Magnetic force and the Lorentz force
Currents make magnetic fields
In 1820, Hans Christian Oersted noticed something strange. A compass needle kept near a wire swung round whenever a current was switched on in the wire.
Place compass needles in a circle around a long straight wire. With the current on, every needle points along the circle, at a tangent to it. Reverse the current, and every needle turns round.
Sprinkle iron filings on a card around the wire, and they line up in concentric circles. So a current, which is moving charge, produces a magnetic field around it.
On paper we draw a current coming out of the page as a dot, and a current going into the page as a cross. Like the tip and the tail of an arrow.
A charge at rest produces an electric field E. A moving charge produces a magnetic field B as well. Both fields obey the superposition principle: the fields of many sources simply add as vectors.
Now turn it round. What does a magnetic field do to a charge q moving through it with velocity v? It pushes it with a force F equals q, v cross B.
Its size is q v B sin θ, where θ is the angle between v and B. So a charge at rest feels no magnetic force, and neither does a charge moving along the field.
The force is perpendicular to both v and B. Point the fingers of your right hand along v, curl them towards B, and your thumb gives the force on a positive charge. On a negative charge, it is the other way.
If there is also an electric field, the total force is F equals q, E plus v cross B. This is the Lorentz force.
If a charge of 1 coulomb, moving at 1 metre per second at right angles to a field, feels a force of 1 newton, the field is 1 tesla. A tesla is a big field. Earth’s field is about 10⁻⁵ tesla, so we also use the gauss: 1 gauss equals 10⁻⁴ tesla.
Force on a current-carrying wire
A current in a wire is a crowd of moving charges. So a wire carrying a current in a magnetic field should feel a force too. Let us add it up.
Take a straight rod of length l and cross-section A. It has n mobile charges per unit volume, each of charge q, drifting with velocity v d.
There are n l A of them, and each feels a force q v d cross B. So the total force is n l A q, v d cross B.
But n q v d is the current density j, and j times A is the current I. So the force becomes F equals I, l cross B, where the vector l points along the current.
Its size is I l B sin θ. Largest when the wire is perpendicular to B, zero when the wire lies along B.
Now an example. A straight wire of mass 200 grams and length 1.5 metres carries a current of 2 amperes. A horizontal magnetic field holds it up in mid-air. How strong must the field be?
The magnetic force must balance the weight: I l B equals m g. So B equals m g over I l: 0.2 times 9.8, divided by 2 times 1.5.
That is 1.96 over 3, or about 0.65 tesla. Earth’s field, about 4 × 10⁻⁵ tesla, is far too weak to do this.
Which way must the field point? If the current flows east, l cross B must point up. East cross north is up, so the field must point north, at right angles to the wire.
Try it: q v × B in 3D
- Drag the scene to look at the three arrows from different sides. F is always perpendicular to both v and B.
- Change θ from 0° to 180°. The force is largest at 90° and zero at 0° and 180°.
- Switch to the electron: the force flips, because q is negative.
- Press Launch to set the charge free. You will study its path in the next lesson.
Which way is the force?
An electron moves towards the east through a magnetic field that points horizontally towards the north. In which direction is the magnetic force on the electron?
- Vertically upwards
- Vertically downwards
- Towards the west
- Zero, because v is perpendicular to B
Answer
B. F = q (v × B). East × north points up, so a positive charge would be pushed upwards. The electron’s charge is negative, so the force is downwards. The force is largest, not zero, when v ⊥ B (sin 90° = 1), and it is never along v, so it cannot point west.
A wire held up by a field
A straight horizontal wire of mass 200 g and length 1.5 m carries a current of 2 A. What horizontal magnetic field, perpendicular to the wire, holds it up in mid-air? (g = 9.8 m/s²)
- 0.067 T
- 0.65 T
- 0.98 T
- 6.5 T
Answer
B. The magnetic force balances the weight: I l B = m g, so B = m g / (I l) = (0.2 × 9.8)/(2 × 1.5) = 1.96/3 ≈ 0.65 T. Leaving out g gives 0.067 T, leaving out the length gives 0.98 T, and using 2 kg for 200 g gives 6.5 T.
Summary
- Magnetic force on a moving charge:
F = q (v × B), F = q v B sin θ - Lorentz force:
F = q [E + v × B] - Force on a current-carrying wire:
F = I (l × B), F = I l B sin θ - Units:
1 T = 1 N s C⁻¹ m⁻¹ = 1 N A⁻¹ m⁻¹; 1 G = 10⁻⁴ T
- A current (moving charge) produces a magnetic field; Oersted saw it deflect a compass needle.
- The magnetic force is perpendicular to both v and B. It is zero for a charge at rest and for a charge moving along B.
- The force on a negative charge is opposite to that on a positive charge with the same velocity.
- A wire carrying current I feels F = I l × B, because the field pushes each drifting charge in it.