Solveeit Logo
Login

Question

Question: What is cyclotron frequency of an electron with energy of \(100eV\) in the earth's magnetic field of...

What is cyclotron frequency of an electron with energy of 100eV100eV in the earth's magnetic field of 1×10−4weber/m21\times {{10}^{-4}}weber/{{m}^{2}} if its velocity is perpendicular to magnetic field?
A) 0.7MHz0.7MHz
B) 2.8MHz2.8MHz
C) 1.4MHz1.4MHz
D) 2.1MHz2.1MHz

Explanation

Solution

A moving electron has energy due to its motion, which is called kinetic energy of the electron. When an electron is moving in a magnetic field then it experiences magnetic force. The frequency is defined as the number rotation/revolution per unit time.

Complete step by step solution:
When an electron is moving in a magnetic field then it experiences magnetic force.
The magnetic force experienced by a charged particle is given as,
Fm=q(v→×B→){{F}_{m}}=q\left( \overrightarrow{v}\times \overrightarrow{B} \right)
Where,
Fm={{F}_{m}}=Magnetic force
q=q= The charge on the particle
v→=\overrightarrow{v}= The velocity vector of the particle
B→=\overrightarrow{B}= The magnetic field of the region
As we know that the kinetic energy is given as,K=mv22K=\dfrac{m{{v}^{2}}}{2}
1ev=1.6×10−19J1ev=1.6\times {{10}^{-19}}J
Then total kinetic energy of the electron=100(1.6×10−19)J=100\left( 1.6\times {{10}^{-19}} \right)J
Now,
mev22=100×1.6×10−19 v=2×100×1.6×10−199.1×10−31 =5.93×106m/s\begin{aligned} & \dfrac{{{m}_{e}}{{v}^{2}}}{2}=100\times 1.6\times {{10}^{-19}} \\\ & v=\sqrt{\dfrac{2\times 100\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}} \\\ & =5.93\times {{10}^{6}}m/s \end{aligned}
When velocity of the charged particle is perpendicular to the magnetic field, then the charged particle moves in a circular path of constant radius. The radius of the circular path of the charged particle in magnetic field is given as,
r=mvBqr=\dfrac{mv}{Bq}
Then, time taken to complete one revolution around the circular path is called time period,
T=2πrv =2πv(mvBq) =2πmBq\begin{aligned} & T=\dfrac{2\pi r}{v} \\\ & =\dfrac{2\pi }{v}\left( \dfrac{mv}{Bq} \right) \\\ & =\dfrac{2\pi m}{Bq} \end{aligned}
Frequency of revolution is given as,
f=1T =Bq2πm\begin{aligned} & f=\dfrac{1}{T} \\\ & =\dfrac{Bq}{2\pi m} \end{aligned}
Therefore, the frequency of the electron can be calculated as,
fe=Bqe2πme =(1×10−4)(1.6×10−19)2π(9.1×10−31)Hz =2.8×106Hz =2.8MHz\begin{aligned} & {{f}_{e}}=\dfrac{B{{q}_{e}}}{2\pi {{m}_{e}}} \\\ & =\dfrac{\left( 1\times {{10}^{-4}} \right)\left( 1.6\times {{10}^{-19}} \right)}{2\pi \left( 9.1\times {{10}^{-31}} \right)}Hz \\\ & =2.8\times {{10}^{6}}Hz \\\ & =2.8MHz \end{aligned}

Hence, option B is the correct option.

Note: If the charge particle enters the magnetic field region at an angle of inclination with the field lines then the path of the charged particle will be helical instead of circular.
The nature of charge on the particle moving in the magnetic field must be considered while finding the direction of the magnetic force.
For the calculation, the units of all physical quantities must be changed to SI units.