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A particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec{E}=E_0 \hat{i}$ and $\vec{B}=B_0 \hat{i}$ with a velocity $\vec{v}=v_0 \hat{j}$. The speed of the particle will become $\frac{\sqrt{5}}{2} v_0$ after a time
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Verified Answer
The correct answer is:
$\frac{m v_0}{2 q E_0}$
Here, $\vec{E}$ and $\vec{B}$ are acting along $x$-axis and $\vec{v}$ is along $y$-axis i.e. perpendicular to both $\vec{E}$ and $\vec{B}$. Therefore, the path of charged particle is a helix with increasing speed. Speed of particle at time $t$ is
$v=\sqrt{v_x^2+v_y^2}$
Here, $v_y=v_0 ; v_x=\frac{q E_0}{m} t$ and $v=\frac{\sqrt{5}}{2} v_0$
Putting values in $(\mathrm{i})$, we get
$t=\frac{m v_0}{2 q E_0}$
$v=\sqrt{v_x^2+v_y^2}$
Here, $v_y=v_0 ; v_x=\frac{q E_0}{m} t$ and $v=\frac{\sqrt{5}}{2} v_0$
Putting values in $(\mathrm{i})$, we get
$t=\frac{m v_0}{2 q E_0}$
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