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Q.

An electron moves with speed $2 \times 10^{5} m / s$ along the positive x-direction in the presence of a magnetic induction
$B = \hat{i} + 4 \hat{j} - 3 \hat{k}$ (in 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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a

$1.18 \times 10^{- 13}$

b

$1.28 \times 10^{- 13}$

c

$1.6 \times 10^{- 13}$

d

$1.72 \times 10^{- 13}$

answer is C.

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Detailed Solution

$\vec{v} = 2 \times 10^{5} \hat{i} and \vec{B} = (\hat{i} + 4 \hat{j} - 3 \hat{k})$

$\vec{F} = q (\vec{v} \times \vec{B}) = - 1.6 \times 10^{- 19} [(2 \times 10^{5} \hat{i}) \times (\hat{i} + 4 \hat{j} - 3 \hat{k})]$

$= - 1.6 \times 10^{- 19} \times 2 \times 10^{5} [\hat{i} \times \hat{i} + 4 (\hat{i} \times \hat{j}) - 3 (\hat{i} \times \hat{k})]$

$= - 3.2 \times 10^{- 14} [0 + 4 \hat{k} + 3 \hat{j}] = 3.2 \times 10^{- 14} (- 4 \hat{k} - 3 \hat{j})$

$\Rightarrow | \vec{F} | = 3.2 \times 10^{- 14} \times 5 = 1.6 \times 10^{- 13} N$

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