XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is l. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L(L>l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is ____ m/s.(g= acceleration due to gravity)

Solution

Physics sir · 4 steps

At terminal speed the magnetic drag on the rod cancels its weight. Use only the length between the rails.

1. Induced current. Only the part of the rod between the rails, of length l, is in the circuit. Falling at speed v, it produces an EMF Blv, so the current through R is I = Blv/R.
2. Magnetic drag. This current flows through the same length l in field B, so the force on it is F = BIl = (B² l² v)/R. By Lenz's law it points upward and opposes the fall.
3. Terminal condition. The drag grows as v grows. The speed stops increasing when the drag equals the weight mg.

   (B² l² vₜ)/R = mg ⇒ vₜ = mgR/(B² l²)

4. So. The terminal speed is mgR/(B² l²), which is option (a).

The idea to keep. For a rod falling on rails, set the magnetic force (B² l² v)/R equal to mg. The length that counts is always the part inside the circuit, not the full rod.

Answer from the NTA final answer key. Solution by PrepWiser.