A particle having a mass of $10^{-2}$ kg carries a charge of $5 imes 10^{-8}$ C. The particle is given an in

Question

A particle having a mass of $10^{-2}$ kg carries a charge of $5 	imes 10^{-8}$ 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?

Accepted Answer

1. **Force Analysis:** The particle experiences three forces: Gravitational force ($mg$) acting downwards, Electric force ($F_e = q\vec{E}$), and Magnetic force ($F_m = q(\vec{v} 	imes \vec{B})$). 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 .
2. **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.
3. **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.
4. **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.
5. **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.
6. **Conclusion:** Statements (3) and (4) represent the possible cases.

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