Why does a current-carrying wire experience a force in a magnetic field, and how is this used in motors?
Calculate the force on a current-carrying conductor in a magnetic field and explain its role in the operation of a motor.
How a current in a magnetic field feels a force F = BIL sin theta, the right-hand rule for its direction, and how this produces the turning effect in an electric motor.
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What this dot point is asking
You need to calculate the force on a current-carrying conductor in a magnetic field and explain how this force drives an electric motor.
The motor effect
A current is just charges in motion, so the same magnetic force that acts on moving charges acts on a current-carrying wire. Summing the force over all the charges in a length of wire gives:
Direction of the force
Use the right-hand rule (treating conventional current as the direction of motion of positive charge): point fingers along the current , curl toward the field ; the thumb/palm gives the force. Equivalently the flat-hand rule: fingers point along , thumb along , palm pushes in the force direction. The force is perpendicular to both the current and the field.
How a DC motor works
A simple motor is a current-carrying coil free to rotate in a magnetic field:
- Current flows through the coil. The two sides of the coil carry current in opposite directions.
- By the motor effect, the two sides feel forces in opposite directions (one up, one down), producing a turning effect (torque) on the coil.
- The coil rotates. As it passes the vertical, the forces would now oppose the rotation.
- A split-ring commutator reverses the current direction in the coil every half-turn, so the torque keeps driving the coil the same way and rotation continues.
The turning effect is largest when the coil is parallel to the field (forces have the longest lever arm) and zero when the coil is perpendicular to the field - but the coil's momentum carries it through this point.
Exam-style practice questions
Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 SACE Stage 23 marksA straight conductor carries a current of 0.120 A and is placed within a uniform magnetic field directed into the plane of the page. The magnitude of the magnetic field is 0.350 T and the length of the conductor within the field is 2.50 x 10^-2 m. Determine the magnitude and direction of the force acting on the conductor due to the magnetic field.Show worked answer →
The force on a current-carrying conductor is F = B I l sin theta. The field is perpendicular to the current (theta = 90 degrees, so sin theta = 1).
F = B I l = (0.350)(0.120)(2.50 x 10^-2) = 1.05 x 10^-3 N.
For the direction, use the right-hand rule (or the right-hand slap rule): point the fingers in the direction of the conventional current, curl toward the field (into the page), and the thumb gives the force direction. With current to the right and field into the page, the force is directed upward (out of the plane toward the top of the page in the diagram).
1 mark for F = B I l, 1 mark for the magnitude of about 1.05 x 10^-3 N, 1 mark for the correct direction from the right-hand rule. This sideways force on current-carrying wires is what produces the turning effect in an electric motor.