Question

A conducting wire $ABC$ has been bent with its two arms $AB$ and $BC$ making ${60}^{\circ }$. Length of each arm is $l$. The bent wire is moving with a velocity $v$ along the bisector of the angle $ABC$ and there exists a uniform magnetic field $\left(B\right)$ in the region, directed perpendicular to plane of the wire. Find the emf between the two ends $A$ and $C$ of wire.

• A. $4Blv$
• B. $3Blv$
• C. $2Blv$
• D. $Blv$

Answer 1

The correct option is D $Blv$

Emf in the arm $BC$ and $AB$ is $E=Blv\sin {30}^{\circ }=\frac{Blv}{2}$



Since both arms have equal emfs and the resultant emf between $A$ and $C$ is = $Blv$