If alpha and beta particles are moving with e | vedclass

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(A) The radius $r$ of the path of a charged particle moving perpendicular to a magnetic field $B$ is given by $r = \frac{mv}{qB}$.Here,$m$ is the mass

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unequal and in the opposite direction

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(A) The radius $r$ of the path of a charged particle moving perpendicular to a magnetic field $B$ is given by $r = \frac{mv}{qB}$.Here,$m$ is the mass,$v$ is the velocity,and $q$ is the charge of the particle.For an $\alpha$ particle,$q_{\alpha} = +2e$ and $m_{\alpha} \approx 4m_p$ (where $m_p$ is the mass of a proton).For a $\beta$ particle (electron),$q_{\beta} = -e$ and $m_{\beta} \approx \frac{1}{1836}m_p$.Since the masses and charges are different,the radii $r_{\alpha} = \frac{m_{\alpha}v}{2eB}$ and $r_{\beta} = \frac{m_{\beta}v}{eB}$ will be unequal.Because the charges have opposite signs ($+2e$ and $-e$),the Lorentz force $\vec{F} = q(\vec{v} 	imes \vec{B})$ will act in opposite directions for the two particles,causing them to curve in opposite directions.

If $\alpha$ and $\beta$ particles are moving with equal velocity perpendicular to the magnetic flux density $B$,then the radii of their paths will be

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