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An electron moves with speed $2 \times 10^5 \mathrm{~m} / \mathrm{s}$ along the positive $x$-direction in the presence of a magnetic field of induction $\mathbf{B}=\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ (in tesla). The magnitude of the force experienced by the electron in newtons is
$\left(\right.$ Charge on the electron $\left.=1.6 \times 10^{-19} \mathrm{C}\right)$
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$\left(\right.$ Charge on the electron $\left.=1.6 \times 10^{-19} \mathrm{C}\right)$
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Verified Answer
The correct answer is:
$1.6 \times 10^{-13}$
$\begin{aligned} & \mathbf{v}=2 \times 10^5 \mathrm{~m} / \mathrm{s} \text { (along } X \text {-axis) } \\ & =\left(2 \times 10^4 \hat{\mathbf{i}}\right) \mathrm{m} / \mathrm{s} \\ & \mathbf{B}=(\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \mathrm{T} \\ & \text { Force } \quad \mathbf{F}=q(\mathbf{v} \times \mathbf{B}) \\ & \text { For electron } q=1.6 \times 10^{-19} \mathrm{C} \\ & \mathbf{V} \times \mathbf{B}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 \times 10^5 & 0 & 0 \\ 1 & 4 & -3\end{array}\right| \\ & =\hat{\mathbf{i}}[(0 \times-3)-(4 \times 10)] \\ & -\hat{\mathbf{j}}\left[\left(2 \times 10^5 \times-3\right)-(1 \times 0)\right] \\ & +\hat{\mathbf{k}}\left[\left(2 \times 10^5 \times 4\right)-(1 \times 0)\right] \\ & =6 \times 10^5 \hat{\mathbf{j}}+8 \times 10^5 \hat{\mathbf{k}} \\ & \mathbf{F}=1.6 \times 10^{-19} \sqrt{\left(6 \times 10^5\right)^2+\left(8 \times 10^5\right)^2}, \\ & =1.6 \times 10^{-19} \times 10 \times 10^5 \\ & =1.6 \times 10^{-18} \mathrm{~N} \\ & \end{aligned}$
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