Q.
An $\angle A O B$ 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 to the plane of the paper.
Electromagnetic Induction
Solution:
The rod $O A$ is equivalent to a cell with emf $v B l \sin \theta / 2$. the positive changes shift towards $A$ due to the force $q v \times B$.
The positive terminal of the equivalent will appears towards $A$. Similarly, the rod $B O$ is equivalent to a cell of emf $v B l \sin \theta / 2$, with the positive terminal towards $O$.
The equivalent circuit is shown in figure. Clearly, the emf between the points $A$ and $B$ is $2 B l v \sin \theta / 2$.
