Charged particles follow circular paths in uniform B-fields

Magnetic Fields & Forces — AQA A-Level Physics

When a charged particle enters a uniform magnetic field at right angles, the magnetic force is always perpendicular to its velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹..
A force perpendicular to motion changes direction without changing speed. That is circular motion.
The magnetic force becomes the centripetal forceThe resultant force directed towards the centre of a circular path that causes an object to move in a circle. It is not a separate force but the net force providing circular motion..

Key Definition

Circular path of a charged particle — A charged particle moving perpendicular to a uniform magnetic field follows a circular path because the magnetic force is always perpendicular to its velocity, providing the centripetal force.

Deriving r = mv / BQ

Set centripetal forceThe resultant force directed towards the centre of a circular path that causes an object to move in a circle. It is not a separate force but the net force providing circular motion. equal to magnetic force:

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

Cancel one v from each side and rearrange:

r = \frac{mv}{BQ}

$r$: radius of circular path (m)
$m$: mass of particle (kg)
$v$: speed of particle (m s⁻¹)
$B$: magnetic flux densityMass per unit volume of a material. Measured in kg m⁻³.The strength of a magnetic field. The force per unit length per unit currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). on a currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).-carrying conductor perpendicular to the field. Measured in teslaThe SI unit of magnetic flux density. One tesla is the flux density when a force of 1 N acts on a 1 m conductor carrying 1 A perpendicular to the field. (T). (T)
$Q$: chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). of particle (C)

r increases with speed (r ∝ v) and mass (r ∝ m).
r decreases with stronger fields (r ∝ 1/B) and larger charge (r ∝ 1/Q).

Worked Example

An electron travels at right angles to a uniform magnetic field of flux density 6.2 mT at a speed of 3.0 × 10⁶ m s⁻¹. Calculate the radius of its circular path.

Show Solution

List known values

From the data sheet:

Specific charge: $\frac{e}{m_e} = 1.76 \times 10^{11} \text{ C kg}^{-1}$
Flux densityMass per unit volume of a material. Measured in kg m⁻³.: $B = 6.2 \times 10^{-3} \text{ T}$
Speed: $v = 3.0 \times 10^{6} \text{ m s}^{-1}$

Rearrange using specific charge

Start with $r = \frac{mv}{BQ}$. Divide top and bottom by $m$:

$$r = \frac{v}{B \times (e/m_e)}$$

Substitute values

$$r = \frac{3.0 \times 10^{6}}{6.2 \times 10^{-3} \times 1.76 \times 10^{11}}$$

Evaluate

Denominator: $6.2 \times 10^{-3} \times 1.76 \times 10^{11} = 1.091 \times 10^{9}$

$$r = \frac{3.0 \times 10^{6}}{1.091 \times 10^{9}} = 2.7 \times 10^{-3} \text{ m (2.7 mm)}$$

Answer

$r = 2.7 \times 10^{-3}$ m (2.7 mm)

Examiner Tips and Tricks

You must be able to derive $r = mv/BQ$ by equating centripetal forceThe resultant force directed towards the centre of a circular path that causes an object to move in a circle. It is not a separate force but the net force providing circular motion. to magnetic force.
This comes up often.
The circular motion equations are on the data sheet under 'Circular Motion', not under 'Magnetic Fields'.