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An electron is moving in a circular path...

An electron is moving in a circular path under the influence fo a transerve 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

1 GHz

B

100 MHz

C

62.8 MHz

D

6.28 MHz

Text Solution

Verified by Experts

The correct Answer is:
A
`f=(qB)/(2pim)`
`=((q)/(m))*(B)/(2pi)`
`=(1.76xx10^(11)xx3.56xx10^(-2))/(2xx3.14)`
`1xx10^(9)Hz`
`=1GHz`
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