Question

A long straight wire is coplanar with a current carrying circular loop of radius $R$ as shown in figure. Current flowing through wire and the loop is $I_0$ and $I_1$ respectively. the distance between center of loop and wire is $r=2R$. If force of attraction between the wire and the loop is $F={\mu }_0{I}_{0}I\left(\frac{X-\sqrt{3}}{\sqrt{3}}\right)$. Find X?

Answer 1

Consider two equal elemental arc lengths $MN$ and $M^{\prime }{N}^{\prime }$ of lengths $Rd\theta$. The distance of each elemental length from straight wire is:
$x=(r-R\cos \theta )$
$=R(2-\cos \theta )........\left(i\right)$

The magnetic induction at these elemental arc lengths due to current: $I_0$ in the wire is given by
$B=\frac{{\mu }_0{I}_0}{2\pi x}=\frac{{\mu }_0{I}_0}{2\pi R(2-\cos \theta )}.........\left(ii\right)$

The magnitude of the force on each elemental length:
$d{F}^{\prime }=BI\left(dl\right)$

$=BI\left(Rd\theta \right)$
$=\frac{{\mu }_0{I}_{0}IRd\theta }{2\pi R(2-\cos \theta )}=\frac{{\mu }_0{I}_{0}Id\theta }{2\pi (2-\cos \theta )}$

The direction of $d{F}^{\prime }$ is shown in figure.
The resultant force $dF$ is given by:
$dF=d{F}^{\prime }\cdot 2\cos \theta$

$=\frac{{\mu }_0{I}_{0}Id\theta }{2\pi (2-\cos \theta )}\times 2\cos \theta$
$=\frac{{\mu }_0{I}_{0}I}{\pi }\left[\frac{\cos \theta }{2-\cos \theta }\right]$

$F=\frac{{\mu }_0{I}_{0}I}{\pi }{\int }_0^{\pi }\left(\frac{\cos \theta }{2-\cos \theta }\right)d\theta ={\mu }_0{I}_{0}I\left(\frac{2-\sqrt{3}}{\sqrt{3}}\right)$