Question

A conductor ABOCD moves along its bisector with a velocity of 1 m/s through a perpendicular magnetic fie
of 1 wblm2, as shown in fig. If all the four sides are of 1m length each, then the induced emf betwee
points A and D is
BM
и А.
O
90°
C
D
0 0
1.41 volt
0 0.71 volt
None of the above​

Answer 1

Answer:

hey mate

Explanation:

refer to the attachment peez

Answer 2

There is no induced e.m.f. in the part AB and kCD because they are moving along their length while e.m.f. induced between B and C i.e., between A and D can be calculate as follows  

induced e.m.f. between B and C= induced e.m.f.  

betwween  

AandB=Bv(2–√l)=1×1×1×2–√=1.41volt.