Question

The magnetic induction of the field at the point $O$ of a loop with current $I$, whose shape is illustrated in figure above is given as $B=\frac{{\mu }_0}{4\pi }i[\frac{x\pi }{2a}+\frac{\sqrt{2}}{b}]$. The radius $a$ and the side $b$ are known. Find the value of $x$.

Answer 1

Here, $B_1=\frac{{\mu }_0}{4\pi }\frac{i}{a}\frac{3\pi }{2}$, $\overrightarrow{B_2}=0$,

$B_3=\frac{{\mu }_0}{4\pi }\frac{i}{b}\sin {45}^{\circ }$,
$B_4=\frac{{\mu }_0}{4\pi }\frac{i}{b}\sin {45}^{\circ }$,
and $B_5=0$
So, $B_0=B_1+B_2+B_3+B_4+B_5$
$=\frac{{\mu }_0}{4\pi }\frac{i}{a}\frac{3\pi }{2}+0+\frac{{\mu }_0}{4\pi }\frac{i}{b}\sin {45}^{\circ }+\frac{{\mu }_0}{4\pi }\frac{i}{b}\sin {45}^{\circ }+0$
$=\frac{{\mu }_0}{4\pi }i[\frac{3\pi }{2a}+\frac{\sqrt{2}}{b}]$