Q. Maximum kinetic energy of the positive ion in the cyclotron is
A) $\frac{q^{2}Br_{0}}{2m}$
B) $\frac{qB^{2}r_{0}}{2m}$
C) $\frac{q^{2}B^{2}r_{0}^{2}}{2m}$
D) $\frac{qBr_{0}}{2m^{2}}$
Explanation
(c) $K_{\max} = \frac{1}{2}mv^{2}$ and $r_{0} = \frac{mv}{qb} \Rightarrow v = \frac{qBr_{0}}{m}$
$\Rightarrow K_{\max} = \frac{1}{2}m\left( \frac{qBr_{0}}{m} \right)^{2} = \frac{q^{2}B^{2}r_{0}^{2}}{2m}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.