Question

A current carrying solenoid is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy:

• A. \(0<a<g\)
• B. \(a>g\)
• C. \(a = 0\)
• D. \(a = g\)

Hint:

The magnetic force \( \vec{F}_m = Q(\vec{v} \times \vec{B}) \) acts only when a charged particle has a velocity component perpendicular to the magnetic field. If the particle moves parallel or anti-parallel to the magnetic field lines, it experiences no magnetic force.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A charged particle is released from rest inside a vertical current-carrying solenoid. We need to determine its acceleration. The key is to analyze the forces acting on the particle.
Step 2: Key Formula or Approach:
The forces on the particle are: 1. Gravitational force: \( \vec{F}_g = m\vec{g} \) (acting vertically downwards). 2. Magnetic force (Lorentz force): \( \vec{F}_m = Q(\vec{v} \times \vec{B}) \). The net force determines the acceleration: \( \vec{F}_{\text{net}} = m\vec{a} \).
Step 3: Detailed Explanation:
1. Magnetic Field Direction: The magnetic field (\(\vec{B}\)) inside a long solenoid is uniform and directed along its axis. Since the solenoid is placed vertically, the magnetic field \(\vec{B}\) is also vertical.
2. Initial Motion: The particle is released from rest, so its initial velocity \(\vec{v} = 0\). At this instant, the magnetic force is \( \vec{F}_m = Q(0 \times \vec{B}) = 0 \). The only force is gravity, so the particle begins to accelerate downwards due to gravity.
3. Subsequent Motion: As the particle starts moving, its velocity vector \(\vec{v}\) will be directed vertically downwards, along the axis of the solenoid.
4. Calculating Magnetic Force: The velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\) are both along the vertical axis. This means they are parallel to each other. The angle \( \theta \) between \(\vec{v}\) and \(\vec{B}\) is 0° or 180°.
The magnitude of the magnetic force is given by \( F_m = |Q|vB\sin\theta \).
Since \( \sin(0^\circ) = 0 \) and \( \sin(180^\circ) = 0 \), the magnetic force on the particle is always zero as long as it moves along the axis.
5. Net Force and Acceleration: Since the magnetic force is always zero, the only force acting on the particle is the gravitational force, \( F_g = mg \).
From Newton's second law:
\[ F_{\text{net}} = mg = ma \] \[ a = g \] Step 4: Final Answer:
The acceleration of the charged particle is equal to the acceleration due to gravity, g. This corresponds to option (D).