A uniform circular wire loop is connected to the terminals of a battery. The magnetic field induction at the centre due to $A B C$ portion of the wire will be (length of $A B C=l_1$, length of $A D C=l_2$ )
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(a)
Let current in part $A B C$ is $i_1$ and in part $A D C$ is $i_2$
$i=\frac{i l_2}{l_1+l_2}$ (As $A B C$ and $A D C$ part are in parallel connection) and subtended by $A B C$ at centre $O$ will be $=\left(\frac{2 \pi}{l_1+l_2}\right)\left(l_1\right)$
so using $B=\frac{\mu_0 i}{2 a}\left(\frac{\theta}{2 \pi}\right)$
$B=\frac{\mu_0}{2 R}\left(\frac{i l_2}{l_1+l_2}\right) \frac{2 \pi}{\left(l_1+l_2\right)} \frac{\left(l_1\right)}{2 \pi}$
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