An alternating electric field,of frequency f, | vedclass

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(C) The cyclotron frequency $f$ is given by $f = \frac{eB}{2\pi m}$.Rearranging for the magnetic field $B$,we get $B = \frac{2\pi mf}{e}$.The maximum

Vedclass · Accepted Answer

$B = \frac{2\pi mf}{e}$,$K = 2m\pi^2f^2R^2$

Vedclass · Answer

(C) The cyclotron frequency $f$ is given by $f = \frac{eB}{2\pi m}$.Rearranging for the magnetic field $B$,we get $B = \frac{2\pi mf}{e}$.The maximum velocity $v$ of the proton at the exit of the dee (radius $R$) is given by $v = \omega R = (2\pi f)R$.The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.Substituting $v = 2\pi fR$ into the kinetic energy formula:$K = \frac{1}{2}m(2\pi fR)^2 = \frac{1}{2}m(4\pi^2f^2R^2) = 2m\pi^2f^2R^2$.Thus,$B = \frac{2\pi mf}{e}$ and $K = 2m\pi^2f^2R^2$.

Similar Questions

$A$ cyclotron's oscillator frequency is $10 \text{ MHz}$ and the operating magnetic field is $0.66 \text{ T}$. If the radius of its dees is $60 \text{ cm}$, then the kinetic energy of the proton beam produced by the accelerator is: (in $\text{ MeV}$)

Assertion: Cyclotron does not accelerate electrons.
Reason: Mass of the electrons is very small.

Cyclotron is used to accelerate

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

In the cyclotron, as the radius of the circular path of the charged particle increases, ($\omega$ = angular velocity, $V$ = linear velocity)

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