Question

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 centre of the coil will be:

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

Answer 1

Answer: (B)

$n^{2}B$

$\text{Let the radius of the circle formed of one turn be}R\text{and that of}n\text{turns be}r.$

$\text{Since the length of the wire remains same,}$

$2\pi R=n\left(2\pi r\right)$

$\Rightarrow r=\frac{R}{n}$

$\text{According to Ampere's Circuital Law,}$

${\mu }_{0}I=B\left(2\pi R\right)\text{applies to the loop with one turn.}$

$\text{For the loop with 'n' turns, current}nI\text{flows in the overlapping loop.}$

$\text{Thus,}$
${\mu }_0\left(nI\right)=B^{\prime }\left(2\pi r\right)=B^{\prime }\left(2\pi \frac{R}{n}\right)$

$\text{Therefore,}{B}^{\prime }=n^{2}B$