Question

An angle $\angle AOB=\theta$ made of a conducting wire of length $l$, moves with speed $u$ along its angle bisector through a magnetic field $B$ as shown. Find the induced emf between the two free ends if the magnetic field $B$ is perpendicular to the plane.

• A. $2Bul\sin \theta$
• B. $\frac{Bul\sin \theta }{2}$
• C. $\frac{Bul\sin \left(\theta /2\right)}{2}$
• D. $Bul\sin \left(\theta /2\right)$

Answer 1

The correct option is D $Bul\sin \left(\theta /2\right)$

Emf induced in the conductor,

$E=Bu{l}^{\prime }$,

Where, $l^{\prime }=\text{effective length}$ i.e, length of the line joining the end points of the conductor.

From diagram, it is clear that

$l^{\prime }=2\times \frac{l}{2}\sin \left(\theta /2\right)$

$\therefore E=Bu\times \left(l\sin \theta /2\right)=Bul\sin \left(\theta /2\right)$

Hence, option $\left(D\right)$ is correct.