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(D) The electric field $\vec{E} = E_0 \hat{i}$ exerts a force $\vec{F}_e = qE_0 \hat{i}$,causing acceleration $a_x = \frac{qE_0}{m}$ along the $x$-axi

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$t = \frac{\sqrt{3}mv_0}{qE_0}$

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(D) The electric field $\vec{E} = E_0 \hat{i}$ exerts a force $\vec{F}_e = qE_0 \hat{i}$,causing acceleration $a_x = \frac{qE_0}{m}$ along the $x$-axis.The magnetic field $\vec{B} = B_0 \hat{i}$ is parallel to the electric field and the $x$-axis. Since the initial velocity $\vec{v} = v_0 \hat{j}$ is perpendicular to the magnetic field,the particle undergoes helical motion in the $yz$-plane while accelerating along the $x$-axis.The velocity components at time $t$ are $v_x = a_x t = \frac{qE_0}{m} t$ and the speed in the $yz$-plane remains constant at $v_{\perp} = v_0$.The total speed $v$ at time $t$ is given by $v = \sqrt{v_x^2 + v_{\perp}^2} = \sqrt{(\frac{qE_0}{m} t)^2 + v_0^2}$.We are given that the speed becomes $2v_0$ at time $t$:$2v_0 = \sqrt{(\frac{qE_0}{m} t)^2 + v_0^2}$Squaring both sides:$4v_0^2 = (\frac{qE_0}{m} t)^2 + v_0^2$$3v_0^2 = (\frac{qE_0}{m} t)^2$Taking the square root:$\sqrt{3}v_0 = \frac{qE_0}{m} t$Solving for $t$:$t = \frac{\sqrt{3}mv_0}{qE_0}$

$A$ 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 a magnetic field $\vec{B} = B_0 \hat{i}$ with an initial velocity $\vec{v} = v_0 \hat{j}$. The speed of the particle will become $2v_0$ after a time:

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