Find the mutual inductance between the straight wire and the square loop of figure.

Explanation

We know `(d phi)/(dt)=E=Lxx(di)/(dt)`
From the question
(di)/(dt)=(d)/(dt) (i_(0) sin(omega) t=i_(0) (omega)cos (omega)t)`
(d phi)/(dt)=E=(mu_(0)ai_(0)(omega)cos(omega)t)/(2 pi)+In (1+(a)/(b))`
`Now, E =M.(d phi)/(dt)`
`Now, E=M.(dphi)/(dt)`
`or (mu_(0)ai_(0)(omega)cos(omega)t)/(2 pi) In (1+(a)/(b))`
`Mxxi_(0) (omega)cos (omega)t`
`impliesM=(mu_(0)a)/(2 pi) In (1+(a)/(b))`.