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MHT CET 2026 (15 Apr, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2026 (15 Apr, Shift 2)

Q10MHT CET 2026 · 15 APR · SHIFT 2Medium

Biot-Savart Law & Circular Coils

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 is passed in both the cases, the ratio of magnetic field produced at the centre for one turn to that of n turns is

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Answer: (c)

Same length: $2\pi R = n(2\pi r) \Rightarrow r = R/n$ .
$B_n = \frac{\mu_0 n I}{2r} = n^2 \frac{\mu_0 I}{2R} = n^2 B_1$
So the answer is (c) $1 : n^2$ .

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