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An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2}$ T. If the value of e/m is $1.76 \times 10^{11}$ C/kg, the frequency of revolution of the electron is:
A
62.8 MHz
B
6.28 MHz
C
1 GHz
D
100 MHz
Explanation
$f = (e/m)B/2\pi$.
Detailed Solution
$f = \frac{eB}{2\pi m}$
$f = \frac{1.76\times10^{11}\times3.57\times10^{-2}}{2\times3.14}$
$f = 10^{9}$ Hz or 1 GHz
$f = \frac{1.76\times10^{11}\times3.57\times10^{-2}}{2\times3.14}$
$f = 10^{9}$ Hz or 1 GHz