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An electron moves with speed of $2 \times 10^{5} \mathrm{~m} / \mathrm{s}$ along the positive $x$-direction in a magnetic field $\mathbf{B}=(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \mathrm{T}$. The magnitude of the force (in N) experienced by the electron is
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The correct answer is:
$1.6 \times 10^{-13} \quad$
Given,
$v=2 \times 10^{5} \mid \mathrm{im} / \mathrm{s}$
$\mathbf{B}-(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \mathrm{T}$
Charge on electron, $4=1.6 \times 10^{-14} \mathrm{C}$
$\therefore$ Foree, $\mathbf{F}=\hbar(v \times \mathbf{B})$
$=1.6 \times 10^{-19}\left[\left(2 \times 10^{5} \mathrm{i}\right\rangle \times\langle\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{k}\rangle\right]$
$=1.6 \times 10^{-10}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 \times 10^{5} & 0 & 0 \\ 1 & -4 & -3\end{array}\right|$
$\mathrm{P}=1.6 \times 10^{-10}\left(6 \times 10^{5} \mathrm{j}-8 \times 10^{5} \hat{k}\right)$
$=9.6 \times 10^{-14} \mathrm{j}-128 \times 10^{-14} \mathrm{k}$
$\Rightarrow \quad|\mathbf{F}|=\sqrt{\left(9.6 \times 10^{-14}\right)^{2}+\left(-128 \times 10^{-14}\right)^{2}}$
$=10^{-14} \sqrt{(9.6)^{3}+(126)^{2}}$
$-16 \times 10^{-14} \mathrm{~N}$
$=1.6 \times 10^{-13} \mathrm{~N}$
$v=2 \times 10^{5} \mid \mathrm{im} / \mathrm{s}$
$\mathbf{B}-(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \mathrm{T}$
Charge on electron, $4=1.6 \times 10^{-14} \mathrm{C}$
$\therefore$ Foree, $\mathbf{F}=\hbar(v \times \mathbf{B})$
$=1.6 \times 10^{-19}\left[\left(2 \times 10^{5} \mathrm{i}\right\rangle \times\langle\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{k}\rangle\right]$
$=1.6 \times 10^{-10}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 \times 10^{5} & 0 & 0 \\ 1 & -4 & -3\end{array}\right|$
$\mathrm{P}=1.6 \times 10^{-10}\left(6 \times 10^{5} \mathrm{j}-8 \times 10^{5} \hat{k}\right)$
$=9.6 \times 10^{-14} \mathrm{j}-128 \times 10^{-14} \mathrm{k}$
$\Rightarrow \quad|\mathbf{F}|=\sqrt{\left(9.6 \times 10^{-14}\right)^{2}+\left(-128 \times 10^{-14}\right)^{2}}$
$=10^{-14} \sqrt{(9.6)^{3}+(126)^{2}}$
$-16 \times 10^{-14} \mathrm{~N}$
$=1.6 \times 10^{-13} \mathrm{~N}$
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