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Q.

A conductor ABOCD moves along its bisector with a velocity of 1 m/s through a perpendicular magnetic field of 1 Wb/m2, as shown in fig. If all the four sides are of 1 m length each, then the induced emf between points A and D is

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a

0

b

1.41 volt

c

0.71 volt

d

None of these

answer is B.

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Detailed Solution

There is no induced e.m.f. in the part AB and CD because they are moving along their length while e.m.f. induced between B and C, i.e., between A and D can be calculated as follows:Induced e.m.f. between B and C = Induced e.m.f. between A and D=Bv(2l)=1×1×1×2=1.41 volts
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