Physics 9702 · Paper 4

F = BQv & Circular Paths

The BIL force is really billions of forces on the moving electrons inside the wire. Free the charge from the wire and the same force bends its path into a circle.

The force on one moving charge

F = BQv\sin\theta

Symbols

$F$ = force on the charge (N)
$Q$ = charge of the particle (C)
$v$ = speed of the particle (m s⁻¹)
$\theta$ = angle between velocity and field (—)

It is BIL written for one particle: a charge $Q$ moving at $v$ is a current $Q/t$ along a length $vt$ , and $BIL = B(Q/t)(vt) = BQv$ . Direction from Fleming's left-hand rule — but for electrons, point the second finger against the motion, because conventional current runs opposite to a negative charge (9702/41/O/N/23 Q6(b)(i)).

A charge crossing a field at right angles: F = BQv is always perpendicular to v, so the speed never changes and the path is a circle of radius r = mv/BQ. The entry and exit lines of a semicircle are separated by d = 2r, which is a common exam geometry step.

The force is always perpendicular to the velocity, so it never changes the speed — no work is done — it only turns the path. Perpendicular entry gives a circle, with $BQv = \dfrac{mv^2}{r}$ so $r = \dfrac{mv}{BQ}$ : faster or heavier particles make bigger circles; stronger fields make tighter ones.

Worked example

The 2023 electron semicircle

Electrons at $1.7\times10^{7}$ m s−¹ enter a 4.8 mT field and leave after a semicircle. Find the separation $d$ of the entry and exit paths (9702/41/O/N/23 Q6(b)).

$F = BQv = 4.8\times10^{-3} \times 1.6\times10^{-19} \times 1.7\times10^{7} = 1.3\times10^{-14}\ \text{N}$ .
$r = \dfrac{mv}{BQ} = \dfrac{9.11\times10^{-31} \times 1.7\times10^{7}}{4.8\times10^{-3} \times 1.6\times10^{-19}} = 0.020\ \text{m}$ .
Entry and exit lines of a semicircle sit a diameter apart: $d = 2r = 0.040\ \text{m}$ .

Answer

d = 4.0\ \text{cm}

The follow-up swapped in positrons at double the speed: opposite charge flips the curve's direction, double speed doubles the radius — exit at $2d$ .

The period does not care about the radius

Combine

r = mv/BQ

with

v = 2\pi r/T

and the speed cancels:

T = \dfrac{2\pi m}{BQ}

. A faster particle rides a bigger circle in exactly the same time. A 2024 show-that built this result and then asked directly: a second alpha particle at twice the radius has the same period (9702/42/F/M/24 Q5(b)). (This independence is what makes cyclotrons work.)

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