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(A) $1$. The kinetic energy $K$ gained by a charged particle $q$ accelerated through a potential difference $V$ is given by $K = qV$.
$2$. The kinetic

Vedclass · Accepted Answer

$q B^2 r^2 / 2V$

Vedclass · Answer

(A) $1$. The kinetic energy $K$ gained by a charged particle $q$ accelerated through a potential difference $V$ is given by $K = qV$.
$2$. The kinetic energy is also expressed as $K = p^2 / 2m$, where $p$ is the momentum and $m$ is the mass. Thus, $qV = p^2 / 2m$, which implies $p = \sqrt{2mqV}$.
$3$. When a charged particle enters a uniform magnetic field $B$ at right angles, it follows a circular path with radius $r = p / (qB)$.
$4$. Substituting the expression for $p$, we get $r = \sqrt{2mqV} / (qB)$.
$5$. Squaring both sides, $r^2 = 2mqV / (q^2 B^2)$.
$6$. Solving for mass $m$, we get $m = (q^2 B^2 r^2) / (2qV) = (q B^2 r^2) / (2V)$.

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