Question

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 $\frac{e}{m}$ is $1.76\times {10}^{11}C/kg$, the frequency of revolution of the electron is

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

Answer 1

The correct option is C $1GHz$

If a charge particle move in magnetic field then it will move in circular orbit.
So, the magnetic force acting on the electron will be equal to the centripetal force
$\frac{m{v}^2}{r}=qvB$
$r=\frac{mv}{qB}$
and the time period is given as, $t=\frac{2\pi r}{v}=\frac{2\pi mv}{qBv}$
$t=\frac{2\pi m}{qB}$
as, $f=\frac{1}{t}$
$f=\frac{qB}{2\pi m}=\frac{1.76\times {10}^{11}\times 3.57\times {10}^{-2}}{2\times 3.14}={10}^{9}Hzor1GHz$