A proton (mass m,charge q) accelerated by a p | vedclass

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(D) From the geometry of the path,the proton moves in a circular arc of radius $R$ within the magnetic field. From the figure,we have $\sin \alpha = \

Vedclass · Accepted Answer

$Bd \sqrt{\frac{q}{2mV}}$

Vedclass · Answer

(D) From the geometry of the path,the proton moves in a circular arc of radius $R$ within the magnetic field. From the figure,we have $\sin \alpha = \frac{d}{R}$.The magnetic force provides the necessary centripetal force,so $\frac{mv^2}{R} = qvB$,which gives the radius $R = \frac{mv}{qB}$.Substituting $R$ into the expression for $\sin \alpha$,we get $\sin \alpha = \frac{d}{mv/qB} = \frac{dqB}{mv}$.The kinetic energy gained by the proton is $\frac{1}{2}mv^2 = qV$,which implies $v = \sqrt{\frac{2qV}{m}}$.Substituting the value of $v$ into the expression for $\sin \alpha$:$\sin \alpha = \frac{dqB}{m \sqrt{\frac{2qV}{m}}} = \frac{dqB}{\sqrt{2mqV}} = Bd \sqrt{\frac{q^2}{2mqV}} = Bd \sqrt{\frac{q}{2mV}}$.Thus,the correct option is $D$.

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