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 e/m is \[1.76 \times {10^{11}}C/kg\] .The frequency of revolution of the electron is
(A) $6.28MHz$
(B) $1GHz$
(C) $100MHz$
(D) $62.8MHz$

Answer 1

Hint: We know that when a charged particle travels in the perpendicular magnetic field, then the force on the particle due to the magnetic field will cause the trajectory of the particle to be a circular path.

Formula used:
$F = qvB$

Complete step by step answer
A charged particle moving with a velocity in magnetic field follows a curved path whose radius remains same everywhere, hence in a plane the curve is just a circle
 In the question, it is given that the particle in the magnetic field has circular motion so the particle will experience a magnetic force $\overrightarrow F = q(\overrightarrow v \times \overrightarrow B )$ where v is the instantaneous velocity and B is the magnetic field.
and the centripetal force will be equal to the magnetic force in this condition.
$\dfrac{{m{v^2}}}{r} = qvB$ where r, is the radius of the circular path, v is the velocity of the particle, B is the magnetic field and q is the charge on the particle
 $ \Rightarrow \dfrac{{mv}}{{qB}} = r$
We also know that the time period is given by $T = \dfrac{{2\pi r}}{v}$
So on substituting the value of r, time period becomes
$T = \dfrac{{2\pi mv}}{{qvB}} = \dfrac{{2\pi m}}{{qB}}$
Frequency is reciprocal of time hence,
$f = \dfrac{1}{T} = \left( {\dfrac{q}{m}} \right) \times \dfrac{B}{{2\pi }} = \dfrac{{1.76 \times {{10}^{11}} \times 3.57 \times {{10}^{ - 2}}}}{{2 \times 3.24}} \approx {10^9}Hz$

Hence the correct option is B.

Note:
The magnetic force experienced on the particle is perpendicular to the velocity and magnetic field. So, there is power on a charged particle due to the magnetic field. Also, the work done by this force is zero at every point. This means that the magnetic force cannot change the magnitude of speed, it can only change direction. If the charged particle moves with a velocity which is not perpendicular to magnetic field, then the velocity will have two components one parallel to the field and second perpendicular to field. Hence the path of such particles will not be circular but helical.