A charged particle of charge 'q' is accelerat | vedclass

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(A) When a charged particle enters a uniform magnetic field perpendicularly,it follows a circular path.The radius of this path is given by:$R = \frac{

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$\frac{r^2 q B^2}{2 V}$

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(A) When a charged particle enters a uniform magnetic field perpendicularly,it follows a circular path.The radius of this path is given by:$R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$where $K$ is the kinetic energy of the particle.Since the particle is accelerated through a potential difference $V$,its kinetic energy is $K = qV$.Substituting this into the radius formula:$R = \frac{\sqrt{2m(qV)}}{qB} = \frac{\sqrt{2mqV}}{qB}$Squaring both sides:$R^2 = \frac{2mqV}{q^2 B^2} = \frac{2mV}{qB^2}$Given that the radius is $r$,we have $r^2 = \frac{2mV}{qB^2}$.Solving for mass $m$:$m = \frac{r^2 q B^2}{2V}$

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