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^(11)C/(kg)`, the frequency of revolution of the electron is

Options

A. 62.8 MHz
B. 6.28 MHz
C. 1 GHz
D. 100 MHz

Answer

C

Explanation

For circular orbit, `(mv^(2))/r=Bev`
`thereforev=(Ber)/m`
`thereforeromega=(Ber)/m" "[becausev=romega]`
`thereforeomega=(Be)/m`
`therefore2pif=(Be)/m" "[becauseomega=2pif]`
`thereforef=B/(2pi)xxe/m=(3.57xx10^(-2)xx1.76xx10^(11))/(2xx3.14)`
`thereforef=10^(9)Hz=1GHz`