Question

A thin conducting wire in the shape of a 'figure of eight' is situated with its circular loops in two planes making an angle of ${120}^{\circ }$ with each other, if the current in the loop is $I$ and the radius is $R$ the magnetic induction at a point of intersection $P$ to axes in the loops is $\frac{N{\mu }_{0}i}{48R}$. Find $N$.

Answer 1

Let $P$ be point of intersection of axes of circular loops, $A,B$ be centre of the loops, $C$ be point of intersection of loops





As angle between plane of loops is ${120}^{\circ },\angle APB={60}^{\circ },\angle APC={30}^{\circ }$

From right angled triangle $\Delta APC,$
$\tan {30}^{\circ }=\frac{AC}{PA}PA=\sqrt{3}R$
From symmetry $PA=PB=\sqrt{3}R$
Magnitude of magnetic field due to each loop is $B_1=\frac{\mu I}{2}\frac{R^2}{(R^2+(\sqrt{3}R{)}^2{)}^{3/2}}=\frac{\mu I}{16R}$
As magnetic field due to each loop is along the axes, hence angle between magnetic field is ${120}^{\circ }$

Net Magnetic field at point $P$ is
$B=\sqrt{B_1^2+B_1^2+2B_1{B}_1\cos {120}^{\circ }}$
$\Rightarrow B=B_1=\frac{\mu I}{16R}=\frac{3{\mu }_{0}I}{48R}$