Correct Answer - A
`B=B_(0)cos omegat`
Net flux `-BA_(2)-BA_(1)`
So, induced `emf`
`e=(A_(2)-A_(1))(dB)/(dt)`
So, `i=(A_(2)-A_(1))/R(dB)/(dt)=pi(r_(2)^(2)-r_(1)^(2))/R , B_(0)omega sin omega t`
So, `i=(pi.(400-100)xx10^(-4))/(2pixx10^(-1)xx30xx10^(-2))xx20xx10^(-3)xx100pi=pi` ampere