Explanation

(b) For a current flowing into a circular arc, magnetic induction at the centre
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$B = \frac{\mu_{0}i}{4\pi}\ \int_{}^{}{\frac{dl \times r}{r^{3}} = \frac{\mu_{0}}{4\pi}\ \int_{}^{}{\frac{r^{2}d\theta}{r^{3}} = (}}\frac{\mu_{0}i}{4\pi})\theta$
The total current is divided into two arcs
$i_{1} = \frac{E}{R_{1}} = \frac{E}{(R/2\pi r)l_{1}} = \frac{E}{\left( (R/2\pi r) \right)(r\ \theta)} = \frac{2\pi E}{R\theta}$
$i_{1}\theta = \frac{2\pi E}{R} = costant$
Similarly, $i_{2} = \frac{E}{R_{2}} = \frac{E}{(R/2\pi r)l_{2}}$
= $\frac{E}{\left( (R/2\pi r) \right)\{ r(2\pi - \theta)\}}$ = $\frac{2\pi E}{R(\pi - \theta)} =$constant
$B = B_{1} - B_{2} = \frac{\mu_{0}}{4\pi r}\ \left( \frac{2\pi E}{R} - \frac{2\pi E}{R} \right) = 0$