A long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having n turns. If the same current passes in both the cases, then the ratio of the magnetic fields produced at the centers in two cases will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $1:n^2$
Magnetic field at the center of a circular coil:
$B = \frac{\mu_0 N I}{2 R}$
Case 1: one turn, radius $R$: $B_1 = \frac{\mu_0 I}{2 R}$
Case 2: $n$ turns, smaller radius $r$: total wire length same: $2 \pi R = n \cdot 2 \pi r \text{ implies } r = \frac{R}{n}$
Magnetic field at center of $n$ turns:
$B_2 = \frac{\mu_0 n I}{2 r} = \frac{\mu_0 n I}{2 (R/n)} = \frac{\mu_0 n^2 I}{2 R}$
Ratio:
$\frac{B_1}{B_2} = \frac{\mu_0 I / 2 R}{\mu_0 n^2 I / 2 R} = \frac{1}{n^2}$
Final Answer: $B_1 : B_2 = 1 : n^2$