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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 $(v)$?

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