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`1.5A``1.0A``0.5A``2.0A`

Answer :

CSolution :

Induced emf's in the two loops will oppose each other. So net induced emf: <br> `e=(d)/(dt)[(b^(2)-a^(2))B]=(b^(2)-a^(2))(dB)/(dt)` <br> `=(0.20^(2)-0.10^(2)) (omega B_(0)cos (omega)t` <br> `=(0.04-0.01)xx100xx10xx10^(-3) cos (omega)t` <br> `=30xx10^(-3) cos (omega)t` <br> Resistance : `R=50xx10^(-3)[4xx20+4xx10]=60xx10^(-3) (Omega)` <br> `I=e/R 0.5(omega)t` <br> so aamplitude of current `=0.5A`.