Question

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 to the plane of the paper.

• A. $Bl\sin (\theta /2)$
• B. $Bv\sin (\theta /2)$
• C. $2Blv\sin (\theta /2)$
• D. $Blv\sin (\theta /4)$

Answer 1

Answer: (C)

The rod $OA$ is equivalent to a cell with emf $vBl\sin \theta /2$. the positive changes shift towards $A$ due to the force $qv\times B$.
The positive terminal of the equivalent will appears towards $A$. Similarly, the rod $BO$ is equivalent to a cell of emf $vBl\sin \theta /2$, with the positive terminal towards $O$.
The equivalent circuit is shown in figure. Clearly, the emf between the points $A$ and $B$ is $2Blv\sin \theta /2$.