Two concentric coplanar circular loops of radii r₁ and r₂ carry currents i₁ and i₂, respectively, in opposite directions (one clockwise and the other anticlockwise.) The magnetic induction at the centre of the loops is half of that due to i₁ alone at the centre. If r₂ = 2r₁ , the value of i₂/i₁ is

Magnetic field at the centre due to smaller loop

${\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{2{\mathrm{\pi i}}_1}{{\mathrm{r}}_1}$                 ….(i)

Due to bigger loop, ${\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{2{\mathrm{\pi i}}_2}{{\mathrm{r}}_2}$

So, net magnetic field at centre

$\mathrm{B}={\mathrm{B}}_1-{\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times 2\mathrm{\pi }\left(\frac{{\mathrm{i}}_1}{{\mathrm{r}}_1}-\frac{{\mathrm{i}}_2}{{\mathrm{r}}_2}\right)$

According to question, $\mathrm{B}=\frac{1}{2}\times {\mathrm{B}}_1$

$\begin{array}{l}\Rightarrow \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.2\mathrm{\pi }\left(\frac{{\mathrm{i}}_1}{{\mathrm{r}}_1}-\frac{{\mathrm{i}}_2}{{\mathrm{r}}_2}\right)=\frac{1}{2}\times \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{2{\mathrm{\pi i}}_1}{{\mathrm{r}}_1}\\ \frac{{\mathrm{i}}_1}{{\mathrm{r}}_1}-\frac{{\mathrm{i}}_2}{{\mathrm{r}}_2}=\frac{{\mathrm{i}}_1}{2{\mathrm{r}}_1}\text{\hspace{0.17em}⇒\hspace{0.17em}}\frac{{\mathrm{i}}_2}{{\mathrm{r}}_2}\Rightarrow \frac{{\mathrm{i}}_1}{2r_1}\text{ ⇒}\frac{i_1}{i_2}=1 \left\{{\mathrm{r}}_2=2{\mathrm{r}}_1\right\}\end{array}$