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Question

(C) The radius $r$ of a charged particle moving in a uniform magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $T = \frac{1}{2}mv^2$

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$\sqrt{\frac{2}{9}} R$

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(C) The radius $r$ of a charged particle moving in a uniform magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $T = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mT}$.Substituting this into the radius formula,we get $r = \frac{\sqrt{2mT}}{qB}$.Given the initial state: $R = \frac{\sqrt{2mT}}{qB}$.For the new state,kinetic energy $T' = 2T$ and magnetic field $B' = 3B$.The new radius $R'$ is given by $R' = \frac{\sqrt{2m(2T)}}{q(3B)}$.$R' = \frac{\sqrt{2} \cdot \sqrt{2mT}}{3qB} = \frac{\sqrt{2}}{3} \left( \frac{\sqrt{2mT}}{qB} \right)$.Substituting $R$ for the term in the parentheses,we get $R' = \frac{\sqrt{2}}{3} R = \sqrt{\frac{2}{9}} R$.

An electron having kinetic energy $T$ is moving in a circular orbit of radius $R$ perpendicular to a uniform magnetic induction $\vec{B}$. If kinetic energy is doubled and magnetic induction is tripled,the radius will become

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