Q 12-04-112JEE MainJEE Main 2023 (31 Jan, Shift 2)Medium
A long conducting wire having a current $I$ flowing through it, is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is
Answer: (C) $N^2:n^2$
For a wire of length $l$: $2\pi rN=l\Rightarrow r=\dfrac{l}{2\pi N}$.
$$B=\frac{\mu_0NI}{2r}=\frac{\pi\mu_0IN^2}{l}\propto N^2$$
So $B_1:B_2=N^2:n^2$.
Solution by Sreeraj P, M.Sc Physics