Let r₁ and r₂ are the radius of coil 1 and 2. If B₁ and B₂ are magnetic induction at their centre, then

${B_1} = {{{\mu _0}{I_1}} \over {2{r_1}}}$ and ${B_2} = {{{\mu _0}{I_2}} \over {2{r_2}}}$

Since B₁ = B₂ ; and r₁ = 2r₂ therefore I₁ = 2I₂.

Again if R₁ and R₂ are resistance of the coil 1 and 2 then R₁ = 2R₂ (as R $\propto$ length = 2$\pi$r) and if V₁ and V₂ are the potential difference across them respectively, then

${{{V_1}} \over {{V_2}}} = {{{I_1}{R_1}} \over {{I_2}{R_2}}} = {{\left( {2{I_2}} \right)\left( {2{R_2}} \right)} \over {{I_2}{R_2}}} = 4$