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MHT CET 2025 (21 Apr, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2025 (21 Apr, Shift 2)

Q22MHT CET 2025 · 21 APR · SHIFT 2Medium

Biot-Savart Law & Circular Coils

A coil having 9 turns carrying current produces magnetic field $B_1$ at the centre. Now that coil is rewounded into 3 turns carrying same current. Then magnetic field at the centre $B_2$ is

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Answer: (a)

For a fixed length of wire, $B = \frac{\mu_0 n I}{2r}$ with $r \propto \frac{1}{n}$ , so $B \propto n^2$ .
$\frac{B_2}{B_1} = \left(\frac{3}{9}\right)^2 = \frac{1}{9}$
So the answer is (a) $\frac{B_1}{9}$ .

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