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

Question

Dr. Abubacker · Accepted Answer

[Solution] Radius becomes r/n with n turns: $B' = \frac{\mu_0 nI}{2(r/n)} = n^2B$. Answer (2).

Q. 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

Explanation

Reviewed by Dr. Abubacker

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