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NEET 2016 (Phase 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from NEET 2016 (Phase 2)

Q93NEET 2016 · PHASE 2Medium

Biot-Savart Law & Circular Coils

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

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Answer: (b)

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

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