Question

1--A proton is accelerated from rest by a potential difference of 350v. It then enters a uniform magnetic field.the orbit radius is 25cm. Find (a) the protons,s speed, (b) the magnetic field strength (c) the period of the motion

Answer 1

given,

potential difference = 350 V

electric potential energy of proton = charge * potential difference

electric potential energy of proton = 1.6 * 10^-19 * 350

electric potential energy of proton = 5.6 * 10^-17 J

by conservation of energy

initial energy = final energy

5.6 * 10^-17 = 0.5 * m * v^2

5.6 * 10^-17 = 0.5 * 1.672 * 10^-27 * v^2

proton's speed = 258815.853 m/s

radius = 0.25 m

radius = mv / qb

0.25 = 1.672 * 10^-27 * 258815.853 / (1.6 * 10^-19 * B)

magnetic field strength = 0.01082 T

period of motion = 2 * pi * m / qB

period of motion = 2 * pi * 1.672 * 10^-27 / (1.6 * 10^-19 * 0.01082)

period of motion = 0.0000061 sec