1. n²B 2. 2nB 3. 2n²B 4. nB

Answer: 1. n²B Field at the centre of a circular coil: B=μ₀ N I/2r . One turn: wire length L=2π r , so B=μ₀ I/2r . n turns from the same wire: 2π r'n=2π r⇒ r'=r/n . B'=μ₀ nI/2(r/n)=n²μ₀I/2r=n²B Option (1).

Topic: Biot-Savart law, field of circular loop· Magnetic Effects of Current and Magnetism

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

1. $n^2B$

2. $2nB$

3. $2n^2B$

4. $nB$

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Answer: 1. $n^2B$

Field at the centre of a circular coil: $B=\dfrac{\mu_0 N I}{2r}$.

One turn: wire length $L=2\pi r$, so $B=\dfrac{\mu_0 I}{2r}$.
$n$ turns from the same wire: $2\pi r'n=2\pi r\Rightarrow r'=r/n$.

$$B'=\frac{\mu_0 nI}{2(r/n)}=n^2\frac{\mu_0I}{2r}=n^2B$$

Option (1).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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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