Question

An ideal solenoid is kept with its axis vertical. Current $I_0$ is flowing in the solenoid. A charge $Q$ is thrown downward inside the solenoid. The acceleration of the charged particle is then

• A. $a > g$
• B. $a = g$
• C. $a < g$
• D. $a = 0$

Hint:

A magnetic field cannot change the speed of a charged particle if its velocity is parallel to the field. Magnetic force only acts on the perpendicular component of velocity.

Answer 1

The Correct Option is B

Concept: An ideal solenoid generates a uniform magnetic field inside it, oriented along its axis. A magnetic field can exert a force on a charged particle only when the particle’s velocity has a component perpendicular to the magnetic field direction. The magnetic force on a moving charge is given by: \[ \vec{F}_B = q\,\vec{v} \times \vec{B} \]
Step 1: Directions of magnetic field and particle velocity Inside the ideal solenoid, the magnetic field $\vec{B}$ is directed along the axis of the solenoid. Since the axis of the solenoid is vertical, the magnetic field is also vertical. The charged particle is projected vertically downward, so its velocity $\vec{v}$ is vertical as well.
Step 2: Magnetic force on the charged particle \[ \vec{F}_B = q\,\vec{v} \times \vec{B} \] As the velocity $\vec{v}$ is parallel to the magnetic field $\vec{B}$, \[ \vec{v} \times \vec{B} = 0 \] Thus, no magnetic force acts on the charged particle.
Step 3: Resultant force and acceleration With zero magnetic force, the only force acting on the particle is its weight: \[ F = mg \] Hence, the acceleration of the particle is: \[ a = g \] Conclusion: The charged particle continues to accelerate solely due to gravity. \[ \boxed{a = g} \]