An electron and a proton having same momenta | vedclass

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(B) The radius of curvature $r$ of a charged particle of mass $m$ and charge $q$ moving in a region with a uniform perpendicular magnetic field $B$ is

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curved path of electron and proton will be same. (Neglect sense of revolution)

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(B) The radius of curvature $r$ of a charged particle of mass $m$ and charge $q$ moving in a region with a uniform perpendicular magnetic field $B$ is given by the formula: $r = \frac{mv}{Bq} = \frac{p}{Bq}$.Here,$p$ is the momentum of the charged particle.Since the momenta $p$ are the same for both the electron and the proton,and the magnetic field $B$ is also the same,the radius of the circular path depends only on the charge $q$ of the particle.However,the magnitude of the charge for both an electron and a proton is equal $(|q_e| = |q_p| = e)$.Therefore,$r_e = \frac{p}{Be}$ and $r_p = \frac{p}{Be}$,which implies $r_e = r_p$.Thus,both particles will follow a curved path of the same radius,neglecting the sense of revolution (direction of curvature).

An electron and a proton having same momenta enter perpendicular into a magnetic field then

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