The wire loop shown in the figure lies in uniform magnetic induction `B=B_(0) cos omega t ` perpendicular to its plane.(Given `r_(1)=10 cm` and `r_(2)=20 cm` and `B_(0)=20 mT` and `omega=100 pi`).Find the amplitude of the current induced in the the loop if its resistance is `0.1 Omega//m`.

Answer

A

Explanation

`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