Question

Find the magnetic induction at the point $O$ if the wire carrying a current $I=8.0$ A has the shaped shown in fig. The radius curved part of the wire is $R=100mm$, the linear parts of the wire are very long.

Answer 1

(a) ${\overrightarrow{B}}_0={\overrightarrow{B}}_1+{\overrightarrow{B}}_2+{\overrightarrow{B}}_3$
$=\frac{{\mu }_0}{4\pi }\frac{i}{R}(-\overrightarrow{k})+\frac{{\mu }_0}{4\pi }\frac{i}{R}\pi (-\overrightarrow{i})+\frac{{\mu }_0}{4\pi }\frac{i}{R}(-\overrightarrow{k})$
$=-\frac{{\mu }_0}{4\pi }\frac{i}{R}[2\overrightarrow{k}+\pi \overrightarrow{i}]$

So, $|{\overrightarrow{B}}_0|=\frac{{\mu }_0}{4\pi }\frac{i}{R}\sqrt{{\pi }^2+4}=0.30\mu T$

(b) ${\overrightarrow{B}}_0+{\overrightarrow{B}}_1+{\overrightarrow{B}}_2+{\overrightarrow{B}}_3$
$=\frac{{\mu }_0}{4\pi }\frac{i}{R}(-\overrightarrow{k})+\frac{{\mu }_0}{4\pi }\frac{i}{R}\pi (-\overrightarrow{i})+\frac{{\mu }_0}{4\pi }\frac{i}{R}(-\overrightarrow{i})$
$=-\frac{{\mu }_0}{4\pi }\frac{i}{R}[\overrightarrow{k}-(\pi +1\left)\overrightarrow{i}\right]$
So,
$|{\overrightarrow{B}}_0|=\frac{{\mu }_0}{4}\frac{i}{R}\sqrt{1}+(\pi +1{)}^2=0.34\mu T$

(c) Here using the law of parallel resistances
$i_1+i_2=i$ and $\frac{i_1}{i_2}=\frac{1}{3}$,
So, $\frac{i_1+i_2}{i_2}=\frac{4}{3}$

Hence $i_2=\frac{3}{4}i$, and $i_1=\frac{1}{4}i$
Thus ${\overrightarrow{B}}_0=\frac{{\mu }_0}{4\pi }\frac{i}{R}(-\overrightarrow{k})+\frac{{\mu }_0}{4}\frac{i}{R}(-\overrightarrow{j})+\left[\frac{{\mu }_0}{4}\left(\frac{3\pi }{2}\right)\frac{i_1}{R}\right(-\overrightarrow{i})+\frac{{\mu }_0}{4\pi }\frac{\left(\pi /2\right)i_2}{R}\overrightarrow{i}]$
$=\frac{-{\mu }_0}{4\pi }\frac{i}{R}(\overrightarrow{j}+\overrightarrow{k})+0$
Thus, $|{\overrightarrow{B}}_0|=\frac{{\mu }_0}{4\pi }\frac{\sqrt{2}i}{R}=0.11\mu T$