Q 12-04-200JEE MainJEE Main 2019 (9 Jan, Shift 2)Easy
One of the two identical conducting wires of length $L$ is bent in the form of a circular loop and the other one into a circular coil of $N$ identical turns. If the same current is passed in both, the ratio of the magnetic field at the centre of the loop $(B_L)$ to that at the centre of the coil $(B_C)$, i.e. $\dfrac{B_L}{B_C}$ will be
Answer: (A) $\dfrac{1}{N^2}$
Loop: radius $R = \dfrac{L}{2\pi}$, $B_L = \dfrac{\mu_0I}{2R}$.
Coil: radius $r = \dfrac{L}{2\pi N} = \dfrac RN$, $B_C = \dfrac{\mu_0NI}{2r} = N^2\dfrac{\mu_0I}{2R}$.
$$\frac{B_L}{B_C} = \frac{1}{N^2}$$
Solution by Sreeraj P, M.Sc Physics