Force on a Moving Charge, Circular Paths and Velocity Selection

A current is a stream of moving charges, so if a magnetic field pushes on a current, it must push on each moving charge. That force, $F = BQv\sin\theta$, bends the paths of electrons, protons and ions into circles. It is used to steer particle beams, to measure the masses of ions in a mass spectrometer and, combined with an electric field, to pick out particles of a single speed. Paper 4 regularly sets multi-part questions linking this force to circular motion and to electric fields.

From current to charges

Consider a wire of cross-sectional area $A$ containing $n$ charge carriers per unit volume, each with charge $q$ and drift speed $v$. From AS, the current is $I = nAvq$. A length $L$ of the wire contains $nAL$ charge carriers.

Put the wire at right angles to a field $B$. The force on the length $L$ is

$$F = BIL = B(nAvq)L$$

Share this force equally between the $nAL$ carriers:

$$F_{\text{one charge}} = \frac{BnAvqL}{nAL} = Bqv$$

The force on the wire is simply the sum of the forces on all the moving charges inside it. For a charge moving at an angle $\theta$ to the field, only the velocity component $v\sin\theta$ perpendicular to the field counts.

Direction of the force

Use Fleming's left-hand rule with the second finger pointing in the direction of the conventional current that the moving charge represents.

• A positive charge moving to the right is a current to the right. Point the second finger along the velocity.
• A negative charge (an electron) moving to the right is a conventional current to the left. Point the second finger opposite to the velocity, or find the force on a positive charge and reverse it.

The force is always perpendicular to both the velocity and the field.

Magnetic forces do no work

Because the magnetic force is always perpendicular to the velocity, it has no component along the direction of motion. It cannot speed the particle up or slow it down; it only changes the direction.

This is the opposite of an electric field, which accelerates charges along the field lines and changes their kinetic energy.

Circular motion in a uniform magnetic field

Suppose a charged particle enters a uniform field at right angles to the field lines. The force has constant size $BQv$ (the speed does not change) and is always perpendicular to the velocity. A constant-size force always perpendicular to the velocity is exactly the condition for uniform circular motion: the magnetic force provides the centripetal force.

Equating the magnetic force to the centripetal force:

$$BQv = \frac{mv^{2}}{r}$$

What this tells you:

Change	Effect on radius
Faster particle (larger $v$)	larger circle
Heavier particle (larger $m$), same speed and charge	larger circle
Larger charge	smaller circle
Stronger field	smaller circle
Opposite sign of charge	same size circle, curving the opposite way

Period and specific charge

The time for one orbit is $T = 2\pi r/v$. Substituting $r = mv/BQ$:

$$T = \frac{2\pi m}{BQ}$$

The period does not depend on the speed: a faster particle travels a larger circle in exactly the same time. (This is the principle of the cyclotron, a circular particle accelerator; the cyclotron itself is not on the syllabus.)

The ratio $Q/m$ is called the specific charge of the particle (unit $\text{C kg}^{-1}$). Rearranging $r = mv/BQ$ gives $Q/m = v/Br$, so measuring the radius of a beam's path in a known field, at a known speed, gives the specific charge. This is how the electron's specific charge, $e/m_e = 1.76 \times 10^{11}\ \text{C kg}^{-1}$, was first measured.

A common context: the particle is first accelerated from rest through a potential difference $V$, so that $\tfrac{1}{2}mv^{2} = QV$. Combining this with $r = mv/BQ$ eliminates $v$:

$$r = \frac{1}{B}\sqrt{\frac{2mV}{Q}}$$

Electric and magnetic deflection compared

	Uniform electric field	Uniform magnetic field
Force on a stationary charge	$EQ$	zero
Force on a moving charge	$EQ$, same whatever the velocity	$BQv\sin\theta$, depends on speed and direction
Direction of force	along the field lines (positive charge)	perpendicular to both field and velocity
Path when entering at right angles	parabola	circular arc
Effect on speed	changes it (work is done)	none (no work done)

Velocity selection

A velocity selector lets through only those charged particles that have one particular speed. It uses an electric field and a magnetic field at right angles to each other and to the beam (often called crossed fields).

For a positive ion moving to the right with the top plate positive, the electric force $EQ$ acts downwards (towards the negative plate). With the magnetic field into the page, Fleming's left-hand rule gives an upwards magnetic force $BQv$. If the two forces are equal, the resultant force is zero and the ion travels in a straight line:

$$EQ = BQv$$

The charge cancels, so the selected speed is the same for every particle, whatever its charge, sign or mass.

• A particle that is faster than $E/B$ has $BQv > EQ$: the magnetic force wins and it is deflected one way (upwards in the diagram, for a positive ion).
• A particle that is slower has $BQv < EQ$: the electric force wins and it is deflected the other way.

A slit at the exit lets through only the undeflected particles.

The mass spectrometer

A mass spectrometer combines both ideas. Ions first pass through a velocity selector so they all have the same speed $v$. They then enter a region with only a magnetic field $B$ and travel in semicircles of radius $r = mv/BQ$. Ions with the same charge but different masses (for example different isotopes) follow semicircles of different radii and land at different points on a detector. With $v$, $B$ and $Q$ known, measuring $r$ gives the mass.

Worked examples

Practice questions

Answers

1. When the charge is stationary; when it moves parallel (or antiparallel) to the field lines.
2. $F = BQv\sin\theta = 0.45 \times 1.60 \times 10^{-19} \times 2.0 \times 10^{7} \times \sin 30^\circ = 7.2 \times 10^{-13}\ \text{N}$.
3. (a) $r = mv/BQ = (1.67 \times 10^{-27} \times 4.0 \times 10^{6})/(0.30 \times 1.60 \times 10^{-19}) = 0.14\ \text{m}$. (b) $T = 2\pi m/BQ = 2\pi \times 1.67 \times 10^{-27}/(0.30 \times 1.60 \times 10^{-19}) = 2.2 \times 10^{-7}\ \text{s}$ (equivalently $2\pi r/v$).
4. (a) $v = E/B = 2.5 \times 10^{4}/0.020 = 1.25 \times 10^{6}\ \text{m s}^{-1}$. (b) $V = Ed = 2.5 \times 10^{4} \times 0.020 = 500\ \text{V}$.
5. $B = m_e v/(er) = (9.11 \times 10^{-31} \times 1.2 \times 10^{7})/(1.60 \times 10^{-19} \times 0.050) = 1.4 \times 10^{-3}\ \text{T}$.
6. $r = mv/BQ$: same $v$, $B$ and $Q$, so $r \propto m$ and the deuteron's radius is twice the proton's. $T = 2\pi m/BQ$, so the deuteron also takes twice as long per orbit.
7. (a) $v = \sqrt{2eV/m_e} = \sqrt{2 \times 1.60 \times 10^{-19} \times 300/9.11 \times 10^{-31}} = 1.03 \times 10^{7}\ \text{m s}^{-1}$; $r = m_ev/(Be) = 9.11 \times 10^{-31} \times 1.03 \times 10^{7}/(2.0 \times 10^{-3} \times 1.60 \times 10^{-19}) = 0.029\ \text{m}$. (b) $v \propto \sqrt{V}$, so quadrupling $V$ doubles $v$; $r \propto v$, so the radius doubles to $0.058\ \text{m}$.
8. (a) $v = E/B_1 = 1.8 \times 10^{5}/0.12 = 1.5 \times 10^{6}\ \text{m s}^{-1}$. (b) $r_6 = (6.0 \times 1.66 \times 10^{-27} \times 1.5 \times 10^{6})/(0.30 \times 1.60 \times 10^{-19}) = 0.311\ \text{m}$; $r_7 = \tfrac{7}{6} \times 0.311 = 0.363\ \text{m}$. Each lands a diameter from the entry point, so the separation is $2(0.363 - 0.311) = 0.10\ \text{m}$. (c) $r = mv/BQ$ depends on both $m$ and $v$. If the ions had a range of speeds, ions of the same mass would land at different points and the isotopes could not be separated. Selecting a single speed makes $r$ depend only on mass (for the same charge).