Charged particle radius
Electromagnetism - OCR A-Level Physics
Worked Example
A proton (m = 1.67 × 10⁻²⁷ kg, Q = 1.60 × 10⁻¹⁹ C) enters a uniform magnetic field of flux densityMass per unit volume of a material. Measured in kg m⁻³. 0.50 T at right angles with a speed of 2.0 × 10⁷ m s⁻¹. Calculate the radius of its circular path.
Show Solution
1
The magnetic force provides 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.
$BQv =\;\text{m}v^{2}/r.$
2
Rearrange
$r =\;\text{m}v / (BQ).$
3
Substitute
$r = (1.67 \times 10⁻^{2}⁷ \times 2.0 \times 10⁷) / (0.50 \times 1.60 \times 10⁻¹⁹).$
4
Numerator
3.34 × 10⁻²⁰. Denominator: 8.0 × 10⁻²⁰.
5
$r = 3.34 \times 10⁻^{2}⁰ / 8.0 \times 10⁻^{2}⁰ = 0.42\;\text{m}.$
Answer
r = 0.42 m. The proton moves in a circle of radius 42 cm.