An electron having mass $m$ and kinetic energy $K$ enter in uniform magnetic field $B$ perpendicularly, then its frequency will be
A$\frac{e K}{q v B}$
B$\frac{e B}{2 \pi m}$
C$\frac{2 \pi m}{e B}$
D$\frac{2 m}{e B K}$
AIPMT 2001, Medium
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B$\frac{e B}{2 \pi m}$
b $\frac{ mv ^{2}}{ r }= qV B$
$\frac{ mv }{ r }= qB$
$m\omega = qB$
$\omega = qB / m$
$f =\frac{ qB }{2 \pi m }$
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