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An electron moves with speed $2 \times 10^{5} {ms}^{- 1}$ along the positive x-direction in the presence of a magnetic induction $\bar{B} = \hat{i} + 4 \hat{j} - 3 \hat{k}$ tesla. The magnitude of the force experienced by the electron in newton is (charge on the electron = $1.6 \times 10^{- 19} C$ )

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$1.18 \times 10^{- 13}$

$1.28 \times 10^{- 13}$

$1.6 \times 10^{- 13}$

$1.72 \times 10^{- 13}$

answer is C.

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Detailed Solution

$\vec{F} = q (\bar{V} \times \bar{B}) = 1.6 \times 10^{- 13} N$

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