Write the resonance condition for a cyclotron.
(N/A) The resonance condition for a cyclotron occurs when the frequency of the applied alternating electric field $(f_a)$ matches the cyclotron frequency $(f_c)$ of the charged particle moving in a magnetic field.
The cyclotron frequency is given by the formula:
$f_c = \frac{qB}{2\pi m}$
Where:
$q$ = charge of the particle
$B$ = magnetic field strength
$m$ = mass of the particle
Therefore,the resonance condition is:
$f_a = f_c = \frac{qB}{2\pi m}$
The cyclotron frequency is given by the formula:
$f_c = \frac{qB}{2\pi m}$
Where:
$q$ = charge of the particle
$B$ = magnetic field strength
$m$ = mass of the particle
Therefore,the resonance condition is:
$f_a = f_c = \frac{qB}{2\pi m}$
Explore More
Similar Questions
In a cyclotron experiment, if we assume that the maximum radius attained by any charged particle is equal to the radius of the dees, then the minimum kinetic energy gained by the particles will be for
Easy
View Solution$A$ cyclotron's oscillator frequency is $20 MHz$. The operating magnetic field for accelerating protons is (charge of proton $= 1.6 \times 10^{-19} C$,mass of proton $= 1.67 \times 10^{-27} kg$). (in $T$)
What is a cyclotron? Discuss the principle of a cyclotron.
Medium
View SolutionIn the cyclotron,as the radius of the circular path of the charged particle increases,what happens to the angular velocity $(\omega)$ and linear velocity $(v)$?
MediumKCET 2016
View Solution$A$ particle of mass $m$ and charge $q$ is moving in a cyclotron with magnetic field $B$. The frequency of the circular motion of the particle is proportional to
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo