An infinitely long, straight wire carrying current I, one side opened rectangular loop and a conductor C with a sliding connector are located in the same plane, as shown, in the figure. The connector has length l and resistance R. It slides to the right with a velocity v. The resistance of the conductor and the self inductance of the loop are negligible. The induced current in the loop, as a function of separation r, between the connector and the straight wire is :

Solution

Physics sir · 4 steps

The connector stays parallel to the wire, so every point on it feels the same field at distance r.

1. Field at the connector. The connector of length l runs parallel to the long wire, so every point on it is the same distance r from the wire. The field there is uniform along its length and points perpendicular to the plane of the loop.

   B = (μ₀ I)/(2π r)

2. Motional EMF. The connector moves with velocity v, perpendicular to both its length and B. So the EMF across it is ε = Blv = (μ₀ I v l)/(2π r).
3. Current in the loop. The only resistance in the circuit is R, and self-inductance is negligible. So i = ε/R = μ₀/2πIvl/Rr.
4. So. The induced current is μ₀/2πIvl/Rr, which is option (c).

The idea to keep. When a rod moves parallel to a long current-carrying wire, B is the same along the rod, so the EMF is just Blv with B = (μ₀ I)/(2π r). Integration is needed only when the rod points toward the wire.

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