A proton enters a magnetic field of flux density $1.5\,weber/{m^2}$ with a velocity of $2 \times {10^7}\,m/\sec $ at an angle of $30^\circ $ with the field. The force on the proton will be
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(a)$F = qvB\sin \theta $
$ = 1.6 \times {10^{ - 19}} \times 2 \times {10^7} \times 1.5\;\sin \;{30^o}$
$ = 1.6 \times {10^{ - 19}} \times 2 \times {10^7} \times 1.5 \times \frac{1}{2} = 2.4 \times {10^{ - 12}}N$
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