Question

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

• A. $0$
• B. $0.71volt$
• C. $1.41volt$
• D. $2.82volt$

Answer 1

The correct option is C $1.41volt$

There is no induced emf in the part AB and CD because they are moving along their length while emf induced between B and C i.e. between A and D can be calculated as follows :
Induced emf between B and C = Induced emf between A and B is
$Bv\sqrt{2}l=1\times 1\times 1\times \sqrt{2}=1.41volt$.