Force on a Moving Charge

A charge $Q$ moving at speed $v$ through a field $B$ feels $F=BQv\sin\theta$ , always perpendicular to its velocity. So it does no work: the speed stays constant and, in a uniform perpendicular field, the charge moves in a circle of radius $r=\dfrac{mv}{BQ}$ . Crossed electric and magnetic fields let through only charges with $v=\dfrac EB$ (a velocity selector). In a conductor, the magnetic force pushes carriers to one side, creating the Hall voltage $V_H=\dfrac{BI}{ntq}$ .

Cambridge 9702 Paper 4Edexcel YPH11 WPH14AQA 7408 Paper 2≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use $F=BQv\sin\theta$ and find its direction.
Derive and use $r=\dfrac{mv}{BQ}$ for circular motion.
Explain velocity selection with crossed fields, $v=\dfrac EB$ .
Explain the Hall voltage and use $V_H=\dfrac{BI}{ntq}$ .

Pause and think

The aurora glows near the poles, not at the equator. What do Earth's magnetic field and charged particles from the Sun have to do with it?

Section 2 of 14

Core idea

Start hereThe aurora

Charged particles from the Sun spiral along Earth's magnetic field towards the poles, where they light up the sky. A magnetic field can't speed a charge up or slow it down, only bend its path into circles and spirals.

The big idea

$F=Bqv\sin\theta$ , perpendicular to $v$ : circular motion, $r=\frac{mv}{Bq}$ . Velocity selector $v=\frac EB$ . Hall voltage $V_H=\frac{BI}{ntq}$ .

1. Circular paths

A positive ion circling in a field into the page

position = 0.0

× means field into the page. The force is always towards the centre: a centripetal force,

Bqv=\dfrac{mv^2}{r}

Radius and period

$Bqv=\frac{mv^2}r$
Magnetic force is the centripetal force.
$r=\frac{mv}{Bq}$
Rearrange.
$T=\frac{2\pi r}v=\frac{2\pi m}{Bq}$
Independent of speed.

Worked exampleProton in 0.50 T

$v=2.0\times10^6$ m s $^{-1}$ .

Goal: $r$ ; period.
Strategy: Formulas.

$r=\frac{1.67\times10^{-27}\times2.0\times10^6}{0.50\times1.6\times10^{-19}}=4.2\text{ cm}$
Radius.
$T=\frac{2\pi(1.67\times10^{-27})}{0.50\times1.6\times10^{-19}}=1.3\times10^{-7}\text{ s}$
Period.

Answer

4.2

cm;

1.3\times10^{-7}

✓ Check

2\pi r/v=2\pi(0.042)/(2\times10^6)=1.3\times10^{-7}

✓.

Radius of the circle increases with speed (r = mv/Bq)

Faster particles curve less; heavier ones curve less; stronger fields curve more. Mass spectrometers use this to separate isotopes.

2. Crossed fields and the Hall effect

Velocity selector and Hall voltage

$qE=Bqv\Rightarrow v=\frac EB$
Undeflected when forces balance.
$q\frac{V_H}d=Bqv$
Hall: sideways field builds until it balances the magnetic force ( $d$ = width).
$v=\frac I{nAq}=\frac I{ndtq}$
Drift speed; $A=dt$ .
$V_H=Bvd=\frac{BI}{ntq}$
Combine.

Small

n

(semiconductors) gives a large Hall voltage: used in Hall probes.

Common misconceptionMagnetic force does work It's always perpendicular to velocity: no work, no change in speed.

Summary

$r=mv/(Bq)$ .
$v=E/B$ ; $V_H=BI/(ntq)$ .

Hall voltage: Transverse p.d. in a conductor in a magnetic field.

HistoryHall's student discovery. Edwin Hall discovered the effect in

1879

as a PhD student at Johns Hopkins, using a thin gold leaf and a galvanometer. The sign of the Hall voltage later proved that some semiconductors conduct with positive carriers ("holes").

Fun factYour phone's compass. Smartphones contain tiny Hall-effect sensors that measure Earth's

50

μT field in three directions. That's how maps know which way you're facing, and how flip covers can tell when they're closed (a small magnet in the cover).

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