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(B) The radius of a circular path for a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2

Vedclass · Accepted Answer

$r_e < r_p$

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(B) The radius of a circular path for a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.Substituting this into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB}$.Given that kinetic energy $K$ and magnetic field $B$ are the same for both particles,we have $r \propto \frac{\sqrt{m}}{q}$.Since the magnitude of charge for an electron and a proton is the same $(|q_e| = |q_p| = e)$,the ratio of the radii is $\frac{r_e}{r_p} = \sqrt{\frac{m_e}{m_p}}$.Because the mass of an electron $m_e$ is much smaller than the mass of a proton $m_p$ $(m_e < m_p)$,it follows that $r_e < r_p$.

An electron and a proton enter a region of uniform magnetic field in a direction at right angles to the field with the same kinetic energy. They describe circular paths of radius $r_e$ and $r_p$ respectively. Then:

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