Question

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

• 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 1

The correct option is C $1.6\times {10}^{-13}$

Force on charge particle in magnetic field is given by,
$\overline{F}=q(\overrightarrow{V}\times \overrightarrow{B})$

$V=2\times {10}^5\hat{i}$

=$1.6\times {10}^{-19}(V\hat{i}\times (\hat{i}+4\hat{j}-3\hat{k}\left)\right)$

=$1.6\times {10}^{-19}(4V\hat{k}+3V\hat{j})$

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

$|F|=1.6\times {10}^{-13}N$