The maximum kinetic energy of a positive ion | vedclass

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Question

(C) The kinetic energy is given by $K = \frac{1}{2}mv^2$.In a cyclotron,the radius of the circular path of a charged particle is $r_0 = \frac{mv}{qB}$

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$\frac{q^2 B^2 r_0^2}{2m}$

Vedclass · Answer

(C) The kinetic energy is given by $K = \frac{1}{2}mv^2$.In a cyclotron,the radius of the circular path of a charged particle is $r_0 = \frac{mv}{qB}$.Rearranging this for velocity,we get $v = \frac{qB r_0}{m}$.Substituting this value of $v$ into the kinetic energy formula:$K_{\max} = \frac{1}{2}m \left( \frac{qB r_0}{m} \right)^2$$K_{\max} = \frac{1}{2}m \left( \frac{q^2 B^2 r_0^2}{m^2} \right)$$K_{\max} = \frac{q^2 B^2 r_0^2}{2m}$.

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

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