Question

A proton (mass $m$ , charge $e$ ) projected with a velocity $v$ passes undeviated through a region of crossed electric and magnetic fields. The velocity with which an alpha particle (mass $4m$ , charge $2e$ ) should be projected so that it passes undeviated through the same region is:
(A) $v$
(B) $2v$
(C) $4v$
(D) $8v$

Answer 1

In order to solve the above question, we will be using the concept, the effects of magnetic and electric fields on a moving charge. We will also use the concept of Lorentz force.
Lorentz force: $\vec{F}=q[\vec{E}+(\vec{v}\times \vec{B})]$
Where $\vec{F}$ is the force, $q$ is the charge, $\vec{v}$ is the velocity, $\vec{E}$ is the electric field and $\vec{B}$ is the magnetic field.

Complete step by step answer:
First of all, from the details mentioned in the above question, we can assume that there is a charged particle that is moving in a region of crossed electric and magnetic fields. Now, we all are aware that the force experienced by a charged particle when passing through an electric field or a magnetic field or both is given by the Lorentz force.
We will be dividing the question into two parts:
PART A
In this part, a proton with mass $m$ and charge $e$ is moving in the region of crossed electric and magnetic fields. Therefore, using the formula for Lorentz force, we get
$\vec{F}=q[\vec{E}+(\vec{v}\times \vec{B})]$
$\Rightarrow \vec{F}=q\vec{E}+q(\vec{v}\times \vec{B})$
Now, it is also mentioned in the question that the proton passes the region without getting deviated from its direction of projection. Therefore the force acting on the particle due to the electric and magnetic field is zero. And hence, the above equation becomes
\Rightarrow 0=q\vec{E}+q(\vec{v}\times \vec{B}) \\\
$\Rightarrow q\vec{E}=q(\vec{v}\times \vec{B}) \\\$
$\Rightarrow \vec{E}=\vec{v}\times \vec{B} \\\$
We can cancel the charge $q$ on both sides as it is constant in both the fields.
As we all know that the magnetic field is always perpendicular to the direction of the velocity with which the charge is moving. Therefore $\vec{v}\times \vec{B}=vB\sin \theta =vB\sin {{90}^{0}}=vB$ . And also the electric field and the magnetic field are perpendicular to each other.
$\Rightarrow E=vB \\\$
$\Rightarrow v=\dfrac{E}{B} \\\$
The velocity of the proton is given by the above equation.
The important thing to note here is that the velocity of the charged particle depends only upon the magnitudes of the electric and magnetic fields in whose effect it is moving.
PART B
In this part, we will be considering the alpha particle with mass $4m$ and charge $2e$ moving in the same conditions as in part A.
As we have learned from the first part, the velocity of the charged particle depends only upon the magnitudes of the electric and magnetic fields in whose effect it is moving. Here, the alpha particle is moving in the same region of fields and same magnitude of fields.
Therefore, even though the mass and charge of the particle is different from the part A, the velocity will be of the same magnitude as of in the part A and is given by
$\Rightarrow v=\dfrac{E}{B}$
Finally, the correct answer to our question is option (A).

Note:
It is very important to note here that it was not necessary to calculate the velocity in the second part as we already derived that the velocity of a charged particle only depends upon the value of electric and magnetic field applied in the region and not on the charge or the mass of the particle.