recall and use the equation F = BIL sin θ, with directions as interpreted by Fleming’s left-hand rule

1. Introduction – linking to the magnetic‑field topic (Syllabus 20.1)

In the study of magnetic fields (syllabus 20.1) we learned that a magnetic field exerts a force on moving charges. When many charges move together in a conductor, the field produces a mechanical force on the whole wire. This effect underpins electric motors, galvanometers, loudspeakers and many other devices covered in syllabus 20.2.

2. Magnetic flux density (B)

The magnetic‑flux‑density vector B is defined as the force per unit current per unit length on a straight conductor that is perpendicular to the field:

• SI unit: tesla (T) (1 T = 1 N A⁻¹ m⁻¹)
• Direction of B is the direction a north‑pole of a compass needle would point.

3. The fundamental equation for the force on a straight conductor

3.1 Scalar form

The magnitude of the force F on a straight piece of wire of length L carrying a conventional current I in a uniform magnetic field of flux density B is

where θ is the angle between the direction of the current (or the line of the conductor) and the magnetic‑field direction.

3.2 Vector (cross‑product) form

Using vectors the force is expressed as

• 𝑳 – vector of magnitude L in the direction of conventional current (positive‑to‑negative).
• 𝑩 – magnetic‑flux‑density vector.
• “×” denotes the vector cross‑product; its magnitude is 𝐿𝐵 sin θ and its direction is given by the right‑hand rule for cross‑products (but the physical direction of 𝐹 for a current‑carrying conductor is obtained with Fleming’s left‑hand rule – see §4).

4. Determining the direction – Fleming’s left‑hand rule

1. Stretch the thumb, fore‑finger and middle finger of the left hand so that they are mutually perpendicular.
2. Point the fore‑finger in the direction of the magnetic field B (from north to south).
3. Point the middle finger** in the direction of the conventional current I (positive to negative).
4. The thumb then points in the direction of the force F on the conductor.

5. Effective length for non‑perpendicular or curved conductors

• If the conductor is not perpendicular to the field, only the component of L that is perpendicular contributes: L_eff = L sin θ. The scalar formula then becomes F = B I L_eff.
• For a curved wire, treat each infinitesimal segment d L as straight and integrate:

  \$\mathbf{F}=I\int_{\text{wire}} \mathrm{d}\mathbf{L}\times\mathbf{B}\$

  In many exam questions the result can be obtained by recognising the effective perpendicular length of the whole shape (e.g. a semicircle has L_eff = πR/2).

6. Torque on a current‑carrying loop – the motor principle (Syllabus 20.2)

A rectangular (or any planar) loop of N turns, carrying a current I, placed in a uniform magnetic field B, experiences a torque that tends to rotate the loop. The torque magnitude is

• A – area of one turn of the loop (for a rectangle, A = length × width).
• θ – angle between the normal to the plane of the loop and the magnetic‑field direction.
• The direction of the torque is given by the right‑hand rule for the vector product τ = N I A (𝐧 × 𝐁), where n is a unit vector normal to the loop (given by the right‑hand rule applied to the direction of current around the loop).

This relationship explains how a simple DC motor converts electrical energy into mechanical rotation.

7. Worked example (straight conductor)

Problem: A straight horizontal wire 0.25 m long carries a current of 3.0 A to the east. It is placed in a uniform magnetic field of 0.40 T directed vertically upwards. Find the magnitude and direction of the force on the wire.

1. Identify vectors:

   • Current (middle finger) – east.
   • Magnetic field (fore‑finger) – up.

2. Direction – Fleming’s left‑hand rule gives the thumb pointing to the south; therefore the force is towards the south.
3. Angle – the wire is perpendicular to the field, so θ = 90° and sin θ = 1.
4. Magnitude:

   \$F = B I L \sin\theta = (0.40\ \text{T})(3.0\ \text{A})(0.25\ \text{m})(1) = 0.30\ \text{N}\$

Result: 0.30 N towards the south.

8. Common pitfalls

• Using the direction of electron flow instead of conventional current when applying Fleming’s left‑hand rule.
• Omitting the sin θ factor when the conductor is not perpendicular to the field.
• Confusing Fleming’s left‑hand rule (current‑carrying conductors) with the right‑hand rule (moving charges).
• For curved conductors, forgetting to consider the effective perpendicular length or to perform the integration of d L × B.
• Neglecting the factor N (number of turns) when calculating torque on a coil.

9. Practice questions

10. Summary checklist (exam‑ready)

• Write down all known quantities; note the directions of B and I.
• Determine the angle θ between the conductor (or current direction) and the magnetic field.
• Calculate the magnitude:

  • Straight conductor: F = B I L sin θ.
  • Curved conductor: use F = B I L_eff or integrate ∫dL × B.
  • Coil torque: τ = N I A B sin θ.

• Apply Fleming’s left‑hand rule to obtain the direction of the force (or the right‑hand rule for the torque vector τ = N I A (𝐧 × 𝐁)).
• Check units: 1 T·A·m = 1 N; for torque, 1 N·m.
• If the field is non‑uniform, remember that the force on each element is dF = I dL × B(r) and the total force is the vector sum (integral) of all elements.