Question

An electron with energy 1keV is entering a uniform magnetic field of 0.04T at an angle 60° with the field.Predict the path of the electron and find the characteristics of the path.

Answer 1

Answer:

Explanation:

Given An electron with energy 1keV is entering a uniform magnetic field of 0.04 T at an angle 60° with the field.Predict the path of the electron and find the characteristics of the path.

We know that

1 Kev = 1 x 1000 x 1.6 x 10^-19

         = 1.6 x 10^-16

Now this is equal to kinetic energy.

So ½ mv^2 = 1.6 x 10^-16 J

 m v^2 = 3.2 x 10^-16

     v^2 = 3.2 x 10 ^-16 / 9.1 x 10^-31 (since mass = 9.1 x 10^-31)

so v ^2 = 3.51 x 10^14

so v = 1.87 x 10^7

Now radius of the path is given by

           = mv sin θ / qB (where B is magnetic field)

          = 9.1 x 10^-31 x 1.87 x 10^7 sin 60 / 1.6 x 10^-19 x 0.04

        = 17.087 x 10^-24 x √3 / 2 / 0.064 x 10^-19

            = 2.31 x 10^-3 m

So pitch can be calculated using

Pitch = 2 π m v cos θ / qB

      = 2 x 3.14 x 9.1 x 10^-31 x 1.87 x 10^7 x 0.5 / 1.6 x 10^-19 x 0.04 (cos 60 = ½)

        = 8.39 x 10^-3 m