Magnetic Forces on Current-Carrying Wires Revision Notes | A Level AQA Physics

1. Why Wires Move in Magnetic Fields

Picture a straight wire sitting between the poles of a horseshoe magnet. Switch on the current — and the wire flicks sideways. No contact, no mechanical push. That is the magnetic force on a current-carrying conductor, and it is the principle behind every electric motor ever built.

This lesson unpacks exactly why that force exists, how to calculate its magnitude using $F = BIl$, and how to predict its direction every time using Fleming's left-hand rule. These three ideas are tightly linked, so the lesson builds them in sequence.

2. Magnetic Flux Density and the Tesla

Before calculating any force, you need a precise measure of how strong the magnetic field is. That measure is magnetic flux density, symbol $B$. Think of it as the concentration of field lines through a given area — the denser the lines, the stronger the field, and the larger the force on any wire placed in it.

The unit, the tesla, is defined directly from the force equation. One tesla is the magnetic flux density that produces a force of exactly 1 N on a 1 m length of conductor carrying a current of 1 A, when the conductor is perpendicular to the field. This is not an arbitrary definition — it falls straight out of $F = BIl$ when $F = 1\text{ N}$, $I = 1\text{ A}$, and $l = 1\text{ m}$.

Typical values worth knowing: the Earth's magnetic field is around $5 \times 10^{-5}$ T, a strong laboratory electromagnet might reach 1–2 T, and MRI scanners operate at 1.5–3 T. These numbers give you a sense of scale when checking whether a calculated answer is physically reasonable.

3. The Force Equation: F = BIl

Now for the equation that ties everything together. When a straight conductor of length $l$ carries a current $I$ through a uniform magnetic field of flux density $B$, and the conductor is perpendicular to the field, the force on the conductor is:

The perpendicularity condition is critical. The equation $F = BIl$ only holds when the field is at 90° to the current. If the wire runs parallel to the field, the force is zero. The full relationship is $F = BIl\sin\theta$, where $\theta$ is the angle between the current direction and the field — but at A-level, AQA questions almost always set up the perpendicular case, so $\sin 90° = 1$ and the equation reduces to $F = BIl$.

Notice what the equation tells you about each variable. Doubling the current doubles the force. Doubling the length of wire inside the field doubles the force. Doubling the flux density doubles the force. All three variables have a direct, linear relationship with $F$ — a clean result that makes calculations straightforward and predictions intuitive.

4. Worked Calculations Using F = BIl

Theory is only useful once you can apply it under exam conditions. The following worked examples show the method AQA mark schemes reward: state the equation, substitute with units, calculate, and state the final answer with units.

Worked example: Calculating the Force on a Current-Carrying Wire

A straight wire of length 0.25 m carries a current of 4.0 A. It is placed perpendicular to a uniform magnetic field of flux density 0.30 T. Calculate the force on the wire. [3 marks]

1. $F = BIl$ — [M1] Writing the correct equation earns the method mark. Always start here.
2. $B = 0.30\text{ T},\quad I = 4.0\text{ A},\quad l = 0.25\text{ m}$ — List every given quantity before substituting. This prevents errors and shows the examiner your reasoning.
3. $F = 0.30 \times 4.0 \times 0.25$ — [M1] Correct substitution with consistent SI units earns the substitution mark.
4. $0.30 \times 4.0 = 1.20$ $1.20 \times 0.25 = 0.30$ — Breaking the multiplication into two steps reduces arithmetic errors. Verify: $1.20 \times 0.25 = 1.20 \times \frac{1}{4} = 0.30$. Correct.
5. $F = 0.30\text{ N}$ — [A1] The unit (N) must be stated for the accuracy mark. Two significant figures match the data given.

Answer: $F = 0.30\text{ N}$

Worked example: Finding Magnetic Flux Density from a Known Force

A wire of length 0.50 m carries a current of 2.0 A perpendicular to a magnetic field. The force on the wire is 0.060 N. Calculate the magnetic flux density. [3 marks]

1. $F = BIl$ — [M1] Starting with the standard equation is always the correct approach.
2. $B = \frac{F}{Il}$ — Divide both sides by $Il$ to isolate $B$. Rearranging before substituting is cleaner and less error-prone.
3. $B = \frac{0.060}{2.0 \times 0.50}$ — [M1] All values in SI units: F in N, I in A, l in m.
4. $2.0 \times 0.50 = 1.0$ — Compute the denominator first to keep the arithmetic clear.
5. $B = \frac{0.060}{1.0} = 0.060\text{ T}$ — [A1] The unit of $B$ is tesla (T). Two significant figures match the data.

Answer: $B = 0.060\text{ T}$

5. Fleming's Left-Hand Rule: Predicting Force Direction

Calculating the magnitude of the force is only half the story. You also need to know which direction the wire will move. That is where Fleming's left-hand rule comes in — a physical mnemonic that encodes the relationship between field direction, current direction, and force direction.

Hold your left hand with the thumb, index finger, and middle finger all mutually perpendicular — at right angles to each other. Each finger represents one quantity: the index finger points in the direction of the magnetic Field, the middle finger points in the direction of conventional Current, and the thumb points in the direction of the Motion (force) on the conductor.

• Index finger (First finger) → direction of magnetic Field (B)
• Middle finger → direction of conventional Current (I, from + to −)
• Thumb → direction of Motion (force, F)

A common memory aid: FBI — Force (thumb), B field (index), I current (middle). The left hand is used because this rule applies to conventional current (positive charge flow). If you accidentally use your right hand, you will get the opposite direction — a trap that costs marks.

To apply the rule in practice: first establish the field direction from the diagram (field lines run from N to S outside a magnet). Then identify the conventional current direction in the wire. Arrange your left hand so both fingers align simultaneously — your thumb then points in the direction of the force. If the force points out of the page, the wire moves towards you; if into the page, it moves away.

Fleming's left-hand rule also explains why the force is zero when the wire is parallel to the field. If you try to align the index finger along the field and the middle finger along the current when both point in the same direction, the three fingers cannot be mutually perpendicular — there is no unique thumb direction, which physically corresponds to zero force.

6. Connecting Force, Field, and Direction: The Full Picture

At this point you have all three spec requirements in place: the equation $F = BIl$ for magnitude, the definition of the tesla, and Fleming's left-hand rule for direction. Exam questions often combine all three in a single scenario — calculate the force and state its direction.

Worked example: Combined Magnitude and Direction Question

A wire of length 0.15 m is placed horizontally, carrying a current of 6.0 A directed northwards. A uniform magnetic field of flux density 0.40 T acts vertically downwards. Calculate the magnitude of the force on the wire and state its direction. [4 marks]

1. Current direction: North. Field direction: vertically downward. Angle between them = 90°. — The field is perpendicular to the current, so $F = BIl$ applies directly with no $\sin\theta$ correction needed.
2. $F = BIl = 0.40 \times 6.0 \times 0.15$ — [M1] Correct equation and substitution. All values in SI units.
3. $0.40 \times 6.0 = 2.4$ $2.4 \times 0.15 = 0.36$ — [A1] Verify: $2.4 \times 0.15 = 2.4 \times \frac{15}{100} = \frac{36}{100} = 0.36$. Correct.
4. $F = 0.36\text{ N}$ — Two significant figures, consistent with the data given.
5. Index finger: downward (field). Middle finger: northward (current). Thumb: points West. — [M1 + A1] The force acts horizontally towards the West. Stating the direction explicitly earns the remaining marks.

Answer: $F = 0.36\text{ N}$, directed towards the West.

8. Exam Technique and Common Mistakes

AQA questions on this topic follow predictable patterns. Knowing where marks are won and lost gives you a significant advantage. The guidance below is drawn from the structure of real mark schemes.

For extended-response questions asking you to explain the force on a current-carrying conductor, structure your answer as: state what the force depends on, give the equation, explain the direction using Fleming's left-hand rule, and link to a real application such as an electric motor. This approach covers the full range of marks available.

Together, $F = BIl$, the definition of the tesla, and Fleming's left-hand rule form a complete toolkit for analysing the motor effect — the foundation of every rotating electric motor, from a domestic fan to a large industrial drive.