An alternating electric field of frequency f | vedclass

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Question

(A) In a cyclotron,the frequency of the alternating electric field $f$ is equal to the cyclotron frequency $f_c = \frac{eB}{2\pi m}$.From this,the mag

Vedclass · Accepted Answer

$\frac{2 \pi m f}{e}, 2 \pi^2 m f^2 R^2$

Vedclass · Answer

(A) In a cyclotron,the frequency of the alternating electric field $f$ is equal to the cyclotron frequency $f_c = \frac{eB}{2\pi m}$.From this,the magnetic field $B$ is given by $B = \frac{2\pi mf}{e}$.The maximum velocity $v$ of the proton at the exit radius $R$ is given by $v = \omega R = (2\pi f)R$.The kinetic energy $K.E.$ is given by $K.E. = \frac{1}{2}mv^2$.Substituting $v = 2\pi f R$,we get $K.E. = \frac{1}{2}m(2\pi f R)^2 = \frac{1}{2}m(4\pi^2 f^2 R^2) = 2\pi^2 mf^2 R^2$.Thus,the magnetic field is $\frac{2\pi mf}{e}$ and the kinetic energy is $2\pi^2 mf^2 R^2$.

An alternating electric field of frequency $f$ is applied across the radius $R$ of a cyclotron to accelerate protons (mass $m$). The operating magnetic field $B$ used and kinetic energy $(K.E.)$ of the proton beam produced by it are respectively ($e=$ charge on proton).

Similar Questions

The maximum kinetic energy of a positive ion in a cyclotron is given by:

$A$ proton is accelerating in a cyclotron where the applied magnetic field is $2 \,T$. If the potential gap is effectively $100 \,kV$, then how many revolutions does the proton have to make between the "dees" to acquire a kinetic energy of $20 \,MeV$?

The oscillating frequency of a cyclotron is $10 \, MHz$. If the radius of its dee is $0.5 \, m$,the kinetic energy of a proton,which is accelerated by the cyclotron,is ...... $MeV$.

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 $(v)$?

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

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