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MHT CET 2025 (20 Apr, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2025 (20 Apr, Shift 1)

Q23MHT CET 2025 · 20 APR · SHIFT 1Medium

Biot-Savart Law & Circular Coils

A long wire carrying a steady current is bent into a circle of single turn. The magnetic field at the centre of the coil is ' B '. If it is bent into a circular loop of radius ' $r_1$ ' having ' n ' turns, the magnetic field at the centre of the coil for same current is

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Answer: (c)

Same length: $2\pi R = n \cdot 2\pi r_1 \Rightarrow r_1 = R/n$ .
$B' = \frac{\mu_0 n I}{2r_1} = n^2 \frac{\mu_0 I}{2R} = n^2 B$
So the answer is (c) $n^2 B$ .

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