D.3 Motion in electromagnetic fields | IB DP Physics

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🎯What you need to be able to do

Use \( F = qvB\sin\theta \) for a charge moving in a magnetic field, and find its direction.
Explain why a magnetic force does no work, and analyse the resulting circular motion.
Describe the parabolic path of a charge in a uniform electric field.
Analyse crossed electric and magnetic fields, including the velocity selector.

📚The physics

The magnetic force on a moving charge is \( F = qvB\sin\theta \), where \(\theta\) is the angle between the velocity and the field. Two consequences to fix immediately: a charge moving parallel to the field feels no force at all, and a stationary charge feels no magnetic force however strong the field. Magnetic fields only act on charge that is moving across them.

Direction comes from Fleming’s left-hand rule, with the second finger along the conventional current — so for a negative charge, point it opposite to the particle’s actual motion. Forgetting that reversal is the standard slip.

Why a magnetic force does no work. The force is always perpendicular to the velocity. From \( W = Fs\cos\theta \), a perpendicular force does zero work. So a magnetic field can change a particle’s direction but never its speed, and never its kinetic energy. If a question tells you a magnetic field speeds a particle up, something else is going on.

Circular motion follows immediately. A constant-magnitude force always perpendicular to the velocity is exactly the condition for a circle. Setting the magnetic force equal to the centripetal requirement, \( qvB = mv^{2}/r \), gives

\[ r = \frac{mv}{qB} \]

Heavier or faster particles curve less; stronger fields or larger charges curve them more. This single relation is the working principle of the mass spectrometer, the bubble chamber and the particle accelerator.

The period of that circular motion, \( T = 2\pi m/qB \), has a striking feature: it does not depend on the speed. A fast particle traces a bigger circle in the same time. That is what makes a cyclotron possible.

In a uniform electric field the behaviour is quite different. The force \(qE\) is constant in magnitude and direction, independent of velocity, so a charge fired across the field follows a parabola — mathematically identical to projectile motion under gravity, with \( qE/m \) playing the role of \(g\). Unlike the magnetic case, an electric field does work and does change kinetic energy.

Crossed fields and the velocity selector. Arrange \(E\) and \(B\) perpendicular to each other and to the beam so that the electric force \(qE\) opposes the magnetic force \(qvB\). Only particles for which the two balance pass straight through:

\[ qE = qvB \qquad\Longrightarrow\qquad v = \frac{E}{B} \]

Notice that the selected speed depends on neither the charge nor the mass — the device sorts purely by speed, which is exactly what makes it useful as the front end of a mass spectrometer.

✏️Worked example

A proton enters a uniform magnetic field of 0.35 T at right angles, moving at \( 4.6 \times 10^{6} \) m s\(^{-1}\). Take \( q = 1.60 \times 10^{-19} \) C and \( m = 1.67 \times 10^{-27} \) kg.

(a) Magnitude of the force. \( F = qvB = 1.60 \times 10^{-19} \times 4.6 \times 10^{6} \times 0.35 = 2.6 \times 10^{-13} \) N.

(b) Radius of its path. \( r = mv/qB = (1.67 \times 10^{-27} \times 4.6 \times 10^{6})/(1.60 \times 10^{-19} \times 0.35) = (7.68 \times 10^{-21})/(5.60 \times 10^{-20}) = 0.137 \) m.

(c) Period and frequency. \( T = 2\pi m/qB = (2\pi \times 1.67 \times 10^{-27})/(5.60 \times 10^{-20}) = 1.87 \times 10^{-7} \) s, so \( f = 5.3 \) MHz. A cyclotron accelerating these protons would need its alternating supply at exactly that frequency, and it would stay correct as the protons speed up.

(d) An electric field is now applied so the proton travels straight. What field strength is needed? \( qE = qvB \), so \( E = vB = 4.6 \times 10^{6} \times 0.35 = 1.6 \times 10^{6} \) V m\(^{-1}\).

Worth noticing. In (b) the proton’s speed is enormous but the radius is only 14 cm — magnetic fields bend charged particles very tightly, which is why particle detectors can fit in a room. And the answer to (c) is independent of how fast the proton is going, which is the fact the whole cyclotron design rests on.

🔭See it happen

Any bubble chamber photograph is this topic made visible: spiral tracks curving one way for positive particles and the other for negative, tightening as the particles lose energy and \( r = mv/qB \) shrinks. Search for CERN bubble chamber images and read a few tracks — sign of charge from the direction of curvature, momentum from the radius.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. An electron travels at \( 2.0 \times 10^{7} \) m s\(^{-1}\) perpendicular to a magnetic field of 0.020 T. Find the force on it. Take \( q = 1.6 \times 10^{-19} \) C.

\( F = qvB = 1.6 \times 10^{-19} \times 2.0 \times 10^{7} \times 0.020 = 6.4 \times 10^{-14} \) N.

2. Find the radius of that electron’s circular path. Take \( m = 9.11 \times 10^{-31} \) kg.

\( r = \dfrac{mv}{qB} = \dfrac{9.11 \times 10^{-31} \times 2.0 \times 10^{7}}{1.6 \times 10^{-19} \times 0.020} = \dfrac{1.82 \times 10^{-23}}{3.2 \times 10^{-21}} = 5.7 \times 10^{-3} \) m, about 5.7 mm.

3. Find the period of that circular motion, and state what happens to it if the electron enters twice as fast.

\( T = \dfrac{2\pi m}{qB} = \dfrac{2\pi \times 9.11 \times 10^{-31}}{3.2 \times 10^{-21}} = 1.8 \times 10^{-9} \) s. The speed does not appear, so the period is unchanged — a faster electron simply traces a bigger circle in the same time. This is what makes the cyclotron work.

4. Explain why a magnetic field can never change the speed of a charged particle.

The magnetic force \( qvB\sin\theta \) is always perpendicular to the velocity. Work is \( W = Fs\cos\theta \), and for a perpendicular force \( \cos 90^\circ = 0 \), so no work is done. With no work done the kinetic energy, and therefore the speed, cannot change — only the direction does.

5. A velocity selector has \( E = 3.0 \times 10^{5} \) V m\(^{-1}\) and \( B = 0.15 \) T in crossed fields. Find the speed of the particles that pass straight through.

The forces balance: \( qE = qvB \), so \( v = E/B = \dfrac{3.0 \times 10^{5}}{0.15} = 2.0 \times 10^{6} \) m s\(^{-1}\). The charge and mass both cancel, so it selects purely by speed.

6. An electron is accelerated from rest through a potential difference of 500 V. Find its final speed.

All the electrical work becomes kinetic energy: \( qV = \tfrac{1}{2}mv^{2} \), so \( v = \sqrt{\dfrac{2qV}{m}} = \sqrt{\dfrac{2 \times 1.6 \times 10^{-19} \times 500}{9.11 \times 10^{-31}}} = \sqrt{1.76 \times 10^{14}} = 1.3 \times 10^{7} \) m s\(^{-1}\). Unlike a magnetic field, an electric field does change the speed.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

HyperPhysics — magnetic force on a moving charge, cyclotron and mass spectrometer
The Physics Hypertextbook — the magnetic force