A monoenergetic electron beam with electron speed of 5.20 × 10^{6} m s^{−1} is subject to a magnetic field of 1.30 × 10^{−4} T normal to the beam velocity. What is the a radius of the circle traced by the beam, given *e/m *for electron equals 1.76 × 10^{11} C kg^{−1}.

#### Solution

Speed of an electron, *v* = 5.20 × 10^{6} m/s

Magnetic field experienced by the electron, *B* = 1.30 × 10^{−4} T

Specific charge of an electron, *e/m* = 1.76 × 10^{11} C kg^{−1}

Where,

*e* = Charge on the electron = 1.6 × 10^{−19} C

*m* = Mass of the electron = 9.1 × 10^{−31} kg^{−1}

The force exerted on the electron is given as:

`F = e|vecv xx vecB|`

`= evBsin theta`

θ = Angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam.

`:. theta = 90^@`

F = eVB ...(1)

The beam traces a circular path of radius, *r*. It is the magnetic field, due to its bending nature, that provides the centripetal force `(F = (mv^2)/r)` for the beam.

Hence, equation (1) reduces to:

`evB = (mv^2)/r`

`:. r = mv/(eB) = v/((e/m)B)`

`= (5.20 xx 10^6) /(1.76 xx 10^11) xx 1.30 xx 10^(-4) = 0.227 m = 22.7 cm`

Therefore, the radius of the circular path is 22.7 cm.