The Motor Effect

When a wire carrying a current is placed in a magnetic field, something remarkable happens: the wire is pushed by a force, even though nothing touches it. This is the motor effect, and it is one of the most important ideas in all of physics. It turns electrical energy into movement, and it is the principle behind every electric motor — from the tiny one that vibrates a phone to the powerful ones that drive electric trains and cars. The motor effect arises because the current's own magnetic field interacts with the field it is placed in, and the two fields together produce a push on the wire. This lesson, part of Topic P4 (Magnetism and magnetic fields) of OCR Gateway Science A, explains the motor effect, shows how to find the direction of the force with Fleming's left-hand rule, introduces magnetic flux density, and works through the Higher-tier equation $F = BIL$F=BIL.

By the end of this lesson you should be able to describe the motor effect, use Fleming's left-hand rule to find the direction of the force, define magnetic flux density and its unit, recall and use the Higher-tier equation $F = BIL$F=BIL, and state the factors that affect the size of the force.

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The Motor Effect

Place a wire carrying a current at right angles to a magnetic field, and the wire experiences a force that pushes it at right angles to both the current and the field. This is the motor effect. The reason is that the current creates its own magnetic field (the concentric circles from lesson 2), and this combines with the field it is sitting in: on one side of the wire the two fields add to give a strong field, and on the other side they partly cancel to give a weak field. The wire is then pushed from the strong-field side towards the weak-field side.

You can demonstrate the motor effect simply: lay a length of wire between the poles of a horseshoe magnet (or a pair of magnadur magnets on a yoke) and pass a current through it. The wire kicks — jumping up or down — the moment the current flows. Reverse the current, and it kicks the other way.

Two key facts about the direction:

• Reversing the current reverses the direction of the force.
• Reversing the magnetic field (swapping the poles) also reverses the direction of the force.
• If you reverse both at once, the force returns to its original direction.

And one key fact about when there is no force at all: the wire only feels the full force when the current is at right angles (90°) to the field. If the wire lies parallel to the field, there is no force on it.

Exam Tip: The motor effect gives the largest force when the wire is at 90° to the field, and no force when the wire is parallel to the field. Reversing either the current or the field reverses the force; reversing both leaves it unchanged.

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Fleming's Left-Hand Rule

To find the direction of the force in the motor effect, we use Fleming's left-hand rule. Hold the thumb and first two fingers of your left hand so that they are all at right angles to one another, and assign:

• First finger → Field (direction of the magnetic field, north to south);
• seCond finger → Current (direction of the conventional current, + to −);
• thuMb → Motion (the direction of the force / the resulting movement).

The mnemonic is the order F – C – M down from the first finger: Field, Current, Motion. Get the field and current pointing the right way with your first and second fingers, and your thumb shows you which way the wire is pushed.

graph TD
  A["Left hand: fingers at right angles"] --> B["First finger = Field (N to S)"]
  A --> C["Second finger = Current (+ to −)"]
  A --> D["Thumb = Motion / Force"]

Exam Tip: Use your left hand for the motor effect (a current-carrying wire being pushed). The order is Field, Current, Motion — First finger, seCond finger, thuMb. Keep the three at right angles. (Save your right hand for the generator effect in lesson 5.)

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Magnetic Flux Density

To say how strong a magnetic field is, we use a quantity called the magnetic flux density, given the symbol $B$B. You can think of it as a measure of how concentrated the field lines are — how much field is packed into a given area — and so how strong the magnetic field is at that place. The greater the flux density, the stronger the field and the larger the motor-effect force.

Magnetic flux density is measured in tesla (symbol $\text{T}$T). One tesla is a fairly large field: the Earth's field is only a few tens of microtesla, while a strong laboratory magnet might be around $0.1$0.1 to $1 \text{ T}$1 T.

Exam Tip: The magnetic flux density $B$B measures the strength of a magnetic field and is measured in tesla ($\text{T}$T). A bigger $B$B means a stronger field and (for a given current and length) a bigger force.

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The Equation $F = BIL$F=BIL (Higher tier)

(Higher tier) For a wire at right angles to a magnetic field, the size of the force on it is given by:

$F = BIL$F=BIL

where

• $F$F is the force in newtons ($\text{N}$N);
• $B$B is the magnetic flux density in tesla ($\text{T}$T);
• $I$I is the current in amperes ($\text{A}$A);
• $L$L is the length of wire in the field, in metres ($\text{m}$m).

The equation tells you straight away what makes the force bigger: a stronger field, a larger current, or a greater length of wire in the field all increase the force, in direct proportion. (This equation, and all calculations using it, are Higher tier only.)

Worked Example 1 — Finding the force

A wire of length $0.20 \text{ m}$0.20 m carries a current of $3.0 \text{ A}$3.0 A at right angles to a magnetic field of flux density $0.40 \text{ T}$0.40 T. Calculate the force on the wire. (Higher)

Step 1 — write the equation: $F = BIL$F=BIL.

Step 2 — substitute the values: $F = 0.40 \times 3.0 \times 0.20$F=0.40×3.0×0.20.

Step 3 — calculate: $F = 0.24 \text{ N}$F=0.24 N.

Answer: the force on the wire is $0.24 \text{ N}$0.24 N.

Worked Example 2 — Rearranging for the current

The force on a $0.05 \text{ m}$0.05 m length of wire in a field of $0.60 \text{ T}$0.60 T is $0.090 \text{ N}$0.090 N. Calculate the current in the wire. (Higher)

Step 1 — rearrange $F = BIL$F=BIL to make current the subject: $I = \dfrac{F}{BL}$I=BLF​.

Step 2 — substitute: $I = \dfrac{0.090}{0.60 \times 0.05}$I=0.60×0.050.090​.

Step 3 — calculate: $I = \dfrac{0.090}{0.030} = 3.0 \text{ A}$I=0.0300.090​=3.0 A.

Answer: the current in the wire is $3.0 \text{ A}$3.0 A.

Worked Example 3 — Rearranging for the flux density

A $0.10 \text{ m}$0.10 m length of wire carrying $5.0 \text{ A}$5.0 A experiences a force of $0.20 \text{ N}$0.20 N when held at right angles to a magnetic field. Calculate the flux density of the field. (Higher)

Step 1 — rearrange $F = BIL$F=BIL to make $B$B the subject: $B = \dfrac{F}{IL}$B=ILF​.

Step 2 — substitute: $B = \dfrac{0.20}{5.0 \times 0.10}$B=5.0×0.100.20​.

Step 3 — calculate: $B = \dfrac{0.20}{0.50} = 0.40 \text{ T}$B=0.500.20​=0.40 T.

Answer: the magnetic flux density is $0.40 \text{ T}$0.40 T.

Exam Tip: (Higher) In $F = BIL$F=BIL, the length $L$L must be in metres and the current $I$I in amperes to give the force in newtons. Lay calculations out as equation, substitution, answer with unit to secure the method marks.

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