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Q. 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

NEETNEET 2016Moving Charges and Magnetism

Solution:

$f = \frac{qB}{2\pi m}$
$ = \left(\frac{q}{m}\right) . \frac{B}{2\pi} $
$= \frac{1.76 \times10^{11} \times3.57 \times10^{-2}}{2 \times3.14} $
$ = 1 \times10^{9} Hz $
$= 1 \,GHz $