Question

A steady current flows in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is $ B $ . If the same wire is bent into a circular loop of $ n $ turns, the magnetic field at the centre of the coil is

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

Answer 1

The Correct Option is D

The magnetic field a circular coil (for $n$ number of loops) $B=\frac{\mu_{0}\, n i}{2 r}$ When,$n_{1}=1, n_{2}=2 $ $2 \pi r_{1}=n \times 2 \pi r_{2}$ $\Rightarrow \frac{r_{1}}{r_{2}}=\frac{n}{1} $ So $ \frac{B_{1}}{B_{2}}=\frac{\mu_{0} \,n_{1} i / 2 r_{1}}{\mu_{0} \,n_{2} i / 2 r_{2}}$ $\frac{B}{B_{2}}=\frac{n_{1}}{n_{2}} \cdot \frac{r_{1}}{r_{2}}$ $\frac{B}{B_{2}}=\frac{1}{n} \cdot \frac{1}{n}$ $\Rightarrow B_{2}=n^{2} B$