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MHT CET 2026 (17 Apr, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2026 (17 Apr, Shift 1)

Q9MHT CET 2026 · 17 APR · SHIFT 1Medium

Biot-Savart Law & Circular Coils

The magnetic field at the center of circular coil carrying current ' $I$ ' for a single turn of a given length of wire is ' $B$ '. The same wire is bent in a circular coil having two turns. When the same current passes through it, the value of magnetic field becomes

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Answer: (a)

Fixed wire length $L$ : $n$ turns gives $r = \frac{L}{2\pi n}$ , so $B = \frac{\mu_0 n I}{2r} \propto n^2$ .
$n$ goes from 1 to 2, so $B' = 2^2 B = 4B$ .
So the answer is (a) $4B$ .

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