Question

A conductor $ABOCD$ moves along its bisector with a velocity $1\text{m/s}$ through a perpendicular magnetic field of $1{\text{Wb/m}}^2$, as shown in figure. If all the four sides are $1\text{m}$ length each, then the induced emf between A and D (in $\text{Volt}$) is approximately


(upto two decimals)

Answer 1

Induced emf between $A$ and $D$
$V_A-V_D=V\times B\times L...\left(1\right)$
From triangle $BOC$
$\frac{L}{\sin {90}^{\circ }}=\frac{1}{\sin {45}^{\circ }}$
$L=\sqrt{2}\text{m}$
From equation (1),
$V_A-V_D=1\times 1\times \sqrt{2}=1.41\text{Volt}$