Question

An electron enters a magnetic field of 0.43 T with a velocity perpendicular to the direction of the field. At what frequency does the electron traverse a circular path? ( m el = 9.11 × 10-31 kg, e = 1.60 × 10-19 C)

Answer 1

Frequency,
`f=1.2* 10^(10)\ Hz`

Step-by-step explanation:

It is given that,

Magnetic field, B = 0.43 T

We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :

`f=(qB)/(2\pi m)`

`f=(1.6* 10^(-19)\ C* 0.43\ T)/(2\pi * 9.11* 10^(-31)\ Kg)`

`f=1.2* 10^(10)\ Hz`

Hence, this is the required solution.