In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^{–4} T) is maintained. An electron is shot into the field with a speed of 4.8 × 10^{6} m s^{–1} normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (*e *= 1.6 × 10^{–19} C, *m*_{e}= 9.1×10^{–31} kg)

#### Solution

Magnetic field strength, *B* = 6.5 G = 6.5 × 10^{–4} T

Speed of the electron, *v* = 4.8 × 10^{6} m/s

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

Mass of the electron, *m*_{e} = 9.1 × 10^{–31} kg

Angle between the shot electron and magnetic field, *θ* = 90°

Magnetic force exerted on the electron in the magnetic field is given as:

*F* = *evB *sin*θ*

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius *r*.

Hence, centripetal force exerted on the electron,

`F_c=(mv^2)/r`

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

`F_c=F`

`(mv^2)/r=evBsintheta`

`r=(mv)/(Besin theta)`

`=(9.1xx10^-31xx4.8xx10^6)/(6.5xx10^-4xx1.6xx10^-19xxsin90^@)`

`=4.2xx10^-2m=4.2 cm`

Hence, the radius of the circular orbit of the electron is 4.2 cm.