The electric current in a circular coil of 2 turns produces a...

Magnetic field on the centre of a circular current coil is given by $B = \dfrac{N{\mu}_{0}i}{2R}$, here, $N$ is number of turns, $i$ is current and $R$ is radius of loop.

Now, for, $N = 2$, magnetic field, ${B}_{1} = \dfrac{{N}_{1}{\mu}_{0}i}{2{R}_{1}}$

For ${N}_{2} = 5$

Radius of coil $= {R}_{2} = \dfrac{{N}_{1} \times {R}_{1}}{{N}_{2}}$

Magnetic field, ${B}_{2} = \dfrac{{N}_{2}{\mu}_{0}i}{2{R}_{2}}$

Thus, the ratio is $\dfrac{{B}_{2}}{{B}_{1}} = \dfrac{{N}_{2}}{{N}_{1}}\dfrac{{R}_{1}}{{R}_{2}} = \dfrac{{N}_{2}}{{N}_{1}} \times \dfrac{{N}_{2}}{{N}_{1}}$

$\dfrac{{B}_{2}}{{B}_{1}} = \dfrac{25}{4}$