Question

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.

Answer 1

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 × v² = 1.6 × 10⁻¹⁷ J
⇒ v² = 0.35 × 10¹⁴
v = 0.591 × 10⁷ 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 × 10⁸
= 9.4 × 10⁶
 f = 9.4 × 10⁶