Question

The cyclotron frequency of an electron gyrating in a magnetic field of $1T$ is approximately:
$A)\text{ }28MHz$
$B)\text{ }2.8MHz$
$C)\text{ }2.8GHz$
$D)\text{ }28GHz$

Answer 1

Hint : This problem can be solved by using the direct formula for the cyclotron frequency of an electron gyrating in a magnetic field in terms of its charge, its mass and the magnitude of the magnetic field. By plugging in the value of the magnetic field given in the question and the known values of the mass and charge of an electron, we can get the required answer.
Formula used:
$\nu =\dfrac{Bq}{2\pi m}$

Complete step by step solution :
We will use the direct formula for the cyclotron frequency of a charged particle gyrating in a magnetic field.
The cyclotron frequency $\nu$ of a particle with magnitude of charge $q$ and mass $m$ moving in the presence of a perpendicular magnetic field of magnitude $B$ is given by
$\nu =\dfrac{Bq}{2\pi m}$ --(1)
Hence, now let us analyze the question.
The given magnitude of the magnetic field is $B=1T$.
The magnitude of charge on an electron is $e=1.6\times {{10}^{-19}}C$.
The mass of an electron is ${{m}_{e}}=9.1\times {{10}^{-31}}kg$.
Let the required cyclotron frequency be $\nu$.
Hence, using (1), we get
$\nu =\dfrac{Be}{2\pi {{m}_{e}}}$
$\therefore \nu =\dfrac{1\times 1.6\times {{10}^{-19}}}{2\pi \times 9.1\times {{10}^{-31}}}=0.02735\times {{10}^{12}}Hz=27.35\times {{10}^{9}}Hz\approx 28\times {{10}^{9}}Hz=28GHz$ $\left( \because {{10}^{9}}Hz=1GHz \right)$
Hence, the required cyclotron frequency of the electron is $28GHz$.
Therefore, the correct option is $D)\text{ }28GHz$.

Note : Though keeping the direct formula for the cyclotron frequency in mind saves a bit of time, still it is not necessary to memorize it. The formula can always be derived by equating the Lorentz force due to the magnetic field on a charged body with the magnitude of the centripetal force required for circular motion (in terms of the velocity, mass and radius of the particle). Using this, the velocity can be converted to angular frequency, which can be further converted to the frequency for the circular motion, thereby obtaining the direct formula for the cyclotron frequency that we have used in the answer. This approach is always better than memorizing the formula since if the student forgets or writes the wrong formula, he or she will arrive at a completely wrong result, whereas by deriving properly, the student will always obtain the correct formula.