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A proton is projected with velocity $\overrightarrow{ V }=2 \hat{ i }$ in a region where magnetic field $\overrightarrow{ B }=(\hat{i}+3 \hat{j}+4 \hat{k})\; \mu T$ and electric field $\overrightarrow{ E }=10 \hat{ i } \;\mu V / m .$ Then find out the net acceleration of proton (in $m / s ^{2}$)
$1400$
$700$
$1000$
$800$
Solution
The expression of force is given by,
$\overrightarrow{ F }= Q \overrightarrow{ E }+ Q (\overrightarrow{ V } \times \overrightarrow{ B })$
$=\left(1.6 \times 10^{-19}\right)\left(10 \hat{ i } \times 10^{-6}\right)+\left(1.6 \times 10^{-19}\right)[(2 \hat{ i }) \times(\hat{ i }+3 \hat{ j }+4 \hat{ k })] \times 10^{-9}$
$=\left(1.6 \times 10^{-19}\right)[10 \hat{ i }-8 \hat{ j }+6 \hat{ k }] \times 10^{-6} N$
Calculate the acceleration as follows,
$\overrightarrow{ a }=\frac{\left(1.6 \times 10^{-19}\right)[10 \hat{ i }-8 \hat{ j }+6 \hat{ k }] \times 10^{-6} N }{1.6 \times 10^{-27}}$
$=1400 m / s ^{2}$
