Question

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic filed at the centre of the loop is B. It is them bent into circular coil of n turns. The magnetic field at the centre of this cold of n turns will be.

• A. $2n^{2}B$
• B. $nB$
• C. $n^{2}B$
• D. $2n=B$

Answer 1

Answer: (C)

$n^{2}B$

The circumference of the first loop is the length of the wire=$2\pi R$

The magnetic field at the center of this wire=$B=\frac{{\mu }_{0}I}{2R}$

The same wire is bent into n circular coils.

Thus new radius is found out by

$n\times 2\pi r=2\pi R$

$⟹r=\frac{R}{n}$

Thus magnetic field at the center of these loops=$n\frac{{\mu }_{0}I}{2r}$

$=\frac{n^2{\mu }_{0}I}{2R}=n^{2}B$