An electron is moving with a velocity $(2 \bar{i}+3 \bar{j}) \mathrm{ms}^{-1}$ in an electric field $(3…

An electron is moving with a velocity $(2 \bar{i}+3 \bar{j}) \mathrm{ms}^{-1}$ in an electric field $(3 \bar{i}+6 \bar{j}+2 \bar{k}) \mathrm{Vs}^{-1}$ and a magnetic field of $(2 \bar{j}+3 \bar{k}) \mathrm{T}$. Then the magnitude and direction (with $x$-axis) of the Lorentz force acting on the electron is
  1. $9.6 \times 10^{-19} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)$
  2. $9.6 \times 10^{-19} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{5}{\sqrt{2}}\right)$
  3. $2.15 \times 10^{-18} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)$
  4. $2.15 \times 10^{-18} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{5}{3}\right)$

Solution

$v=(2 \hat{i}+3 \hat{j}) \mathrm{m} / \mathrm{s}, E=(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \mathrm{Vs}$ $\overrightarrow{\mathrm{B}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}) \mathrm{T}$ The lorentz force acting on electron is $\begin{aligned} & \overrightarrow{\mathrm{F}}=-\mathrm{e}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{b}})-\mathrm{e} \overrightarrow{\mathrm{E}} \\ & =-\mathrm{e}[(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{k}})+(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})]\end{aligned}$ $\begin{aligned} & =-6 \mathrm{e}(2 \hat{\mathrm{i}}+\hat{\mathrm{k}}) \\ & \therefore \quad \mathrm{F}=6 \times 1.6 \times 10^{-19} \times \sqrt{5}=2.15 \times 10^{-18} \mathrm{~N}\end{aligned}$ The direction of Lorentz force with x -axis is $\begin{aligned} & \cos \theta=\frac{2}{\sqrt{2^2+1^2}}=\frac{2}{\sqrt{5}} \\ & \therefore \theta=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\end{aligned}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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