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:

Question

Consider two arrangements of wires. Find the ratio of magnetic field at the centre of the semi–circular part.

Answer 1

To solve this problem, we need to understand the forces acting on the charged particle when it is placed in a current-carrying solenoid.

A solenoid with a constant current produces a magnetic field within it, directed along the axis of the solenoid. The magnitude of this magnetic field can be given by the formula:

B = \mu n I

where \mu is the permeability of free space, n is the number of turns per unit length of the solenoid, and I is the current flowing through the solenoid.

When a charged particle with charge Q is placed inside this solenoid, it experiences a magnetic force given by:

F = Q(v \times B)

Here, v is the velocity of the particle, and B is the magnetic field. The cross-product indicates that the magnetic force acts perpendicular to the direction of the magnetic field and the velocity of the particle.

Since the particle is released from rest, its initial velocity is zero (v = 0), making the magnetic force also zero initially. Therefore, there is no magnetic force acting on the particle in the direction of motion.

The only force acting on the particle is its weight, which is due to gravity:

F = mg

where m is the mass of the particle, and g is the acceleration due to gravity.

Therefore, the acceleration of the particle will be solely due to gravity, and thus:

a = g

Since there is no magnetic force contributing along the direction of motion, the charged particle moves with acceleration equal to g, experiencing only its weight.

Thus, the correct answer is:

\(a = g\)

Answer 2