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(B) The radius of the circular path of a charged particle moving in a uniform magnetic field is given by $r = \frac{p}{qB}$,where $p$ is the momentum,

Vedclass · Accepted Answer

$0.5$

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(B) The radius of the circular path of a charged particle moving in a uniform magnetic field is given by $r = \frac{p}{qB}$,where $p$ is the momentum,$q$ is the charge,and $B$ is the magnetic field strength.Since the radii $r$ and the magnetic field $B$ are the same for both particles,the momentum $p$ is directly proportional to the charge $q$ $(p \propto q)$.Therefore,the ratio of the momentum of the proton $(p_p)$ to the alpha particle $(p_\alpha)$ is $\frac{p_p}{p_\alpha} = \frac{q_p}{q_\alpha}$.We know that the charge of a proton is $q_p = e$ and the charge of an alpha particle is $q_\alpha = 2e$.Substituting these values,we get $\frac{p_p}{p_\alpha} = \frac{e}{2e} = \frac{1}{2} = 0.5$.

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