Consider a circular loop falling in a uniform magnetic field as shown. By Faradays law, as it falls there is no change in flux, so no induced emf. But, if we apply Motional EMF equations the the 2 semicircles, emf becomes $B(2r)v$. Why is this happening?


1 Answer 1


Sorry I misread the question.

The induced motional emf across the ring will be $B2rv$ as you have stated with the "top" part of the ring and the bottom part of the ring contributing the same magnitude.

The rails together with the termination $PQ$ define another loop with only the loop moving and the rails and termination stationary and the magnetic flux through that loop is changing.

The moving loop by itself itself does not contribute as the magnetic flux through it does not change and so there is no "extra effect" due to the moving conductor across the rail being in two parts ie the rate of change of magnetic flux through the stationary part of the loop and the "top" of the ring is the same as the rate of change of flux through the stationary part of the loop and the "bottom" of the ring.

  • $\begingroup$ but suppose we want to find the current through PQ. the induced emf's will add up, and we will get some current $\endgroup$ Jun 29, 2021 at 12:56
  • $\begingroup$ @nagarkaradi I have rewritten my answer. $\endgroup$
    – Farcher
    Jun 29, 2021 at 16:09

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