Question

A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnetic field travels in a circular path of radius 18.0 mm. What is the charge of the particle

Answer 1

the charge of the particle is 2.47 x 10⁻¹⁹ C

Step-by-step explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

`a = (v^2)/(r) \\\\ma = qvB\\\\q = (ma)/(vB) \\\\q = (mv^2)/(rvB) = (mv)/(rB) \\\\q = ((6.64* 10^(-27) ) * (8.7* 10^5))/((1.8* 10^(-2)) * (1.3)) \\\\q = 2.47 \ * 10^(-19) \ C`

Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C