# Force on current-carrying conductor

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u23-force-on-current-carrying-conductor/

This sub-topic explains how a current-carrying conductor placed in an external magnetic field experiences a net force. You will learn to calculate the magnitude and find the direction of this force, a core concept for CIE A-Level Physics electromagnetism.

**Prerequisites:** [Magnetic flux density and magnetic field basics](https://www.owlsprep.com/study/cie-9702-u23-magnetic-flux-density/)

## Learning objectives

- Explain the origin of force on a current-carrying conductor
- Derive and use the formula for force magnitude
- Determine force direction using Fleming's Left Hand Rule
- Calculate force for conductors at any angle to a magnetic field

## Origin of the Force

An electric current in a conductor is a net flow of charged particles (usually electrons in metallic conductors). When the conductor is placed inside an external magnetic field, each moving charged particle experiences an individual magnetic force. The sum of all these individual forces gives the net force on the entire conductor.

**Motor Effect** — The phenomenon where a current-carrying conductor placed in a magnetic field experiences a net force, which is the working principle of electric motors.

## Magnitude of the Force

The magnitude of the force depends on four factors: magnetic flux density $B$, current $I$, length of conductor $L$ inside the field, and the angle $\theta$ between the direction of current and the magnetic field vector.

**Derivation:** Derive the general force formula for a current-carrying conductor

*Starting from:* Force on a single moving charge: $F = qvB\sin\theta$

1. For a conductor of length $L$, cross-sectional area $A$, number density of charge carriers $n$, total charge carriers = $nAL$
2. Current is defined as $I = nAqv$, where $v$ is drift velocity of charge carriers
3. Total force = number of charge carriers × force per charge carrier
4. $$F = (nAL)(qvB\sin\theta) = (nAqv)LB\sin\theta$$
5. Substitute $I = nAqv$ into the equation:
6. $$F = BIL\sin\theta$$

*Conclusion:* This is the general formula for force on any straight current-carrying conductor in a uniform magnetic field.

> **info**
>
> When conductor is perpendicular to the field, $\theta = 90^\circ$, $\sin 90^\circ = 1$, so $F = BIL$ (maximum force). When parallel to the field, $\theta = 0^\circ$, $\sin 0^\circ = 0$, so force is zero.

**Worked example:** A straight wire of length 0.5 m carries a current of 3 A, placed in a uniform magnetic field of flux density 0.2 T. The angle between current and magnetic field is 30°. Calculate the force on the wire.

1. List known values: $B = 0.2$ T, $I = 3$ A, $L = 0.5$ m, $\theta = 30^\circ$
2. Use the general force formula $F = BIL\sin\theta$
3. Substitute values: $F = 0.2 \times 3 \times 0.5 \times \sin 30^\circ$
4. We know $\sin 30^\circ = 0.5$, so: $F = 0.2 \times 3 \times 0.5 \times 0.5 = 0.15$ N
5. Final answer: Force on the wire is 0.15 N

## Direction of the Force

The force is always perpendicular to both the direction of current and the direction of the magnetic field, following the cross product rule $\vec{F} = I \vec{L} \times \vec{B}$. For exam purposes, we use a simple mnemonic to find the direction.

> **Fleming's Left Hand Rule**
>
> Hold your left hand with first finger, second finger, and thumb all mutually perpendicular:
> - **First finger**: Points in direction of **Field**
> - **seCond finger**: Points in direction of **Current**
> - **THumb**: Points in direction of **Thrust (force)**

**Worked example:** A horizontal wire carries current from left to right, placed in a magnetic field pointing into the page. Find the direction of the force on the wire.

1. Align your left hand: point your first finger into the page (matches magnetic field direction)
2. Point your second finger to the right (matches current direction left to right)
3. Your thumb will point upwards, which is the direction of the force on the wire

**Check your understanding**

Test your understanding:

1. A current-carrying wire is parallel to a uniform magnetic field. What is the force on the wire?

   - Equal to $BIL$
   - Zero
   - Half $BIL$
   - Cannot be calculated

   *Answer:* Zero

   *Why:* When the wire is parallel to the field, $\theta = 0^\circ$, so $\sin\theta = 0$, meaning force is zero.

## Common pitfalls

- **Wrong:** Using Fleming's Right Hand Rule instead of Left Hand Rule
  - Why it fails: Right Hand Rule is for electromagnetic induction (generators), not for finding force on existing current (motor effect)
  - Correct: Always use Left Hand Rule to find the direction of force on a current-carrying conductor
- **Wrong:** Using $F = BIL$ for all angles, omitting $\sin\theta$
  - Why it fails: $F = BIL$ only works when the conductor is perpendicular to the magnetic field. It gives the wrong magnitude for any other angle
  - Correct: Always use the general formula $F = BIL\sin\theta$, and only simplify to $F=BIL$ when you confirm the conductor is perpendicular
- **Wrong:** Using the angle between conductor and force instead of current and field
  - Why it fails: $\theta$ in the formula is defined specifically as the angle between current direction and magnetic field direction, so using the wrong angle gives an incorrect result
  - Correct: Always first identify the direction of current and direction of magnetic field, then calculate the angle between these two vectors
- **Wrong:** Swapping the fingers for current and field in Fleming's Left Hand Rule
  - Why it fails: Swapping these gives the opposite direction of force, leading to wrong answers in multiple choice and written questions
  - Correct: Remember the mnemonic: First = Field, seCond = Current, THumb = Thrust to avoid mixing up fingers

## Cheatsheet

| Concept | Formula / Rule | Key Notes |
| --- | --- | --- |
| General force magnitude | $F = BIL\sin\theta$ | $\theta$ = angle between current and field |
| Conductor perpendicular to field | $F = BIL$ | Maximum possible force |
| Conductor parallel to field | $F = 0$ | No force acts on the conductor |
| Fleming's Left Hand | First=Field, Second=Current, Thumb=Force | Use for direction of force |

## What's next

The force on a current-carrying conductor is a foundational concept for all electromagnetism topics in CIE A-Level Physics. This effect is the basis of electric motors, loudspeakers, and many other electromagnetic devices that are common in exam questions. It also leads directly to the force between parallel current-carrying conductors, which is used to define the SI unit of current, the ampere. Mastering this topic will make it much easier to understand electromagnetic induction, the next major core concept in magnetic fields.

- [Force on moving charged particle](https://www.owlsprep.com/study/cie-9702-u23-force-on-moving-charged-particle/)
- [Charged particle motion in B-fields](https://www.owlsprep.com/study/cie-9702-u23-charged-particle-motion-in-b/)
- [Hall effect](https://www.owlsprep.com/study/cie-9702-u23-hall-effect/)

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