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An electron is moving in a circular path...

An electron is moving in a circular path under the influence of a transverse magnetic field of `3.57xx10^(-2)T`. If the value of `e//m` is `1.76xx10^(141) C//kg`. The frequency of revolution of the electron is

A

`62.8 MHz`

B

`6.28 MHz`

C

`1 GHz`

D

`100 MHz`

Text Solution

Verified by Experts

The correct Answer is:
C
`f = (eB)/(2pi m)`
`f = (1.76xx10^(11)xx3.57xx10^(-2))/(2xx3.14)`
`f = 10^(9) Hz` or `1 GHz`
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