Question

The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on an electron with kinetic energy of 1 keV? Compare with the gravitational force on the electron.

Answer 1

`F = 1.5 * 10^(-16) N`

this force is
`1.68 * 10^(13)` times more than the gravitational force

Step-by-step explanation:

Kinetic Energy of the electron is given as

`KE = 1 keV`

`KE = 1 * 10^3 (1.6 * 10^(-19)) J`

`KE = 1.6 * 10^(-16) J`

now the speed of electron is given as

`KE = (1)/(2)mv^2`

now we have

`v = \sqrt{(2 KE)/(m)}`

`v = 1.87 * 10^7 m/s`

now the maximum force due to magnetic field is given as

`F = qvB`

`F = (1.6* 10^(-19))(1.87 * 10^7)(0.5 * 10^(-4))`

`F = 1.5 * 10^(-16) N`

Now if this force is compared by the gravitational force on the electron then it is

`(F)/(F_g) = (1.5 * 10^(-16))/(9.1 * 10^(-31) (9.8))`

`(F)/(F_g) = 1.68 * 10^(13)`

so this force is
`1.68 * 10^(13)` times more than the gravitational force