Question

A particle of specific charge $\left( \dfrac{q}{m} \right)$ is projected from the origin of coordinate with initial velocity $\left[ u\widehat{i}-v\widehat{j} \right]$ uniform electric and magnetic fields exist in the region along the +y direction, of magnitude E and B. The particle will definitely return to the origin once if.
A. $\dfrac{Bv}{2\pi E}$ is an integer
B.${{\left( {{u}^{2}}+{{v}^{2}} \right)}^{\dfrac{1}{2}}}\left( \dfrac{B}{\pi E} \right)$ is an integer
C. $\dfrac{Bv}{\pi E}$ in an integer
D. $\dfrac{uB}{E}$ is an integer

Answer 1

Hint: By equating the force due the electric field to mass times charge’s acceleration, you could get the expression for acceleration. Now substitute this in the equation of motion to get the time taken for the displacement. Then you could equate this to N times the time period of rotation of charge and hence find the answer.

Formula used:
Force due to electric field,
$F=qE$
Time period of charge moving in circular path,
$t=\dfrac{2\pi m}{qB}$
Equation of motion,
$y={{v}_{y}}t+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
In the question we are given a particle of specific charge $\dfrac{q}{m}$ that is projected from the origin with an initial velocity $u\widehat{i}-v\widehat{j}$. Along the y-direction we have both electric as well as magnetic fields of magnitudes E and B. We are asked to find the condition such that the particle will definitely return to origin.
As we are said that there exists an electric field E along y-direction, we know that there will be a force on the charge due to this field given by,
$F=qE$
But from Newton’s second law,
$F=ma$
$\Rightarrow qE=ma$
$\therefore a=\dfrac{qE}{m}$
Now, we know that the acceleration is directed opposite to the direction of velocity, also, if the charge were to return to origin then, the displacement in y− direction will be zero, that is,
$y={{v}_{y}}t+\dfrac{1}{2}a{{t}^{2}}$
$\Rightarrow 0=-vt+\dfrac{1}{2}\left( \dfrac{qE}{m} \right){{t}^{2}}$
$\therefore t=\dfrac{2vm}{qE}$
Under the magnitude field that charge will move in a circular path with a time period of
$t=\dfrac{2\pi m}{qB}$
Let us assume that it takes n turns to return to the origin, then,
$n\times \dfrac{2\pi m}{qB}=\dfrac{2vm}{qE}$
$\Rightarrow N=\dfrac{Bv}{\pi E}$
As N is the number of rounds taken by the charge, it should definitely be an integer.
Therefore, we found that, $\dfrac{Bv}{\pi E}$ is an integer.

Hence, option C is found to be the correct answer.

Note:
 We should note that in the question the direction of the magnetic field is specified as y- direction. Hence, we have to substitute the y-component of the velocity while dealing with motion of the charge in a circular path. Also, as we are said that the charge returns back to the origin which would obviously mean that the charge completes an integral number of turns.