JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Magnetic Effects of Current

A fixed length of wire $L$ is wound first into $N$ turns and then into $n$ turns. For $N$ turns the radius obeys $L=N(2\pi r_1)$, and the field at the centre is $B_1=\frac{{\mu }_{0}NI}{2r_1}$. Replacing $r_1=\frac{L}{2\pi N}$ gives $B_1=\frac{{\mu }_0\pi N^{2}I}{L}$, and in the same way $B_2=\frac{{\mu }_0\pi n^{2}I}{L}$ for the $n$-turn coil. The central field thus scales as the square of the number of turns, and the ratio of the first case to the second is $N^2:n^2$.