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Question

A conductor ABOCD moves along its bisector with a velocity of 1 ms−1 through a perpendicular magnetic field of 1 (Wb)m−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
0
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B
0.71 volt
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C
1.41 volt
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D
2.82 volt
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Solution

The correct option is C 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 as follows :
Induced emf between B and C = Induced emf between A and B is
Bv√2l=1×1×1×√2=1.41 volt.

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