JEE Main · 2013 · 07 Apr 2013 · Offline · Electromagnetic Induction

Rather than wrestle with the small loop's field spread over the big loop, lean on the symmetry of mutual inductance: the flux through the large loop from a current $i$ in the small one is the same as what the large loop's field would send through the small one. Treating the small loop as lying on the big loop's axis a distance $d$ away, $\varphi =\frac{{\mu }_{0}i\pi r^2{R}^2}{2(d^2+R^2{)}^{3/2}}$. Putting in $r=0.003\text{m}$, $R=0.20\text{m}$, $d=0.15\text{m}$ and $i=2.0\text{A}$ gives a flux of about $9.1\times 10^{-11}\text{Wb}$.