Question

A conductor ABOCD moves along its bisector with a velocity of 1 ms${}^{-1}$ through a perpendicular magnetic field of 1 Wbm${}^{-2}$, as shown in the figure. If all the four sides are of 1 m length each, then the induced emf between point A and D is :

• A. zero
• B. 1.41 V
• C. 0.71 V
• D. none of the above

Answer 1

The correct option is B 1.41 V

Given: $B=1$ $Wb{m}^{-2}$; $v=1$ m/s

From figure, $x=1\times sin{45}^o=\frac{1}{\sqrt{2}}$

Thus, $l=2x=\sqrt{2}$

Induced emf between A and D: $E=Bvl$

$\therefore$ $E=1\times 1\times \sqrt{2}=1.41V$