Question

A conductor ABOCD moves along its bisector with a velocity of 1 m/s through a perpendicular magnetic field of 1 $wb/m^2$, as shown in fig. If all the four sides are of 1m length each, then the induced emf between points A and D is

• A. \N
• B. 1.41 volt
• C. 0.71 volt ​
• D. None of the above

Answer 1

The correct option is B 1.41 volt

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


Induced emf between B and C = Induced emf between $AandB=Bv\left(\sqrt{2}l\right)=1\times 1\times 1\times \sqrt{2}=1.41volt$