An electron is moving in a circular path under the influence of a transverse magnetic field of `3.57 xx 10^(-2)T`. If the value of elm is `1.76 xx 10^(11)C kg^(-1)` 101 C kg!, the frequency of revolution of the electron is

Options

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

Answer

A

Explanation

Here, `B = 3.57 xx 10^(-2) T. (e)/(m) = 1.76 xx 10^(11) C kg^(-1)`
Frequency of revolution of the electron,
`u = 1/T = (v)/(2pir)` ............(i)
Also, `(mv^2)/(r) = evB rArr (v)/(r) = (eB)/(m)` ..........(ii)
From eqns. (i) and (ii)
`u = (1)/(2pi) xx (eB)/(m) = (1)/(2xx 3.14) xx 1.76 xx 10^(11) xx 3.57 xx 10^(-2)`
`10^9 Hx = 1 GHz`