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Question

(C) The kinetic energy $K$ gained by the particle when accelerated through a potential difference $V$ is given by $K = qV$.Since $K = \frac{1}{2}mv^2$

Vedclass · Accepted Answer

$\frac{1}{B} \sqrt{\frac{2mV}{q}}$

Vedclass · Answer

(C) The kinetic energy $K$ gained by the particle when accelerated through a potential difference $V$ is given by $K = qV$.Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = qV$,which implies $v = \sqrt{\frac{2qV}{m}}$.When a charged particle enters a magnetic field $B$ perpendicularly,it moves in a circular path of radius $R = \frac{mv}{qB}$.Substituting the value of $v$ into the radius formula:$R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{m^2 \cdot 2qV}{q^2 \cdot m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Thus,the correct option is $C$.

$A$ charged particle of charge $q$ and mass $m$ is accelerated through a potential difference of $V$ volts. It enters a region of orthogonal magnetic field $B$. The radius of its circular path will be:

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