Question

An electron travels in a circular path of radius 20cm in a magnetic field of 2x10^-3 T. Calculate the speed of the electron. What is the potential difference through which the electron must be accelerated to acquire this speed?

Answer 1

Answer:

Given data:

Radius= 20 cm = 20/100= 0.2m

Magnetic field "B" = 2x10-3T

Mass (m)= 9.1x10-31 kg

q = 1.6x10-19 C

1- Calculate speed by using the formula

Bqr = mv2 /2

V = Bqr /m

V = 2x 10-3 x 1.6x10-19 x 0.2 / 9.1x10-31

  = 0.0703 x 10-3-19+31

  = 0.0703 x 10^9

V = 7.03 x 10 ^ 7 m/s

2- Now calculate potential difference by using th erelation.

eV = 1/2 mv ^ 2

V =1/2 mv ^ 2/e

  = 1 x 9.1x10-31 x (7.03 x 10^7)2 / 2 x 3.2 x 107.03 x 10^19

 = 140.5 x 10^2

 = 14050 V