Question

Find the magnetic induction of the field at the point $O$ of a loop with current $I$, whose shaped is illustrated
(a) in Fig. a, the radii $a$ and $b$, as well as the angle $\phi$ are known;
(b) in Fig. b, the radius $a$ and the side $b$ are known.

Answer 1

(a) From the Biot-Savart law,

$dB=\frac{{\mu }_0}{4\pi }i\frac{(d\overrightarrow{l}\times \overrightarrow{r})}{r^3}$

So, magnetic field induction due to the segment $1$ at $O$,

$B_1=\frac{{\mu }_0}{4\pi }\frac{i}{a}(2\pi -\phi )$

also $B_2=B_4=0$, as $d\overrightarrow{l}\uparrow \uparrow \overrightarrow{r}$

and $B_3=\frac{{\mu }_0}{4\pi }\frac{i}{b}\phi$

Hence, $B_0=B_1+B_2+B_3+B_4$

$=\frac{{\mu }_0}{4\pi }i[\frac{2\pi -\phi }{a}+\frac{\phi }{b}]$,

(b) Here, $B_1=\frac{{\mu }_0}{4\pi }\frac{i}{a}\frac{3\pi }{a},{\overrightarrow{B}}_2=0$,

$B_3=\frac{{\mu }_0}{4\pi }\frac{i}{b}\sin {45}^{\circ }$,

$B_4=\frac{{\mu }_0}{4\pi }\frac{i}{b}\sin {45}^{\circ }$,

and $B_5=0$

So, $B_0=B_1+B_2+B_3+B_4+B_5$

$=\frac{{\mu }_0}{4\pi }\frac{i}{a}\frac{3\pi }{2}+0+\frac{{\mu }_0}{4\pi }\frac{i}{b}\sin {45}^{\circ }+\frac{{\mu }_0}{4\pi }\frac{i}{b}\sin {45}^{\circ }+0$

$=\frac{{\mu }_0}{4\pi }i[\frac{3\pi }{2a}+\frac{\sqrt{2}}{b}]$.