Shows a closed coil `ABCL` moving in a uniform magnetic field `B` with a velocity `v`.
Find
(a) emf induced in the coil.
(b) emf induced in curve part `ACB` and straight `AB` as `vev(L) xx vec(B)`

Explanation

Consider rod `AB`, which is a part of the coil. Emf induced in the `rod = BLv`.
Suppose the emf induced in part `ACB` is `E`, as shown.
Since the emf in the zero, `emf ("in" ACB) + emf (in BA) = 0`
or `-E + vBL = 0`
`E = vBL`
Thus, emf induced in any path joining `A and B` is same, provided the magnetic field is uniform. Also, the equivalent emf between `A and B` is `BLv` (here the two emf's are in parallel).