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) → \( 1 : n^2 \)
The field at the centre of a circular coil of \( N \) turns and radius \( r \) is \( B = \mu_0 N I/(2r) \) (Biot-Savart law for a circular loop). The same wire of length \( L \) is used both times, so the radius depends on the number of turns.
\( \displaystyle L = 2\pi r_1 = n\,(2\pi r_2) \;\Rightarrow\; r_2 = \frac{r_1}{n} \)
\( \displaystyle B_1 = \frac{\mu_0 I}{2 r_1} \qquad B_2 = \frac{\mu_0 n I}{2 r_2} = \frac{\mu_0 n^2 I}{2 r_1} \)
\( \displaystyle \frac{B_1}{B_2} = \frac{1}{n^2} \)
So the ratio is \( 1 : n^2 \).