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JEE Main 2004 (AIEEE) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2004 (AIEEE)

Q107JEE Main 2004 · AIEEEMedium

Biot-Savart Law & Circular Coils

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

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Answer: (b)

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

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