A $2\, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5\, tesla$. The force on the proton is
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(d) $F = qvB = 1.6 \times {10^{ - 19}} \times \left[ {\sqrt {\frac{{2E}}{m}} } \right]\;2.5$
$ = 4 \times {10^{ - 19}}\sqrt {\frac{{2 \times 2 \times 1.6 \times {{10}^{ - 19}} \times {{10}^6}}}{{1.66 \times {{10}^{ - 27}}}}} $$ = 7.6 \times {10^{ - 12}}\,N$
$ = 4 \times {10^{ - 19}}\sqrt {\frac{{2 \times 2 \times 1.6 \times {{10}^{ - 19}} \times {{10}^6}}}{{1.66 \times {{10}^{ - 27}}}}} $$ = 7.6 \times {10^{ - 12}}\,N$
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