Coaching Centre
The motor effect
The force on a current-carrying conductor in a magnetic field, F = lIB sin θ, where it comes from, how its size depends on current, length, field and angle, finding its direction, and measuring it with a current balance.
Every loudspeaker, from the one in your earbuds to the stacks at a concert, contains a small coil of wire sitting in the gap of a permanent magnet. Feed the music through the coil as a changing current and the coil is pushed in and out hundreds or thousands of times a second, dragging the speaker cone with it. That push is the motor effect.
A force on a current
In 1821, a year after Ørsted showed that a current makes a magnetic field, Michael Faraday showed the reverse interaction: a wire carrying a current in a magnetic field feels a force. This is the motor effect. Its size is
$$F = lI_{\perp}B = lIB\sin\theta$$
where l is the length of wire in the field (m), I is the current (A), B is the field strength (T) and θ is the angle between the wire (the direction of the current) and the field. The force is a maximum, $lIB$, when the wire is at right angles to the field, and zero when the wire runs parallel to it.
Where does it come from?
A current is just moving charge, and you already know that each moving charge feels qvB. Suppose a wire of length l has n free electrons per cubic metre, cross-sectional area A, and the electrons drift at speed vd. There are nAl charges in the field, each feeling $qv_dB$, so the total force is $F = nAlqv_dB$. But the current is $I = nAqv_d$, so
$$F = (nAqv_d)\,lB = lIB$$
The magnetic forces on the electrons push them sideways against the metal lattice, and that push is passed on to the wire as a whole. Another way to picture it: the wire's own circular field adds to the external field on one side and partly cancels it on the other, and the wire is pushed from the stronger field towards the weaker.
Which way does it push?
Use the same right-hand palm rule as for a moving positive charge: fingers along B, thumb along the conventional current I, and the palm pushes in the direction of F. Conventional current runs from + to − around a circuit, opposite to the electrons' drift, so you never need to reverse anything for a current in a wire. (Some books use Fleming's left-hand rule instead; it always gives the same answer.)
Reversing either the current or the field reverses the force. Reversing both leaves it unchanged.
A 30 cm length of wire carries 4.0 A through a uniform 0.20 T field. Find the force on it when the wire is (a) at right angles to the field, (b) at 30° to the field, (c) parallel to the field.
$F = lIB\sin 90^\circ = (0.30)(4.0)(0.20)(1) = 0.24$ N.
$F = (0.30)(4.0)(0.20)\sin 30^\circ = 0.12$ N. Only the component of the current perpendicular to the field, $I\sin\theta$, counts.
$\sin 0^\circ = 0$, so $F = 0$. A wire lying along the field lines feels no force.
Measuring the force: the current balance
The forces involved are small, often a few hundredths of a newton, so the neatest way to measure them is with an electronic balance. Sit a U-shaped magnet on the balance and clamp a straight wire (or a rigid copper rod) so it runs through the gap between the poles without touching the magnet. When a current flows, the field pushes the wire one way; by Newton's third law the wire pushes the magnet the opposite way with an equal force, and the balance reading changes.
- 4 cm of wire, 100 mT, 5.0 A at 90°. Check the force (0.0200 N) and the change in reading (F/g = 2.04 g).
- Press Reverse current. Why does the reading now go up? Use Newton's third law.
- Record a point at every 1.0 A from 1 A to 6 A. Use the gradient of the F–I graph to find B, and compare with the slider value.
- Set I = 5.0 A and turn the wire from 90° down to 0°. Record F every 15° and plot F against sin θ. What shape do you expect?
- Which single change doubles the force: doubling the length in the field, or doubling the angle from 30° to 60°? Test it.
In a current balance experiment, 5.0 cm of rod lies at right angles to the field. The gradient of the line of best fit of force against current is 4.2 × 10−3 N A−1. Find B.
$F = (lB\sin\theta)\,I$, so the gradient is $lB\sin\theta = lB$ at 90°.
$B = \dfrac{4.2 \times 10^{-3}}{0.050} = 0.084$ T.
At 3.0 A the predicted force is 0.050 × 3.0 × 0.084 = 0.0126 N, a change in reading of 0.0126 ÷ 9.8 = 1.29 g. Readings this small are why a 0.01 g balance is needed.
Test yourself
A wire carrying a current lies in a uniform magnetic field. The force on it is greatest when the wire is
F = lIB sin θ is a maximum at θ = 90°.
A horizontal wire runs east–west and carries a current to the east. The magnetic field is vertically downwards. The force on the wire is towards the
Fingers down, thumb east: the palm faces north.
The current through a wire in a magnetic field is tripled and the length of wire in the field is halved. The force becomes
F ∝ lI: 3 × ½ = 1.5.
In a current balance, the balance reading increases by 2.0 g when the current is switched on. The force on the wire is
The magnet is pushed down with 0.0020 × 9.8 = 0.020 N, so the wire is pushed up with an equal and opposite force.
A 0.50 m wire carries 2.0 A at 60° to a 0.40 T field. The force on it is
F = 0.50 × 2.0 × 0.40 × sin 60° = 0.35 N.
Label the current balance.
What does each change do to the force on the wire?
A current-carrying conductor in a magnetic field experiences a force called the effect. The force is a maximum when the wire is to the field. It is directly proportional to the current, the field strength and the of wire in the field. In the right-hand palm rule the thumb points along the conventional . In a current balance the force on the magnet is equal and opposite to the force on the wire, by Newton's law.
A student uses a current balance to investigate the relationship between the force on a conductor and the current in it. Outline a valid method, and explain how the results could be used to determine the magnetic field strength.
Place a U-shaped magnet on an electronic balance and tare it. Clamp a straight, rigid copper rod to a stand so that a measured length (say 5.0 cm) passes through the gap between the poles at right angles to the field, without touching the magnet. Connect the rod in series with a DC supply, rheostat and ammeter. Vary the current from 0.5 A to 5.0 A in 0.5 A steps and record the balance reading for each; repeat each reading three times and again with the current reversed, averaging the magnitudes. Keep the length in the field, the angle and the magnet the same throughout, and switch off between readings so the rod doesn't heat up. Convert each change in reading to a force using F = Δm × g, with Δm in kilograms. Plot force against current: the result should be a straight line through the origin, showing F ∝ I. Since F = lIB sin θ with θ = 90°, the gradient of the line equals lB, so B = gradient ÷ l.