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(C) The particle has an initial velocity $\vec{v}_{i} = v_{0} \hat{j}$.The electric field $\vec{E} = E_{0} \hat{i}$ exerts a force $\vec{F}_{E} = q E_

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$\frac{\sqrt{3} m v_{0}}{q E_{0}}$

Vedclass · Answer

(C) The particle has an initial velocity $\vec{v}_{i} = v_{0} \hat{j}$.The electric field $\vec{E} = E_{0} \hat{i}$ exerts a force $\vec{F}_{E} = q E_{0} \hat{i}$,causing acceleration $a_{x} = \frac{q E_{0}}{m}$ along the $x$-axis.The magnetic field $\vec{B} = B_{0} \hat{i}$ is parallel to the $x$-axis. Since the initial velocity is along the $y$-axis,the magnetic force $\vec{F}_{B} = q(\vec{v} 	imes \vec{B})$ will act in the $z$-direction. However,the magnetic force does no work,so the kinetic energy changes only due to the electric field.Let the final speed be $v_{f} = 2 v_{0}$. The final velocity components are $v_{x} = a_{x} t = \frac{q E_{0}}{m} t$ and $v_{y} = v_{0}$.Using $v_{f}^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$,and noting that the speed squared is $v_{f}^{2} = (2 v_{0})^{2} = 4 v_{0}^{2}$,we have:$4 v_{0}^{2} = v_{x}^{2} + v_{0}^{2} + v_{z}^{2}$.Since the magnetic force only rotates the velocity vector in the $yz$-plane,the magnitude of the velocity component in the $yz$-plane remains constant at $v_{0}$. Thus,$v_{x}^{2} = 4 v_{0}^{2} - v_{0}^{2} = 3 v_{0}^{2}$.Therefore,$v_{x} = \sqrt{3} v_{0}$.Substituting $v_{x} = \frac{q E_{0}}{m} t$,we get $\frac{q E_{0}}{m} t = \sqrt{3} v_{0}$,which simplifies to $t = \frac{\sqrt{3} m v_{0}}{q E_{0}}$.

$A$ particle of mass $m$ and charge $q$ has an initial velocity $\vec{v} = v_{0} \hat{j}$. If an electric field $\vec{E} = E_{0} \hat{i}$ and a magnetic field $\vec{B} = B_{0} \hat{i}$ act on the particle,its speed will double after a time:

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