Two circular coils 1 and 2 are made from the same wire but the radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them, so that the magnetic field at their centres is same?

Magnetic field at the center of a circular coil is
\(B=\frac{\mu_0}{4 \pi} \times \frac{2 \pi i}{r}\)
Where is current flowing in the coil and \(r\) is radius of coil.
At the centre of coil 1 ,
\(B_1=\frac{\mu_0}{2} \frac{i_1}{r_1}\)
\(B_2=\frac{\mu_0}{2} \frac{i_2}{r_2}\)
Since, \(B_1=B_2\)
\(\begin{aligned} & \therefore \frac{\mu_0}{2} \frac{i_1}{r_1}=\frac{\mu_0}{2} \frac{i_2}{r_2} \\ & \Rightarrow \frac{i_1}{r_1}=\frac{i_2}{r_2} \\ & \because r_1=2 r_2 \\ & \therefore \frac{i_1}{2 r_2}=\frac{i_2}{r_2} \\ & \Rightarrow i_1=2 i_2\end{aligned}\)
According to Ohm's Law: the potential difference across the loop is proportional to the product of length of the wire of the loop and current flowing through it.
Now, the ratio of the potential differences can be expressed as
\(\frac{V_2}{\mathrm{~V}}=\frac{i_2 \times r_2}{\boldsymbol{i} \times \boldsymbol{r}}=\frac{i_2 \times r_2}{\boldsymbol{i} \times \mathrm{Tr}_r}=\frac{1}{\boldsymbol{\Lambda}}\)