Question

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 $\vec{B}=B_{0}\hat{i}$ acts on the particle, its speed will double after a time,
\[\begin{align}
  & A.\dfrac{\sqrt{3}m{{v}_{0}}}{q{{E}_{0}}} \\
 & B.\dfrac{3m{{v}_{0}}}{q{{E}_{0}}} \\
 & C.\dfrac{\sqrt{2}m{{v}_{0}}}{q{{E}_{0}}} \\
 & D.\dfrac{2m{{v}_{0}}}{q{{E}_{0}}} \\
\end{align}\]

Answer 1

Hint: We know from Lorentz force gives the total force experienced by a moving charge, when both electric and magnetic force acts on it. The total force is the sum of the electrical and the magnetic force applied on the charge.
Formula:
$F=q(E+v\times B)$

Complete answer:
Let us consider a charge $q$ with mass $m$ and velocity$v$. If an electrical force $E$ and magnetic force $B$ are applied on the charge simultaneously, then the net force $F$ on the charge is given from Lorentz force as $F=q(E+v\times B)$
Given that velocity of the charge is $\vec{v}=v_{0}\hat{j}$, the applied electric field is given as $\vec{E}=E_{0}\hat{i}$ and the applied magnetic field is given as $\vec{B}=B_{0}\hat{i}$.
Clearly the electrical field accelerates the particle along the $\hat i$, while the magnetic field will make the particle to trace a circular motion.
Clearly, the velocity had two component, namely$x$,$y$
Since the $v$ is a vector, the magnitude of $v$ at any given time is given as $\sqrt{v_{x}^{2}+v_{y}^{2}+v^{2}_{z}}$
Here, since there is no $z$ component of the velocity, then the magnitude of $v$ at any given time is reduced as $\sqrt{v_{x}^{2}+v_{y}^{2}}$

Also given the initial velocity is $v_{0}$ is along $y$, then we can write, $v_{0}^{2}=v_{y}^{2}$.
Let us assume that it takes $t$ time for the velocity to double itself to $2v_{0}$
Then $(2v_{0})^{2}=v_{x}^{2}+v_{y}^{2}=v_{x}^{2}+v_{0}^{2}$
$\implies v_{x}^{2}=4v_{)}^{2}-v_{0}^{2}=3v_{0}^{2}$
$\implies v_{x}=\sqrt 3v_{0}$
This $v_{x}$ component is due to acceleration of the particle in the presence of an electric field.
Hence from the laws of motion, we can write, $v_{x}=u_{x}+a_{x}t$
Since there was no component of the initial velocity along the x-axis, we get $v_{x}=a_{x}t$
Also the force due to $E$ is given as $F=qE$, but $F=ma$ where $a$ is the acceleration of the particle,
Then, we get, $a=\dfrac{qE}{m}$
Since $E$ is only along $x$ direction, this acceleration is denoted as $a_{x}$
Then on substituting the values, we get, $\sqrt 3v_{0}=\dfrac{qE_{0}}{m}t$
$\implies t=\dfrac{\sqrt 3v_{0}m}{qE_{0}}$

Hence the answer is \[A.\dfrac{\sqrt{3}m{{v}_{0}}}{q{{E}_{0}}}\]

Note:
The electrical field applied on a charge particle accelerates the particle, while the magnetic field alters the direction of the particle. When both are applied, the particle experiences a helical path. Lorentz force is the basic principle used in the cyclotrons.