Question

A negatively charged particle of mass 6.11 x 10-23 kg is accelerated from rest through an electric potential difference of 3.5 x 103 V. After this, it enters a magnetic field of strength 0.80 T and undergoes uniform circular motion with a radius of 25 cm.

1. Find the speed and charge of the particle

2. 2.How many excess electrons does the particle have?

Answer 1

part 1)
given
m = 6.11*10^-23 kg
delta_V = 3.5*10^3 V
B = 0.80 T
R = 25 cm = 0.25 m

let v is the speed of charged particle when it enters magnetic field region.
let q is the magnitude of the charged particle.
use, Workdone on the charge particle = gain in kinetic energy

q*delta_V = (1/2)*m*v^2

q*3.5*10^3 = (1/2)*6.11*10^-23*v^2 ---------(1)

we know,

R = m*v/(B*q)

0.25 = 6.11*10^-23*v/(0.8*q) ---------(2)

on solving the above two equation we get

v = 3.50*10^4 m/s <<<<<<<<<<<--------------------------Answer

q = 1.07*10^-17 C <<<<<<<<<<<--------------------------Answer

2) number of excess electrons, N = q/e

= 1.07*10^-17/(1.6*10^-19)

= 67 <<<<<<<<<<<--------------------------Answer