Question

A particle having a mass of ${10^{ - 2}}kg$ carries a charge of $5 \times {10^{ - 8}}C$. The particle is given an initial horizontal velocity of ${10^5}m{s^{ - 1}}$ in the presence of electric field $\vec E$ and $\vec B$. To keep the particle moving in a horizontal direction, it is necessary that
(i) $\vec B$ should be perpendicular to the direction of velocity and $\vec E$ should be along the direction of velocity.
(ii)Both $\vec B$ and $\vec E$ should be along the direction of velocity.
(iii) Both $\vec B$ and $\vec E$ are mutually perpendicular and perpendicular to the direction of velocity.
(iv) $\vec B$ should be along the direction of velocity and $\vec E$ should be perpendicular to the direction of velocity.
Which one of the following pairs of statements is possible
(i) and (iii)
(iii) and (iv)
(ii) and (iii)
(ii) and (iv)

Answer 1

To answer this question, we need to examine the direction of the force which acts due to the electric and the magnetic fields on a charged particle. Then, for keeping the particle moving in the horizontal direction, the particle must be in equilibrium. SO the electric force and the magnetic force must cancel each other.
Formula used: The formulae used in solving this question are given by
${\vec F_E} = q\vec E$
${\vec F_B} = q\left( {\vec v \times \vec B} \right)$
Here, ${\vec F_E}$ and ${\vec F_B}$ are respectively the electric and the magnetic forces on the charged particle of charge $q$ due to the electric field $\vec E$ and the magnetic field $\vec B$, and $\vec v$ is the velocity of the particle.

Complete step-by-step solution:
We know that the electric force on a charged particle due to an electric field is given by the equation
${\vec F_E} = q\vec E$ ………….. (i)
The above equation shows that the electric force acts along the direction of the electric field.
Also, the magnetic force on a moving charged particle is given by the expression
${\vec F_B} = q\left( {\vec v \times \vec B} \right)$.......... (ii)
From the above expression, the magnetic force on a charged particle acts perpendicular to the plane of the velocity and the magnetic field.
Let us suppose that the initial horizontal velocity of the given direction is along the positive x-axis.
Now, for keeping the charged particle moving with the same velocity, the acceleration of the charged particle in the horizontal direction must be zero. From Newton's second law of motion, the net force on the charged particle in the horizontal direction must be zero.
Now, if the electric field $\vec E$ is along the direction of velocity of the charged particle, then the electric field will also be along the direction of the velocity. So the magnetic force needs to be opposite to the direction of the magnetic field. But from (ii) we have seen that the magnetic field is always perpendicular to the velocity. So in the case when $\vec E$ is along the direction of velocity, the net force on the charged particle cannot be equal to zero.
Therefore, both the statements (1) and (2) are incorrect.
From the statement (3) $\vec B$ and $\vec E$ are mutually perpendicular and perpendicular to the direction of velocity. We can easily see that in this case we can make the magnetic and the electric forces opposite to each other which can cancel each other.
So the statement (3) is correct.
According to the statement (4), $\vec B$ is along the direction of velocity, then the magnetic force will be zero. Also, if $\vec E$ is perpendicular to the velocity of the charged particle, the electric force will act along the direction perpendicular to the velocity. So no force will act along the direction of the velocity, and hence its horizontal velocity will remain unchanged.
So the statement (4) is correct.
Thus, statements (3) and (4) are correct.

Hence, the correct answer is option B.

Note: By carefully reading the question, we have been given the condition that the horizontal velocity should remain unchanged. This doesn’t necessarily mean that net force on the particle should be zero. But this means that the net force along the direction of velocity must be zero.