Recall and use the relative directions of force, magnetic field and current

4.5.4 Force on a Current‑Carrying Conductor

Learning Objective

Recall and use the relative directions of the magnetic field B, the current I, and the magnetic force F on a straight conductor, and describe all factors that affect the magnitude of the force.

Key Formulae

  • Vector form (right‑hand rule):

    𝐅 = I 𝐥 × 𝐁

  • Magnitude:

    F = I l B sin θ

    where θ is the angle between the direction of the current (the line l) and the magnetic field B.

Factors that Influence the Magnitude of the Force

The force is greatest when each factor is maximised. The table shows the proportionality and a quick “What‑if” check for AO2 handling‑information questions.

FactorSymbol in formulaEffect on FWhat‑if?
CurrentIDirectly proportional – double I → double FIf I is halved, F is halved.
Length of wire inside the uniform fieldlDirectly proportional – double l → double FIf l is reduced to one‑quarter, F becomes one‑quarter.
Magnetic field strengthBDirectly proportional – double B → double FIf B is reduced by 30 %, F also falls by 30 %.
Angle between I and Bsin θMaximum when θ = 90° (perpendicular); zero when θ = 0° or 180°If θ = 30°, sin θ ≈ 0.5 → F is half the maximum value.
Uniformity of the field over the length l—If B varies, use the average field B̅ over the portion of wire that lies in the field.For a field that drops linearly from 0.6 T to 0.2 T, B̅ = 0.4 T.

Direction Rules

Right‑Hand Rule (conventional current)

  1. Stretch the right hand so the thumb points in the direction of the conventional current I (positive charge flow).

  2. Point the fingers in the direction of the magnetic field B (from north to south).

  3. The palm pushes in the direction of the magnetic force F on the conductor.

Right‑hand rule: thumb = I, fingers = B, palm = F

Right‑hand rule (conventional current)

Fleming’s Left‑Hand Rule (electron flow)

  1. First finger → magnetic field B (north → south).

  2. Second finger → direction of electron flow (opposite to conventional current).

  3. Thumb → direction of the force on the conductor.

This rule gives the same force direction as the right‑hand rule, but the current direction is opposite to the electron flow.

Charged‑Particle Beams (Supplementary Content)

  • For a positively charged particle (e.g. proton) moving with velocity v in a magnetic field B, use the right‑hand rule for v × B to find the force direction.

  • For an electron beam, reverse the direction obtained from the right‑hand rule (or use Fleming’s left‑hand rule directly).

  • Equation: 𝐅 = q 𝐯 × 𝐁 where q is the charge of the particle.

Experimental Demonstration – Qualitative & Quantitative

Paper‑clip‑in‑magnet experiment (qualitative)

  1. Place a short length of insulated copper wire (≈ 10 cm) horizontally on a sheet of paper.

  2. Connect the ends to a low‑voltage battery (e.g. 1.5 V) using crocodile clips so that a steady current flows from left to right.

  3. Position a strong bar magnet beneath the paper so that the magnetic field points into the page (south pole up).

  4. Place a small paper clip on the wire. It is pushed upwards (or downwards) depending on the current direction.

  5. Reverse the battery connections – the current reverses – and the paper clip moves in the opposite direction.

  6. Flip the magnet over so that the field now points out of the page; the force again reverses.

Making the observation quantitative

  • Replace the paper clip with a lightweight spring balance (or a digital force sensor) attached to the wire via a fine thread.

  • Measure the deflection (or force reading) for a known current I, length l inside the field, and field strength B. Compare the result with the predicted value from F = I l B sin θ.

  • Repeat with different currents or by halving the length of wire in the field to verify the proportionalities.

Safety Precautions

  • Use a low‑voltage battery (≤ 3 V) to avoid hazardous currents.

  • Ensure the copper wire is insulated to prevent accidental short‑circuits.

  • Keep strong magnets away from electronic devices, credit cards, and pacemakers.

  • Do not touch the wire while the current is flowing.

Direction Table (Right‑Hand Rule)

Current IMagnetic field BResulting force F
Into the page (×)To the right (→)Upwards (↑)
To the right (→)Upwards (↑)Out of the page (·)
Downwards (↓)Into the page (×)To the left (←)

Using the Formula

When the conductor is perpendicular to the magnetic field (θ = 90°), sin θ = 1 and the equation simplifies to:

F = I l B

Example

A 0.15 m long straight wire carries a current of 3.0 A in a uniform magnetic field of 0.40 T, the field being perpendicular to the wire.

F = (3.0 A)(0.15 m)(0.40 T) = 0.18 N

Closed Loop – Torque but No Net Translational Force

Consider a rectangular loop placed in a uniform magnetic field, with two sides parallel to B and the other two sides perpendicular.

  • The sides parallel to B experience no force (θ = 0°).

  • Each perpendicular side experiences a force of magnitude F = I l B, but the forces act in opposite directions (one upward, one downward).

  • These opposite forces form a couple, producing a torque that tends to rotate the loop, while their translational components cancel, giving zero net linear force.

Rectangular loop in uniform magnetic field showing forces on opposite sides and resulting torque

Rectangular loop in a uniform field – opposite forces on the vertical sides produce a torque.

Common Mistakes

  • Using the direction of electron flow with the right‑hand rule instead of conventional current.

  • Omitting the sin θ factor when the field is not perpendicular.

  • Confusing magnetic field B (flux density) with magnetic flux Φ.

  • Assuming the field is uniform over the whole wire when only part of the wire lies inside the field.

Summary – How to Find the Force on a Straight Conductor

  1. Identify the direction of the conventional current I and the magnetic field B.

  2. Apply the right‑hand rule (or Fleming’s left‑hand rule) to obtain the direction of the force F.

  3. Determine the length l of wire that is inside the uniform field.

  4. Calculate the magnitude using F = I l B sin θ, remembering that sin θ = 1 when the wire is perpendicular to the field.

Practice Questions

  1. A straight wire 0.20 m long carries a current of 2.5 A. It is placed in a uniform magnetic field of 0.30 T directed into the page. The current flows to the right.

    • State the direction of the magnetic force on the wire.

    • Calculate its magnitude.

  2. A conductor 0.10 m long carries a current of 4.0 A at an angle of 30° to a magnetic field of 0.50 T. Find the magnitude of the force.

  3. Explain why a current‑carrying rectangular loop placed in a uniform magnetic field experiences a torque but no net translational force.

  4. In the paper‑clip experiment described above, predict what happens to the force if the current is doubled while the magnetic field remains unchanged.

  5. (Extension) A proton moving with speed 2.0 × 10⁶ m s⁻¹ enters a uniform magnetic field of 0.20 T directed into the page. The velocity is to the right. Determine the direction and magnitude of the magnetic force on the proton.

See Also

  • 4.5.3 Magnetic effect of a current – the principle behind the right‑hand rule.

  • 4.5.5 d.c. motor – application of the force on a current‑carrying conductor.

  • 4.5.6 Transformers – magnetic fields produced by alternating currents.