A uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5$ $Wb/{m^2}$. If a proton having mass $ = 1.7 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C$ moves in this field vertically downwards with energy $5\, MeV$, then the force acting on it will be
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(b) $F = qvB$ and $K = \frac{1}{2}m{v^2}$ $==>$ $F = qB\sqrt {\frac{{2k}}{m}} $
$ = 1.6 \times {10^{ - 19}} \times 1.5\sqrt {\frac{{2 \times 5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{1.7 \times {{10}^{ - 27}}}}} $
$ = 7.344 \times {10^{ - 12}}\,N$
$ = 1.6 \times {10^{ - 19}} \times 1.5\sqrt {\frac{{2 \times 5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{1.7 \times {{10}^{ - 27}}}}} $
$ = 7.344 \times {10^{ - 12}}\,N$
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