The electric field acts along positive x-axis. A charged particle of charge q and mass m is released from origin and moves with velocity $\vec{v} = v_{0} \hat{j}$ under the action of electric field and magnetic field, $\vec{B} = B_{0} \hat{i}$ The velocity of particle becomes $2 v_{0}$ after time $\frac{\sqrt{3} {mv}_{0}}{\sqrt{2} {qE}_{0}} .$ Find the electric field.

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$\frac{\sqrt{2}}{\sqrt{3}} E_{0} \hat{i}$

$\sqrt{2} E_{0} \hat{i}$

$\frac{\sqrt{3}}{\sqrt{2}} E_{0} \hat{i}$

$\sqrt{3} E_{0} \hat{i}$

answer is D.

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Detailed Solution

$\vec{v}$ is perpendicular to both $\vec{E}$ and $\vec{B}$ .
Path of particle is helix with increasing pitch

$v = {(v_{x}^{2} + v_{y}^{2} + v_{z}^{2})}^{1 / 2}$

$\begin{array}{l} Here, v_{x}^{2} = {(\frac{qE}{m} t)}^{2} and v_{y}^{2} + v_{z}^{2} = v_{0}^{2} \\ Also, v = 2 v_{0} \\ {(2 v_{0})}^{2} = \frac{q^{2} E^{2} t^{2}}{m^{2}} + v_{0}^{2} \\ t = \frac{\sqrt{3} {mv}_{0}}{qE} = \frac{\sqrt{3} {mv}_{0}}{\sqrt{3} {qE}_{0}} (given) \\ \vec{E} = \sqrt{2} E_{0} \hat{i} \end{array}$