A wire of length l is in the form of a circular loop A of one turn. Now loop A is reshaped into loop B having three turns. The ratio of the magnetic fields at the centres of loop A and loop B for the same current through them is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1/9
The field at the centre of a circular coil is \( B = \frac{\mu_0 N I}{2r} \). Reshaping the same length of wire into more turns makes each turn smaller, so both N and r change.
\( \text{Loop A}: N = 1,\ l = 2\pi r_A \)
\( \text{Loop B}: N = 3,\ l = 3 \times 2\pi r_B \)
\( \displaystyle r_B = \frac{r_A}{3} \)
\( \displaystyle \frac{B_A}{B_B} = \frac{\mu_0 (1) I/2r_A}{\mu_0 (3) I/2r_B} \)
\( \displaystyle \frac{B_A}{B_B} = \frac{r_B}{3r_A} = \frac{1}{9} \)
So the field at the centre of B is 9 times that of A: three times the turns, each at one-third the radius.