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(C) The radius $(r)$ of the circular path of a charged particle moving perpendicular to a magnetic field $(B)$ is given by $r = \frac{mv}{qB}$.From th

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$v$ increases,$\omega$ remains constant

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(C) The radius $(r)$ of the circular path of a charged particle moving perpendicular to a magnetic field $(B)$ is given by $r = \frac{mv}{qB}$.From this relation,we can see that $r \propto v$,which means as the radius $(r)$ increases,the linear velocity $(v)$ of the particle must also increase.The angular velocity $(\omega)$ is given by $\omega = \frac{v}{r}$.Substituting $v = \frac{rqB}{m}$ into the expression for $\omega$,we get $\omega = \frac{rqB}{mr} = \frac{qB}{m}$.Since the charge $(q)$,magnetic field $(B)$,and mass $(m)$ are constants,the angular velocity $(\omega)$ remains independent of the radius and velocity.Therefore,as the radius increases,$v$ increases and $\omega$ remains constant.

In the cyclotron,as the radius of the circular path of the charged particle increases,what happens to the angular velocity $(\omega)$ and linear velocity $(v)$?

Similar Questions

Verify that the cyclotron frequency $\omega = \frac{eB}{m}$ has the correct dimensions of $[T]^{-1}$.

What is a cyclotron? Discuss the principle of a cyclotron.

$A$ cyclotron's oscillator frequency is $n$ and the radius of the dees is $r$. The operating magnetic field $(B)$ for accelerating protons of charge $q$ and the kinetic energy of the protons produced by the accelerator are respectively ($m$ and $v$ are the mass and velocity of the proton).

Assertion : Cyclotron is a device which is used to accelerate the positive ion.
Reason : Cyclotron frequency depends upon the velocity.

What do you mean by 'dees' which is used in cyclotron?

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