an apparent mass change. You use: F = Δ m g and F = BIl to find B.

See: Current Balance.

Continue with the next resource in this course.

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