Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $10^5$ ms $^{-1}$ in the presence of electric field $\vec{E}$ and magnetic field $\vec{B}$ . To keep the particle moving in a horizontal direction, it is necessary that: (1) $\vec{B}$ should be perpendicular to the direction of velocity and $\vec{E}$ should be along the direction of velocity. (2) Both $\vec{B}$ and $\vec{E}$ should be along the direction of velocity. (3) Both $\vec{B}$ and $\vec{E}$ are mutually perpendicular and perpendicular to the direction of velocity. (4) $\vec{B}$ 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 are possible?

(1) and (3)

(3) and (4)

(2) and (3)

(2) and (4)

Step-by-Step Solution

Force Analysis: The particle experiences three forces: Gravitational force ( $mg$ ) acting downwards, Electric force ( $F_e = q\vec{E}$ ), and Magnetic force ( $F_m = q(\vec{v} \times \vec{B})$ ). For the particle to move in a constant horizontal direction (straight line), the net vertical force must be zero. Additionally, if the velocity magnitude changes, the magnetic force ( $qvB$ ) would change, potentially disrupting the vertical equilibrium. Thus, a stable horizontal path typically requires constant velocity .
Evaluating Statement (1): $\vec{B} \perp \vec{v}$ and $\vec{E} \parallel \vec{v}$ . The magnetic force $F_m$ is perpendicular to $\vec{v}$ and can be vertical to balance $mg$ . However, the electric force $F_e$ is along $\vec{v}$ , causing the particle to accelerate (change speed $v$ ). Since $F_m \propto v$ , a change in speed will change the vertical magnetic force, destroying the balance with gravity. The particle would eventually deviate. Thus, this is not possible for sustained horizontal motion.
Evaluating Statement (2): Both along $\vec{v}$ . $F_m = 0$ . $F_e$ is horizontal. Gravity ( $mg$ ) acts downwards unbalanced. The path will be parabolic (projectile motion), not horizontal.
Evaluating Statement (3): $\vec{B}, \vec{E}, \vec{v}$ mutually perpendicular. We can orient $\vec{E}$ and $\vec{B}$ such that the net vertical force is zero (e.g., $qE + qvB = mg$ ). Since there is no component of force along the velocity (horizontal), the speed $v$ remains constant, $F_m$ remains constant, and the horizontal path is maintained. This is possible.
Evaluating Statement (4): $\vec{B} \parallel \vec{v}$ implies $F_m = 0$ . $\vec{E} \perp \vec{v}$ implies $F_e$ is perpendicular to velocity. If $\vec{E}$ is vertical and satisfies $qE = mg$ , the net force is zero. The particle moves with constant velocity in a horizontal line. This is possible.
Conclusion: Statements (3) and (4) represent the possible cases.

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