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 produces a uniform magnetic field inside it, directed along its axis. A magnetic field exerts a force on a moving charged particle only if the velocity of the particle has a component perpendicular to the magnetic field. The magnetic force on a charge is given by: \[ \vec{F}_B = q\,\vec{v} \times \vec{B} \]
Step 1: Direction of magnetic field and velocity Inside an ideal solenoid, the magnetic field $\vec{B}$ is along the axis of the solenoid. Since the solenoid’s axis is vertical, $\vec{B}$ is vertical. The charged particle is thrown downward, so its velocity $\vec{v}$ is also vertical. 
Step 2: Magnetic force on the charged particle \[ \vec{F}_B = q\,\vec{v} \times \vec{B} \] Because $\vec{v}$ is parallel to $\vec{B}$, \[ \vec{v} \times \vec{B} = 0 \] Hence, the magnetic force acting on the charge is zero. 
Step 3: Net force and acceleration Since no magnetic force acts on the particle, the only force acting on it is gravity: \[ F = mg \] Therefore, the acceleration of the particle is: \[ a = g \] Conclusion: The acceleration of the charged particle remains equal to gravitational acceleration. \[ \boxed{a = g} \]