An angle aob made of a conducting wire moves along its bisector through a magnetic field B as suggested by figure. Find the emf induced between the two free ends if the magnetic field is perpendicular ot the plane of the angle.

Explanation

Consider the cirucit as being closed externally, as shown in the figure
Fi ad =x , the area of the rectangular part abcd is A = 2xl sin(theta/2)
As the angle moves towards right, the flux of the magnetic field through this rectangular area decreases at the rate `(dPhi)/(dt) = B(dA)/(dt) = 2Blv sin (theta/2)`
This is also the rate of decrease of the flux through the closed circuit shown in the figure
so the induced emf is `epsilon = 2 Blv sin(theta/2)`
as the emf is induced solely because of the movement of the angle, this is the emf induced between its ends