An electron moving with kinetic energy 6.6 x 10^-14 J enters a magnetic field 4 X10^-3 T at right angle to it, The radius of circular path will be nearest to

$\mathrm{evB}={\mathrm{mv}}^2/\mathrm{R}\text{ or }\mathrm{ebR}=\mathrm{mv}$
Squaring on both sides, we get
$\begin{array}{l}{\mathrm{e}}^2{\mathrm{B}}^2{\mathrm{R}}^2={\mathrm{m}}^2{\mathrm{v}}^2=2\mathrm{m}\left[\frac{1}{2}{\mathrm{mv}}^2\right]\\ =2\mathrm{m}\times \text{ K.E. }\\ \therefore \mathrm{R}=\frac{(2\mathrm{m}\times \mathrm{K}.\mathrm{E}.{)}^{1/2}}{\mathrm{eB}}\\ =\frac{{\left(2\times 9\cdot 1\times {10}^{-31}\times 6\cdot 6\times {10}^{-14}\right)}^{1/2}}{1.6\times {10}^{-19}\times 4\times {10}^{-3}}\end{array}$
= 54 cm, which is nearest to 50 cm.