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JEE Main 2023 (31 Jan, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2023 (31 Jan, Shift 2)

Q55JEE Main 2023 · 31 JAN · SHIFT 2Medium

Biot-Savart Law & Circular Coils

A long conducting wire having a current I flowing through it, is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is:

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Answer: (a)

Same wire length: $r \propto \frac{1}{N}$ , so $B = \frac{\mu_0 N I}{2r} \propto N^2$ .
$\frac{B_1}{B_2} = \frac{N^2}{n^2}$
So the answer is (a) $N^2 : n^2$ .

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