JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
A bar magnet is released from rest along the axis of a very long vertical copper tube. After some time the magnet will
- (a) move down with almost constant speed
- (b) move down with an acceleration equal to g
- (c) move down with an acceleration greater than g
- (d) oscillate inside the tube
Answer and solution
Answer (a) move down with almost constant speed
Solution
Physics sir · 4 steps
The faster the magnet falls, the harder the eddy currents push back, until drag equals weight.
- Eddy currents appear. As the magnet falls, the flux through each ring of the copper tube keeps changing. This induces circulating eddy currents in the tube walls, both above and below the magnet.
- They oppose motion. By Lenz's law these currents oppose the cause, which is the magnet's motion. So they exert an upward drag force on the magnet, and this force grows with the magnet's speed: F ∝ v.
ma = mg - kv
- Speed stops rising. At the start v = 0, so a = g. As v grows, the drag kv grows and a falls. When kv becomes equal to mg, the net force is zero and the magnet keeps a steady terminal speed vₜ = mg/k.
- So. After some time the magnet moves down with almost constant speed, which is option (a).
The idea to keep. Eddy-current drag acts like air resistance: it grows with speed and opposes motion. Anything falling against such a drag reaches a terminal speed where drag equals weight.
Answer from the NTA final answer key. Solution by PrepWiser.