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MHT CET 2023 (14 May, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2023 (14 May, Shift 2)

Q53MHT CET 2023 · 14 MAY · 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 passes in both the cases, the ratio of magnetic fields produced at the centre for one turn to that of $n$ turns is

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Answer: (c)

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

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