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Question

(D) The electric field $\vec E = E\hat i$ exerts a force $\vec F_e = qE\hat i$ on the particle,causing acceleration $a_x = \frac{qE}{m}$ along the $x$

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The electric field $\vec E = E\hat i$ exerts a force $\vec F_e = qE\hat i$ on the particle,causing acceleration $a_x = \frac{qE}{m}$ along the $x$-axis.The magnetic field $\vec B = B\hat i$ is parallel to the initial velocity $\vec v = v_0\hat j$. Since the magnetic force $\vec F_m = q(\vec v 	imes \vec B) = 0$ when the velocity is parallel to the magnetic field,the velocity component along the $y$-axis remains constant at $v_y = v_0$.At time $t$,the velocity components are $v_x = a_x t = \frac{qE}{m}t$ and $v_y = v_0$.The speed $v$ at time $t$ is given by $v = \sqrt{v_x^2 + v_y^2}$.We are given that the speed becomes $2v_0$,so:$2v_0 = \sqrt{(\frac{qE}{m}t)^2 + v_0^2}$Squaring both sides:$4v_0^2 = (\frac{qE}{m}t)^2 + v_0^2$$3v_0^2 = (\frac{qE}{m}t)^2$Taking the square root:$\sqrt{3}v_0 = \frac{qE}{m}t$Solving for $t$:$t = \frac{\sqrt{3}mv_0}{qE}$

$A$ particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec E = E\hat i$ and a magnetic field $\vec B = B\hat i$ 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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