Two circular coils made from same wire but radius of \(1^{\text {st }}\) coil is twice that of \(2^{\text {nd }}\) coil. If magnetic field at their centres is same then ratio of potential difference applied across them is \(\left(1^{\text {st }}\right.\) to \(2^{\text {nd }}\) coil)

Magnetic fields at the centre of the coils are equal.
\(\therefore \frac{\mu_0 I_1}{2 r_1}=\frac{\mu_0 I_2}{2 r_2}\)
\(\therefore \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{2 \mathrm{r}}{\mathrm{r}}\)....(given \(r_1=2 r_2\) )
\(\therefore \frac{\mathrm{I}_1}{\mathrm{I}_2}=2\)...(ii)
The resistance through the coil, \(\mathrm{R} \propto l\)
Here, \(l=2 \pi \mathrm{r}\)
\(\therefore \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{2 \pi \mathrm{r}_1}{2 \pi \mathrm{r}_2}=2\)...(ii)
\(\therefore \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_2 \mathrm{R}_2}=2 \times 2=4\)
[From (i) and (ii)]