Question

An angle $aob$ made of a conducting wire moves towards right along its bisector through a uniform magnetic field $B$ as shown in the figure. Find the net emf induced between two free ends if the magnetic field is perpendicular to the plane of the angle. Velocity of the angle is $v$.

• A. $Bvl\sin \theta$
• B. $2Bvl\sin \theta$
• C. $2Bvl\sin \left(\theta /2\right)$
• D. $Bvl\sin \left(\theta /2\right)$

Answer 1

The correct option is C $2Bvl\sin \left(\theta /2\right)$


The rod $oa$ is equivalent to a battery of emf, $Bvl\sin \left(\theta /2\right)$.

From right-hand rule, the positive terminal of the battery appears towards $a$.

Similarly, the rod $ob$ is equivalent to a battery of emf, $Bvl\sin \left(\theta /2\right)$ with the positive terminal towards $o$.

The equivalent circuit is shown in the figure.

Clearly, the net emf between the points $a$ and $b$ is $2Bvl\sin \left(\theta /2\right)$.