A conducting wire is bent in the form of an angle `theta`. The wire moves with velocity v along the bisector of `angleA'OB'` as shown in figure. Find the emf induced between the two ends of the wire if a magnetic field is acting perpendicular to the plane of the paper and directed into it.

Explanation

Let, 2l = length of the wire A'OB'
where, OA' = OB' = l
`therefore vlsin.(theta)/(2)` = area described in unit time by the portion OA'.
`Blvsin.(theta)/(2)` = magnetic flux intercepted by OA' in unit time
= emf induced between O and A'.
Similarly, `Blvsin.(theta)/(2)` = emf induced between O and B'.
Using Flemming's right hand rule, we get that the induced emf is directed from B' to A', i.e., A' is at a higher potential.
`therefore` Net induced emf `=2xxBlvsin.(theta)/(2)=2Blvsin.(theta)/(2)`