JEE Main 2019 — Magnetic Properties of Matter Question with Solution

Since the coils are co-axial, the  flux through the coils can be written as the product of their surface area and the magnetic field produced by the coil along its axis.

Consider the flux passing through Q

${\varphi }_Q=\frac{{\text{μ}}_0{i}_1{R}^2}{2{\left(R^2+x^2\right)}^{\frac{3}{2}}}\times \text{π}{R}^2={10}^{-3} \mathrm{Wb}$

And the flux passing through P:
${\varphi }_P=\frac{{\text{μ}}_0{i}_1{r}^2}{2{\left(R^2+x^2\right)}^{\frac{3}{2}}}\times \text{π}{R}^2$

Taking the ratio
$\frac{{\varphi }_P}{{\varphi }_Q}=\frac{i_2}{i_1}\times \frac{{\left(R^2+x^2\right)}^{\frac{3}{2}}}{{\left(R^2+x^2\right)}^{\frac{3}{2}}}\Rightarrow \frac{{\varphi }_P}{{10}^{-3}}=\frac{i_2}{i_1}$
$\Rightarrow \frac{{\varphi }_P}{{10}^{-3}}=\frac{2}{3}$
$\therefore {\varphi }_P=6.67\times {10}^{-4}\mathrm{W}\mathrm{b}$