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

Options

A. `0`
B. `1.41` volt
C. `0.71` volt
D. none of these

Answer

B

Explanation

There is no induced e.m.f. in the part `AB` and k`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 calculate as follows
induced e.m.f. between `B` and `C`= induced e.m.f.
betwween
`A and B=Bv(sqrt(2)l)=1xx1xx1xxsqrt(2)=1.41`volt.