1The idea in one paragraph

A charged particle in an electric field feels a force qE whether it is moving or not. In a uniform field that force is constant, so the particle either speeds up along the field, gaining energy qV, or follows a parabola across it, exactly like a ball thrown sideways. A charged particle in a magnetic field feels a force only if it is moving across the field, and that force, qvB sin θ, is always at right angles to its velocity. A force at right angles to the motion does no work, so the speed never changes: the particle moves in a circle, with a radius that depends on its mass, charge and speed. That radius is how the charge-to-mass ratio of the electron is measured. Put an electric and a magnetic field at right angles and the two forces can cancel for one speed only. A current is moving charge, so a wire carrying a current in a magnetic field also feels a force, BIL sin θ, and two wires side by side push or pull on each other through their magnetic fields.

2A charged particle in a uniform electric field

In a uniform electric field E, a particle of charge q feels a constant force F = qE, so it has a constant acceleration

a = qE ÷ m

A positive charge accelerates along the field; a negative charge, against it. Gravity is ignored for electrons and ions: D.2 showed the electric force on an electron in an ordinary field is more than 10¹⁴ times its weight.

Along the field: gaining energy. A particle released from rest at one plate and accelerated to the other has had work done on it: W = Fd = qEd = qV, since E = V/d. All of that work becomes kinetic energy:

An electron accelerated from rest through 500 V reaches v = √(2 × 1.60 × 10⁻¹⁹ × 500 ÷ 9.11 × 10⁻³¹) = 1.3 × 10⁷ m s⁻¹. This is what an electron gun does, and it is the first step of most questions in this subtopic. The kinetic energy is 500 eV whatever the mass: in electronvolts, a particle of charge e gains as many eV as the volts it is accelerated through.

Across the field: a parabola. Now fire the particle in at right angles to the field. Along its original direction there is no force, so that velocity stays constant. Across, there is a constant acceleration. This is projectile motion from A.1, with qE/m in place of g, so the path between the plates is a parabola. Figure 1 shows an electron curving towards the positive plate.

Worked example 1. An electron enters the field midway between two plates at 3.0 × 10⁷ m s⁻¹, moving parallel to them. The plates are 5.0 cm long and 2.5 cm apart, with a pd of 500 V. Find how far the electron is deflected by the time it leaves the plates, and the angle at which it leaves.

The deflection of 4.9 mm is less than the 12.5 mm to the plate, so the electron clears it. Once it leaves the field there is no force at all, and it continues in a straight line at 11° to its original direction. Note that the field does change the electron's speed here: it gains a sideways velocity, so its kinetic energy rises. An electric field can always do work on a charge.

3The magnetic force on a moving charge

A charge sitting still in a magnetic field feels nothing. A charge moving through it feels a force whose size is

F = qvB sin θ

where v is the speed, θ is the angle between the velocity and the field, and B is the magnetic field strength. This equation defines B: it is the force per unit charge per unit speed on a charge moving at right angles to the field. The unit is the tesla, T, and 1 T = 1 N s C⁻¹ m⁻¹, which is the same as 1 N A⁻¹ m⁻¹. For scale, the Earth's field is about 5 × 10⁻⁵ T and a strong laboratory magnet about 1 T.

Figure 2(a) shows why sin θ is there. Only the part of the velocity at right angles to the field, v sin θ, counts. A charge moving along the field lines (θ = 0) feels no force at all; one moving straight across them (θ = 90°) feels the largest force, qvB.

The direction is at right angles to both the velocity and the field. Use the right-hand rule of Figure 2(b): hold your right hand flat, fingers pointing along the field B, thumb along the velocity of a positive charge; the force comes out of your palm, the way you would push. For a negative charge, the force is the other way: find it as for a positive charge and reverse it. (If your school taught Fleming's left-hand rule, it gives the same answers; use whichever one you trust, but use it the same way every time.)

Worked example 2. A proton moves at 2.0 × 10⁶ m s⁻¹ at right angles to a uniform field of 0.15 T. Find the force on it. What would the force be if it moved at 30° to the field?

4Circular motion in a uniform magnetic field

The magnetic force is always at right angles to the velocity. A force at right angles to the motion does no work, so it cannot change the particle's speed or its kinetic energy. It can only change the direction. A force of constant size that always acts at right angles to the velocity is exactly the centripetal force of A.2, so a charged particle moving at right angles to a uniform magnetic field moves in a circle at constant speed. Figure 3 shows the two senses of rotation.

The kinetic energy of a charged particle stays constant in a magnetic field. The magnetic force bends the path; it never speeds the particle up or slows it down.

The magnetic force provides the centripetal force:

This is the result to know how to derive, because questions ask you to "show that" it. Read it one variable at a time: a faster or heavier particle is harder to bend, so its circle is bigger; a larger charge or a stronger field bends it more, so its circle is smaller. Since mv is the momentum, the radius measures the momentum per unit charge.

The time for one orbit is

and the speed has cancelled. A faster particle travels a larger circle in exactly the same time. For the proton of worked example 2:

What a path tells you about a particle. This is the guiding question of the subtopic. The sense of the curve gives the sign of the charge: in the same field, positive and negative particles curl opposite ways, as in Figure 3. The radius gives mv/q, so particles of the same speed and charge but different mass separate into circles of different sizes. If the velocity is not at right angles to the field, its component along the field is untouched while the component across it goes round in a circle, so the particle spirals along the field lines in a helix.

5Measuring the charge-to-mass ratio

The guide requires you to determine the charge-to-mass ratio q/m of a particle from its path in a uniform magnetic field. For the electron, the classic arrangement is the fine-beam tube of Figure 4(a): an electron gun accelerates electrons through a pd V, and a uniform magnetic field (from a pair of coils outside the tube) bends the beam into a circle. A trace of low-pressure gas in the tube glows where the electrons pass, so the circle can be seen and its radius measured.

Two equations do all the work: the gun gives the speed, and the circle gives the radius.

Worked example 3. Electrons accelerated through 250 V move in a circle of radius 4.5 cm in a field of 1.2 mT. Find q/m for the electron.

That is within 3% of the accepted value, 1.76 × 10¹¹ C kg⁻¹. A single reading carries all its uncertainty into the answer, so a better experiment varies V at fixed B and plots r² against V, as in Figure 4(b). The model predicts a straight line through the origin, and

The radius is squared, so its percentage uncertainty doubles; the radius measurement is usually the weakest part of the experiment.

This is how the electron was identified. In 1897 J. J. Thomson measured the charge-to-mass ratio of cathode rays, using electric and magnetic deflection, and found it far larger than for any ion. The particles had to be far lighter than any atom. Combine the ratio with Millikan's value of e from D.2 and the mass of the electron follows: m = 1.60 × 10⁻¹⁹ ÷ 1.76 × 10¹¹ = 9.1 × 10⁻³¹ kg.

6Crossed electric and magnetic fields

Put a uniform electric field and a uniform magnetic field at right angles to each other and to the particle's velocity. Figure 5 shows a positive ion moving to the right, with E pointing down and B into the page. The electric force qE points down. By the right-hand rule the magnetic force qvB points up.

The two forces cancel when

A particle with exactly this speed passes through in a straight line. A faster one feels a larger magnetic force, which wins, and it curves one way; a slower one feels a smaller magnetic force, the electric force wins, and it curves the other way. The charge and the mass have cancelled, so the arrangement picks out one speed regardless of what the particle is. Used with a narrow exit slit, it is called a velocity selector.

Its use is to feed particles of one known speed into a second region, where a magnetic field alone bends them into circles. Then r = mv/qB, with v and B known, measures m/q directly. That combination is a mass spectrometer.

Worked example 4. A velocity selector has E = 1.0 × 10⁴ V m⁻¹ and B = 0.10 T. The selected singly charged ions then enter a region where the same 0.10 T field acts alone, and move on a circle of radius 0.207 m. Identify the ions. (Take 1 u = 1.661 × 10⁻²⁷ kg.)

A singly charged ion of mass 20 u is neon-20. Neon-22 ions in the same apparatus would move on a circle of radius 0.228 m, so after a half-turn the two isotopes land about 4 cm apart. That separation is how isotopes are counted.

7The force on a current-carrying conductor

A current is a flow of charge, so a wire carrying a current through a magnetic field feels the sum of the forces on all its moving charges. The result is

F = BIL sin θ

where I is the current, L is the length of wire in the field and θ is the angle between the wire (the current direction) and the field. To see where it comes from, think of a charge q that takes a time t to travel the length L: its speed is v = L/t, and the current it makes is I = q/t, so qv = q × L/t = IL. Replace qv by IL in F = qvB sin θ and you have F = BIL sin θ.

The direction comes from the same right-hand rule, with the thumb along the conventional current. Figure 6(a) shows a wire between the poles of a magnet: the field runs from N to S, the current comes out of the page, and the force is upwards. Reverse the current or the field and the force reverses. Figure 6(b) shows why sin θ appears: only the length of wire across the field feels a force, and a wire lying along the field feels none.

Worked example 5. A straight wire carries 3.0 A through a uniform 0.20 T field. The length of wire in the field is 5.0 cm. Find the force when the wire is at right angles to the field and when it is at 30° to it.

This force is what turns an electric motor: a coil carrying a current sits in a magnetic field, and the forces on its two sides push in opposite directions.

8The force between two parallel wires

Two long straight wires carrying currents exert forces on each other, although neither is a magnet. Each wire sets up the circular field of D.2 around itself, and the other wire, carrying a current, sits in that field and feels a force BIL.

Follow it through in Figure 7(a), where both currents come out of the page. By the right-hand grip rule, the field of wire 1 circles anticlockwise, so at wire 2, on its right, the field B₁ points up the page. Now apply the right-hand rule to wire 2: current out of the page, field up, so the force on wire 2 points to the left, towards wire 1. The same argument applied to wire 1 in the field of wire 2 gives a force to the right. The wires attract. Reverse one current, as in Figure 7(b), and the forces reverse: the wires repel.

Currents in the same direction attract; currents in opposite directions repel.

This is the opposite of the rule for charges, which is exactly why it is a favourite multiple-choice trap. The size of the force on each metre of either wire is

F ÷ L = μ₀I₁I₂ ÷ 2πr

where r is the separation of the wires and μ₀ = 4π × 10⁻⁷ T m A⁻¹ is the permeability of free space. The forces on the two wires are always equal in size and opposite in direction, even when the currents are different, because they are a Newton's third law pair: I₁ and I₂ appear symmetrically in the formula.

Worked example 6. Two long parallel wires 5.0 cm apart carry 10 A and 20 A in the same direction. Find the force per metre on each wire.

Both wires feel 8.0 × 10⁻⁴ N on every metre, towards each other. Until 2019 the ampere itself was defined through this force: one ampere was the current that, in two very long wires one metre apart, gave a force of 2 × 10⁻⁷ N on each metre. Since 2019 the ampere has been defined by fixing the value of e instead.