Force on current-carrying conductors, F = BIL, Fleming's left-hand rule, and the DC motor
AQA GCSE Physics 4.7 | Year 11 | Higher Tier
⚡ Explain why a current-carrying conductor in a magnetic field experiences a force
🧮 Use F = BIL to calculate the force on a conductor
✋ Apply Fleming's left-hand rule to determine the direction of the force
🔄 Describe the construction and operation of a simple DC motor
📐 Understand what factors affect the size of the force on a conductor
🔬 Explain the role of the split-ring commutator in keeping a DC motor rotating
The Motor Effect
When a current-carrying conductor is placed inside a magnetic field, it experiences a force. This is called the motor effect. The force arises because the magnetic field of the current interacts with the external magnetic field — the two fields either reinforce or oppose each other on either side of the wire, creating a net push.
Motor Effect: The force experienced by a current-carrying conductor placed in an external magnetic field, due to the interaction of the two magnetic fields.
For the motor effect to occur, the current must have a component that is perpendicular to the magnetic field. If the wire is parallel to the field, there is no force. The maximum force occurs when the current is at 90° to the field.
The force is always perpendicular to both the current direction and the magnetic field direction.
The size of the force depends on three factors:
Magnetic flux density (B) — a stronger magnet creates a larger force
Current (I) — a larger current creates a larger force
Length of conductor in the field (L) — a longer wire in the field creates a larger force
The Equation: F = BIL
The force on a current-carrying conductor can be calculated using:
F = B × I × L
Symbol
Quantity
SI Unit
F
Force
Newtons (N)
B
Magnetic flux density
Tesla (T)
I
Current
Amperes (A)
L
Length of conductor in the field
Metres (m)
This equation can be rearranged to find any of the four quantities:
B = F ÷ (I × L) I = F ÷ (B × L) L = F ÷ (B × I)
Magnetic Flux Density (B): A measure of the strength of a magnetic field, measured in Tesla (T). 1 T = 1 N A⁻¹ m⁻¹. It represents how much magnetic force acts per unit length per unit current.
Note: This equation assumes the current is perfectly perpendicular to the magnetic field. At the GCSE level, you will always be given situations where this condition is met (the maximum case).
Fleming's Left-Hand Rule
Fleming's left-hand rule is used to determine the direction of the force on a current-carrying conductor in a magnetic field. Hold your left hand with the thumb, index finger, and middle finger all at right angles to each other:
FBI:
👆 First finger → direction of magnetic Field (N to S)
🖕 seCond finger (middle) → direction of Conventional Current (+ to −)
👍 Thumb → direction of Motion / Force on the conductor
A simple memory aid: FBI — Field (first finger), Current (seCond finger), Motion/Force (thuMb).
Always use your LEFT hand for the motor effect. (The right hand is used for the generator effect / Faraday's law.)
To use the rule effectively:
Identify the direction of the magnetic field (from N to S pole)
Identify the direction of conventional current (from + to − terminal)
Point your first finger in the field direction and your second finger in the current direction simultaneously
Your thumb will point in the direction of the force on the conductor
If you reverse the current direction OR reverse the field direction, the force direction also reverses. If you reverse both, the force stays the same.
The Simple DC Motor
A DC motor uses the motor effect to convert electrical energy into kinetic (rotational) energy. A simple DC motor consists of:
A coil of wire (the armature) placed between the poles of a permanent magnet
A split-ring commutator — a conducting ring split into two halves that rotates with the coil
Carbon brushes — stationary contacts that press against the commutator to maintain electrical contact
A DC power supply to drive the current
Split-Ring Commutator: A device that reverses the direction of the current through the coil every half turn, ensuring the coil always rotates in the same direction.
How it works: Current flows through the coil. The two sides of the coil in the magnetic field carry current in opposite directions, so they experience forces in opposite directions (by Fleming's left-hand rule). This creates a turning effect (torque) that rotates the coil.
After half a turn, without the commutator, the forces would reverse and the coil would rotate back. The split-ring commutator swaps the connections to the power supply every half turn, reversing the current direction at exactly the right moment so the forces always act in the same rotational direction.
The motor effect is the basis of all electric motors — from tiny motors in smartphones to huge motors in electric vehicles and industrial machinery.
Increasing motor speed/force: You can increase the turning effect of a DC motor by:
Increasing the current
Using a stronger magnet (greater B)
Using more turns of wire on the coil (increases effective length L)
Summary of Key Relationships
Understanding how changing each variable affects the force is essential for exam questions.
Change Made
Effect on Force
Double the magnetic flux density B
Force doubles
Double the current I
Force doubles
Double the length L in the field
Force doubles
Reverse the current direction
Force reverses direction
Reverse the magnetic field direction
Force reverses direction
Wire parallel to field (0°)
No force (F = 0)
Wire perpendicular to field (90°)
Maximum force
F = BIL gives the maximum force — when the wire is at exactly 90° to the magnetic field. This is always assumed in GCSE calculations.
Example 1: A conductor 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 conductor.
1Identify the equation and known values:
F = B × I × L
B = 0.30 T, I = 4.0 A, L = 0.25 m
2Substitute values into the equation:
F = 0.30 × 4.0 × 0.25
3Calculate:
F = 0.30 × 4.0 = 1.20
F = 1.20 × 0.25 = 0.30 N
Force on the conductor = 0.30 N
Example 2: A wire experiences a force of 0.096 N. The wire has a length of 0.12 m in the field and carries a current of 2.0 A. Calculate the magnetic flux density of the field.
1Write down what is known and what is unknown:
F = 0.096 N, L = 0.12 m, I = 2.0 A, B = ?
2Rearrange F = BIL to make B the subject:
B = F ÷ (I × L)
3Substitute and calculate:
B = 0.096 ÷ (2.0 × 0.12)
B = 0.096 ÷ 0.24
B = 0.40 T
Magnetic flux density = 0.40 T
Example 3: A current of 5.0 A flows through a wire of length 0.08 m in a magnetic field. The force on the wire is 0.20 N. The current is then doubled and the magnetic flux density is halved. Calculate the new force on the wire.
1First, find the original magnetic flux density:
B = F ÷ (I × L) = 0.20 ÷ (5.0 × 0.08) = 0.20 ÷ 0.40 = 0.50 T
2Find the new values of I and B:
New I = 5.0 × 2 = 10.0 A
New B = 0.50 ÷ 2 = 0.25 T
L remains = 0.08 m
3Calculate the new force:
F = B × I × L = 0.25 × 10.0 × 0.08
F = 0.25 × 10.0 = 2.5
F = 2.5 × 0.08 = 0.20 N
4Interpret the result:
Doubling I and halving B have opposite effects that cancel out — the force remains the same. This illustrates that F is directly proportional to both B and I.
New force = 0.20 N (unchanged, since doubling I and halving B cancel out)
Example 4 (DC Motor): In a simple DC motor, a rectangular coil has 50 turns, each side of length 0.06 m carrying current perpendicular to a magnetic field of flux density 0.15 T. The current in the coil is 3.0 A. Calculate the force on one current-carrying side of the coil.
1Identify the equation and values for one side:
F = B × I × L
B = 0.15 T, I = 3.0 A, L = 0.06 m per turn × 50 turns = 3.0 m
(50 turns means the effective length is 50 × 0.06 m)
2Calculate the force on one side:
F = 0.15 × 3.0 × 3.0
F = 0.15 × 9.0 = 1.35 N
3Interpret for the motor:
The opposite side of the coil carries current in the opposite direction, so the force on it is 1.35 N in the opposite direction. Together they produce a couple (turning effect) that rotates the coil.
Force on one side of the coil = 1.35 N
Question 1: A wire carrying a current is placed in a magnetic field. What happens to the force on the wire if the current is tripled?
Question 2: When using Fleming's left-hand rule, what does the thumb represent?
Question 3: A conductor of length 0.40 m carries a current of 3.0 A in a magnetic field of flux density 0.25 T. Calculate the force on the conductor in Newtons.
Question 4: In a simple DC motor, what is the purpose of the split-ring commutator?
Question 5: A wire experiences a force of 0.45 N. It is 0.15 m long and sits in a magnetic field of flux density 0.60 T. Calculate the current in the wire in Amperes.
Challenge 1 (6 marks): A rectangular coil in a DC motor has 80 turns. Each side of the coil that is perpendicular to the magnetic field has a length of 0.05 m. The magnetic flux density is 0.20 T and the current through the coil is 2.5 A.
(a) Calculate the force on one of the perpendicular sides of the coil.
(b) Explain why the coil experiences a turning effect rather than a linear force.
(c) State and explain one change that could be made to increase the speed of rotation of the motor.
(a) Effective length = 80 × 0.05 = 4.0 m
F = B × I × L = 0.20 × 2.5 × 4.0 = 2.0 N
(b) The two perpendicular sides of the coil carry current in opposite directions. By Fleming's left-hand rule, the forces on these two sides are in opposite directions. Because these forces act on opposite sides of the coil, they create a couple (a pair of equal and opposite forces not acting through the same point), producing a turning/rotational effect (torque).
(c) Possible answers (any one with explanation):
• Increase the current — greater I increases F = BIL, so larger force on each side, greater turning effect.
• Use a stronger magnet — greater B increases F = BIL, so larger turning force.
• Add more turns to the coil — increases effective length L, so greater force and turning effect.
Challenge 2 (5 marks): A student sets up an experiment with a wire of length 0.10 m in a magnetic field of flux density 0.50 T. The wire carries a current of 4.0 A. The student then:
Halves the magnetic flux density
Doubles the current
Doubles the length of wire in the field
Calculate the original force and the new force, and determine the ratio new force : original force.
Original force:
F = B × I × L = 0.50 × 4.0 × 0.10 = 0.20 N
New values:
B = 0.50 ÷ 2 = 0.25 T
I = 4.0 × 2 = 8.0 A
L = 0.10 × 2 = 0.20 m
New force:
F = 0.25 × 8.0 × 0.20 = 0.40 N
Ratio: new : original = 0.40 : 0.20 = 2 : 1
The force doubles overall, because halving B is more than compensated by doubling both I and L (net effect: × ½ × 2 × 2 = × 2).
Challenge 3 (4 marks): A wire carrying a conventional current is directed from left to right on the page. A uniform magnetic field is directed into the page.
(a) Using Fleming's left-hand rule, determine the direction of the force on the wire.
(b) The current direction is reversed. Describe and explain the new direction of the force.
(a) Using Fleming's left-hand rule:
• First finger (field): into the page
• Second finger (current): to the right (left to right)
• Thumb (force/motion): upward (towards the top of the page)
(b) If the current is reversed (now flowing from right to left), the second finger now points to the left. Keeping the first finger pointing into the page, the thumb now points downward (towards the bottom of the page).
The force reverses direction because reversing the current direction reverses the interaction between the current's magnetic field and the external field. F = BIL still holds, but the sign of the force (direction) depends on the current direction.
Challenge 4 — Extended Response (6 marks): Explain in detail how a simple DC motor works, including the roles of the split-ring commutator and carbon brushes. In your answer, refer to the motor effect and Fleming's left-hand rule.
Model Answer (award marks for the following points):
1. A current flows from the DC supply through the carbon brushes and into the coil via the split-ring commutator.
2. The coil is placed between the poles of a permanent magnet, so it sits in a magnetic field.
3. The motor effect applies: the current-carrying sides of the coil experience forces due to the interaction of the current's magnetic field with the external field (F = BIL).
4. By Fleming's left-hand rule, the two sides of the coil perpendicular to the field carry current in opposite directions, so they experience forces in opposite directions — creating a couple that rotates the coil.
5. After the coil has rotated half a turn, without intervention the forces would now oppose the rotation. However, the split-ring commutator (two half-rings that rotate with the coil) swaps which brush contacts which half of the ring, reversing the current direction through the coil at this exact moment.
6. This current reversal means the forces on the coil sides remain in the same rotational direction, so the coil continues to rotate in the same sense continuously.
7. The carbon brushes are stationary and maintain electrical contact with the rotating commutator without tangling the supply wires.
Award: 1 mark each for any 6 clearly explained points from the above.