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 of $1\text{m}$ length each, then the induced emf between A and D (in $\text{volts}$) 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 $BOC$,
$L=\frac{1}{\cos {45}^{\circ }}=\sqrt{2}\text{m}$
From equation (1),
$V_A-V_D=1\times 1\times \sqrt{2}=1.41\text{volts}$