Answer
406.2k+ views
Hint: A cyclotron is a gadget used to accelerate particles using electric and magnetic fields. In order to answer this question, you need to know the working of the cyclotron and the whole setup. After knowing the whole setup, apply the conditions according to how the fields operate the charges and derive the frequency and kinetic energy of the charged particles. Also make a qualitative analysis on the trajectory of the particles and draw a schematic sketch of the path.
Complete step by step answer:
A cyclotron consists of two metal halves (Dees) insulated from each other. This makes the vacuum chamber flat and cylindrical. A high frequency alternating voltage is applied between these two Dees which in turn accelerate the charged particles or ions. These Dees are located in between the poles of strong magnet which give a static magnetic field. The ions are injected from the center. The direction of the magnetic field is perpendicular to the plane of Dees.
As the magnetic field acts on the ion, the path of the particles is bent in a circle due to Lorentz force. The particles will move in a circular path inside the Dees as their velocities are constant. To change the velocity, or say, to accelerate the particle, a high frequency voltage is applied between the Dees. This voltage creates an oscillating electric field which will in turn apply force on the ions and the ions will accelerate. As a result, the particles will follow a path with a spiral trajectory.
(a) To calculate the frequency, consider the force on the ion. The cyclotron frequency is the frequency of the charged particle moving in a direction which is perpendicular to the direction of the magnetic field. The centripetal force is balanced by the magnetic force and mathematically, we have,
$
\dfrac{{m{v^2}}}{r} = qvB \\
\Rightarrow\omega = \dfrac{v}{r} = \dfrac{{qB}}{m} \\
$
Where $\omega $ is the angular frequency. The frequency in terms of angular frequency in terms of angular frequency is given as $f = \dfrac{\omega }{{2\pi }}$. Substituting, we get,
$f = \dfrac{{qB}}{{2\pi m}}$
Now,
$\dfrac{{m{v^2}}}{r} = qvB\\
\Rightarrow v = \dfrac{{qrB}}{m}$
(b) As you know that the kinetic energy of a particle is given as $\dfrac{1}{2}m{v^2}$, we substitute the value of the velocity of the ion in the formula for kinetic energy,
$
K.E = \dfrac{1}{2}m{\left( {\dfrac{{qrB}}{m}} \right)^2} \\
\Rightarrow K.E = \dfrac{{{q^2}{r^2}{B^2}}}{{2m}} \\ $
$
\Rightarrow K.E = \dfrac{1}{2}m{\left( {\dfrac{{qrB}}{m}} \right)^2} \\
\therefore K.E = \dfrac{{{q^2}{r^2}{B^2}}}{{2m}} \\ $
Hence, the trajectory of an ion in cyclotron is spiral, the cyclotron frequency is given as $f = \dfrac{{qB}}{{2\pi m}}$ and the kinetic energy of the ions is given as $\dfrac{{{q^2}{r^2}{B^2}}}{{2m}}$.
Note: Remember the structure of the cyclotron, it contains two metal halves (Dees) which are in between magnetic poles. The magnetic field bends the path of the ion and the oscillating electric field will accelerate the particle. Also remember the values of frequency and kinetic energy of the ions or try to remember the method which we used to derive them.
Complete step by step answer:
A cyclotron consists of two metal halves (Dees) insulated from each other. This makes the vacuum chamber flat and cylindrical. A high frequency alternating voltage is applied between these two Dees which in turn accelerate the charged particles or ions. These Dees are located in between the poles of strong magnet which give a static magnetic field. The ions are injected from the center. The direction of the magnetic field is perpendicular to the plane of Dees.
As the magnetic field acts on the ion, the path of the particles is bent in a circle due to Lorentz force. The particles will move in a circular path inside the Dees as their velocities are constant. To change the velocity, or say, to accelerate the particle, a high frequency voltage is applied between the Dees. This voltage creates an oscillating electric field which will in turn apply force on the ions and the ions will accelerate. As a result, the particles will follow a path with a spiral trajectory.
![seo images](https://www.vedantu.com/question-sets/468ce9cb-e1c7-4980-8ed4-36448c78d7232105541874695103195.png)
(a) To calculate the frequency, consider the force on the ion. The cyclotron frequency is the frequency of the charged particle moving in a direction which is perpendicular to the direction of the magnetic field. The centripetal force is balanced by the magnetic force and mathematically, we have,
$
\dfrac{{m{v^2}}}{r} = qvB \\
\Rightarrow\omega = \dfrac{v}{r} = \dfrac{{qB}}{m} \\
$
Where $\omega $ is the angular frequency. The frequency in terms of angular frequency in terms of angular frequency is given as $f = \dfrac{\omega }{{2\pi }}$. Substituting, we get,
$f = \dfrac{{qB}}{{2\pi m}}$
Now,
$\dfrac{{m{v^2}}}{r} = qvB\\
\Rightarrow v = \dfrac{{qrB}}{m}$
(b) As you know that the kinetic energy of a particle is given as $\dfrac{1}{2}m{v^2}$, we substitute the value of the velocity of the ion in the formula for kinetic energy,
$
K.E = \dfrac{1}{2}m{\left( {\dfrac{{qrB}}{m}} \right)^2} \\
\Rightarrow K.E = \dfrac{{{q^2}{r^2}{B^2}}}{{2m}} \\ $
$
\Rightarrow K.E = \dfrac{1}{2}m{\left( {\dfrac{{qrB}}{m}} \right)^2} \\
\therefore K.E = \dfrac{{{q^2}{r^2}{B^2}}}{{2m}} \\ $
Hence, the trajectory of an ion in cyclotron is spiral, the cyclotron frequency is given as $f = \dfrac{{qB}}{{2\pi m}}$ and the kinetic energy of the ions is given as $\dfrac{{{q^2}{r^2}{B^2}}}{{2m}}$.
Note: Remember the structure of the cyclotron, it contains two metal halves (Dees) which are in between magnetic poles. The magnetic field bends the path of the ion and the oscillating electric field will accelerate the particle. Also remember the values of frequency and kinetic energy of the ions or try to remember the method which we used to derive them.
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