An electron of kinetic energy 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.

#### Solution

Given:

Kinetic energy of an electron = 100 eV

Radius of the circle = 10 cm

`1/2mv^2` = 100 eV = 1.6 × 10^{−17} J (1 eV = 1.6 × 10^{−19} J)

Here,*m* is the mass of an electron and *v* is the speed of an electron. Thus,

1/2 × 9.1 × 10^{−31} × *v*^{2} = 1.6 × 10^{−17} J

⇒ *v*^{2} = 0.35 × 10^{14}*v* = 0.591 × 10^{7} m/s

Now,

`r = (mv)/(eB)`

`⇒ B = (mv)/(er)`

`= (9.1xx10^-31xx0.591xx10^7)/(1.6xx10^-19xx0.1)`*B* = 3.3613 × 10^{−4} T

Therefore, the applied magnetic field = 3.4 × 10^{−4} T

Number of revolutions per second of the electron,

`f = 1/T`

`T = (2pir)/v= 2pim/(eB)`

`T = (2pim)/Be`

`f = (Be)/(2pim)`

`=( 3.4xx10^-4xx1.6xx10^-19)/(2xx3.14xx9.1xx10^-31`

= 0.094 × 10^{8}

= 9.4 × 10^{6}* f* = 9.4 × 10^{6}