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A particle of mass $1 \times 10^{-26} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ travelling with a velocity $1.28 \times 10^6 \mathrm{~ms}^{-1}$ along the positive $X$-axis enters a region in which a uniform electric field $\mathbf{E}$ and a uniform magnetic field of induction $\mathbf{B}$ are present. If $\mathbf{E}=-102.4 \times 10^3 \hat{\mathbf{k}} \mathrm{NC}^{-1}$ and $\mathbf{B}=8 \times 10^{-2} \hat{\mathbf{j}} \mathrm{Wbm}^{-2}$ the direction of motion of the particles is
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Verified Answer
The correct answer is:
along the positive $X$-axis
$\begin{aligned}
m & =1 \times 10^{-26} \mathrm{~kg}, q=1.6 \times 10^{-19} \mathrm{C} \\
v & =1.28 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Electric field, $\mathbf{E}=-102.4 \times 10^3 \hat{\mathbf{k}} \mathrm{N} / \mathrm{C}$
Magnetic field, $\mathbf{B}=8 \times 10^{-2} \hat{\mathbf{j}} \mathrm{Wbm}^{-2}$
$\begin{aligned}
\frac{|\mathbf{E}|}{|\mathbf{B}|} & =\frac{102.4 \times 10^3}{8 \times 10^{-2}} \\
& =\frac{10.24 \times 10^6}{8} \\
& =1.28 \times 10^6
\end{aligned}$
Hence,
$|\mathbf{v}|=\frac{|\mathbf{E}|}{|\mathbf{B}|}$
So, particle will remain undeflected, hence direction of motion of particle is along the positive $X$-axis.
m & =1 \times 10^{-26} \mathrm{~kg}, q=1.6 \times 10^{-19} \mathrm{C} \\
v & =1.28 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Electric field, $\mathbf{E}=-102.4 \times 10^3 \hat{\mathbf{k}} \mathrm{N} / \mathrm{C}$
Magnetic field, $\mathbf{B}=8 \times 10^{-2} \hat{\mathbf{j}} \mathrm{Wbm}^{-2}$
$\begin{aligned}
\frac{|\mathbf{E}|}{|\mathbf{B}|} & =\frac{102.4 \times 10^3}{8 \times 10^{-2}} \\
& =\frac{10.24 \times 10^6}{8} \\
& =1.28 \times 10^6
\end{aligned}$
Hence,
$|\mathbf{v}|=\frac{|\mathbf{E}|}{|\mathbf{B}|}$
So, particle will remain undeflected, hence direction of motion of particle is along the positive $X$-axis.
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