Question

A proton is accelerated from rest through a potential differences of 1.0 kV. It enters a uniform magnetic field of 4.5 mT that is initially perpendicular to its velocity. (a) Find the radius of the proton's circular path (b). Calculate the period of revolution of the proton.

Answer 1

Here

Potential , V = 1 kV = 1000 V

magnetic fiedld , B = 4.5 mT

B = 0.0045 T

a) for the redius

radius of path = sqrt(2 * m * e * V)/(B * e)

radius of path = sqrt(2 * 1.67 *10^-27 * 1.602 *10^-19 * 1000)/(0.0045 * 1.602 *10^-19 )

radius of path = 1.015 m

the radius of path is 1.015 m

b)

for the time period

time period = 2*pi * m/(B * q)

time period = 2pi *1.67 *10^-27/(0.0045 * 1.602 *10^-19)

time period = 1.46 *10^-5 s

the time period of each revolution is 1.46 *10^-5 s