Question

$\text{BACD}$ is a fixed conducting smooth rail placed in a vertical plane. $\text{PQ}$ is a conducting rod which is free to slide on the rails. A horizontal uniform magnetic field exists in space as shown. If the rod $\text{PQ}$ is released from rest then,

• A. The rod $\text{PQ}$ will move downward with constant acceleration.
• B. The rod $\text{PQ}$ will move upward with constant acceleration.
• C. The rod will move downward with decreasing acceleration and finally acquire a constant velocity.
• D. either $\text{A}$ or $\text{B}$.

Answer 1

The correct option is C The rod will move downward with decreasing acceleration and finally acquire a constant velocity.

When the rod $\text{PQ}$ is released, the force acting on it is only the force of gravity. Therefore, the rod will start to move downwards.

As the rod starts moving, an emf will generate across the rod due to which a current will begin to flow in the circuit.

According to the Lenz`s law, the current will flow such that it will produce a magnetic force $F_m$ which opposes the downwards motion of the rod.

Therefore, the acceleration of rod will decrease with time until both the force balances out each other. Therefore, the rod will attain a constant velocity.


Therefore, The rod will move downward with decreasing acceleration and finally acquire a constant velocity.

Hence, option $\left(\text{C}\right)$ is correct.