An electron and a proton enter a magnetic fie | vedclass

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Question

(B) The magnetic force acting on a charged particle moving perpendicularly to a magnetic field provides the centripetal force: $qvB = \frac{mv^2}{r}$.

Vedclass · Accepted Answer

Trajectory of proton is less curved

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(B) The magnetic force acting on a charged particle moving perpendicularly to a magnetic field provides the centripetal force: $qvB = \frac{mv^2}{r}$.Rearranging for the radius of the path,we get $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we can write $mv = \sqrt{2mK}$.Substituting this into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB}$.For the same kinetic energy $K$ and same charge magnitude $q$,the radius is proportional to the square root of the mass: $r \propto \sqrt{m}$.Since the mass of a proton $(m_p)$ is much greater than the mass of an electron $(m_e)$,the radius of the proton's path $(r_p)$ will be greater than the radius of the electron's path $(r_e)$.$A$ larger radius of curvature implies that the path is less curved. Therefore,the trajectory of the proton is less curved.

An electron and a proton enter a magnetic field perpendicularly. Both have the same kinetic energy. Which of the following is true?

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