Charged Particles in Circles | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Charged Particles in Circles

When a charged particle enters a uniform magnetic field at right angles, the magnetic force is always perpendicular to its velocity.

The force changes the direction of motion, but it does not change the speed.

The magnetic force becomes the centripetal force

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A charge crossing a field at right angles: F = B|q|v is always perpendicular to v, so the speed never changes and the path is a circle of radius r = mv/(B|q|). The entry and exit lines of a semicircle are separated by d = 2r.

For perpendicular motion, the magnetic force magnitude is:
$F=B|q|v$
A force of constant size that stays perpendicular to the velocity provides centripetal acceleration:
$B|q|v=\dfrac{mv^2}{r}$
Cancel one factor of speed and rearrange:
$r=\dfrac{mv}{B|q|}$

Greater mass or speed gives a larger radius.
Greater magnetic flux density or charge magnitude gives a smaller radius.
Changing the charge sign reverses the direction of curvature.

The magnetic force is perpendicular to the motion, so it does no work. Kinetic energy and speed stay constant.

Read a visible circular path

A fine-beam tube shows a glowing circular electron path between two large magnetic coils. — The glowing trace makes the electron path visible. The magnetic force stays perpendicular to the velocity, so it changes direction but not speed.

Worked example

An electron follows a semicircle

Electrons of speed 1.7 × 10⁷ m s⁻¹ enter a 4.8 mT field. They leave after a semicircle. Find the separation between the entry and exit paths (9702/41/O/N/23 Q6(b)).

Calculate the radius.
$r=\dfrac{mv}{B|q|}$
$r=0.020\ \mathrm{m}$
The separation is the diameter of the semicircle.
$d=2r=0.040\ \mathrm{m}$

Answer

d=4.0\ \mathrm{cm}

Your turnquick check

With speed, charge and field fixed, what happens to orbit radius when particle mass doubles?

Show worked answer

The orbit radius doubles.