Magnetic flux and magnetic forces

Flux density, flux and flux linkage

• Magnetic flux density B is the strength of a magnetic field, in tesla (T).
• Magnetic flux through an area A perpendicular to the field: φ = BA, in weber (Wb).
• Flux linkage for a coil of N turns: Nφ, in Wb turns.

If the area is at an angle to the field, use the component of B perpendicular to it.

Force on a current

A wire of length l carrying current I at angle θ to a field B feels a force

F = BIl sin θ

It is greatest when the wire is perpendicular to the field and zero when parallel. This defines the tesla: 1 T gives a force of 1 N on 1 m of wire carrying 1 A at 90° to the field.

Force on a moving charge

A charge q moving at speed v at angle θ to the field feels

F = Bqv sin θ

The force is perpendicular to both v and B, so it does no work: it changes the direction, not the speed.

Fleming's left-hand rule

Hold the thumb and first two fingers of the left hand at right angles:

• First finger: Field (north to south)
• seCond finger: Current (conventional, + to -)
• thuMb: Motion (force)

For a moving negative charge, the 'current' points opposite to its velocity.

Worked example

A wire 0.25 m long carries 3.0 A at right angles to a 0.080 T field. Find the force on it.

1. F = BIl sin θ = 0.080 x 3.0 x 0.25 x sin 90°

Answer: F = 0.060 N

Check yourself

1. A uniform 0.40 T field passes at right angles through an area of 2.5 x 10^-3 m². Find the flux in mWb.
   Show answer

   1 mWb. φ = BA = 0.40 x 2.5 x 10^-3 = 1.0 x 10^-3 Wb = 1.0 mWb.

2. A coil of 200 turns and area 4.0 x 10^-4 m² is perpendicular to a 0.15 T field. Find the flux linkage in Wb turns.
   Show answer

   0.012 Wb turns. Nφ = 200 x 0.15 x 4.0 x 10^-4 = 0.012 Wb turns.

3. A 0.40 m wire carrying 2.5 A is perpendicular to a 0.12 T field. Find the force in N.
   Show answer

   0.12 N. F = BIl = 0.12 x 2.5 x 0.40 = 0.12 N.

4. A 0.50 m wire carrying 4.0 A is at 30° to a 0.20 T field. Find the force in N.
   Show answer

   0.2 N. F = BIl sin θ = 0.20 x 4.0 x 0.50 x 0.5 = 0.20 N.

5. An electron moves at 2.0 x 10⁶ m/s at right angles to a 0.050 T field. Find the force on it, in units of 10^-14 N.
   Show answer

   1.6 x 10^-14 N. F = Bqv = 0.050 x 1.60 x 10^-19 x 2.0 x 10⁶ = 1.6 x 10^-14 N.

6. In Fleming's left-hand rule, what does the second finger represent?
   • Magnetic field
   • Force (motion)
   • Electron flow
   • Conventional current
   Show answer

   Conventional current. First finger field, second finger current, thumb motion.

7. A charged particle moves parallel to a uniform magnetic field. What force does the field exert on it?
   • A force along its direction of motion
   • No force
   • A force at right angles to its motion
   • A force opposite to its motion
   Show answer

   No force. θ = 0°, so F = Bqv sin 0° = 0.

8. Why does a magnetic field not change the speed of a moving charge?
   • The force is always perpendicular to the velocity, so it does no work
   • The force is too small
   • The force acts along the velocity
   • Magnetic fields only act on stationary charges
   Show answer

   The force is always perpendicular to the velocity, so it does no work. A force at 90° to the motion changes direction only.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson Edexcel IAL Physics Unit 4 (WPH14) question paper, June 2024: list of data, formulae and relationships

Last updated 7 November 2026. Spotted a mistake? Tell us.