Question

shows a horizontal solenoid connected to a battery and a switch. A coper ring is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will

• A. remain stationary
• B. move towards the solenoid
• C. move away from the solenoid
• D. oscillate about its initial position

Answer 1

Answer: (B)

move towards the solenoid

Answer 2

When the switch is closed, the current in the solenoid starts to flow and increases from zero to a steady value. This increasing current produces an increasing magnetic field along the axis of the solenoid.

1. Change in Magnetic Flux: As the magnetic field produced by the solenoid increases, the magnetic flux passing through the copper ring also increases.

   $\Phi_B = \int \vec{B} \cdot d\vec{A}$

   Since $\vec{B}$ is increasing, $\Phi_B$ through the ring increases.

2. Faraday's Law of Induction: According to Faraday's law, this change in magnetic flux induces an electromotive force (EMF) in the copper ring, which in turn drives an induced current in the ring (since it's a closed conducting loop).

   $\mathcal{E} = -\frac{d\Phi_B}{dt}$

3. Lenz's Law: Lenz's law states that the direction of the induced current will be such that it opposes the change in magnetic flux that produced it.

   • Let's assume the current in the solenoid flows in such a way that the right end of the solenoid acts as a North pole (magnetic field lines emerge from it, pointing to the right).
   • As the current in the solenoid increases, the magnetic field pointing to the right, passing through the ring, increases.
   • To oppose this increase in the rightward magnetic field, the induced current in the copper ring must produce its own magnetic field pointing to the left.
   • For the ring to produce a magnetic field pointing to the left, its right face must act as a South pole (and its left face as a North pole).

4. Force between Solenoid and Ring:

   • The right end of the solenoid is an N-pole.
   • The right face of the copper ring (due to induced current) is an S-pole.
   • Since opposite poles attract, the N-pole of the solenoid will attract the S-pole of the copper ring.

Therefore, the copper ring will move towards the solenoid.