A straight conductor carrying a direct current i amp. is split into circular loop as shown. Then the

Question

A straight conductor carrying a direct current i amp. is split into circular loop as shown. Then the magnetic induction at the centre of the circular loop of radius $r$ metre is\

Professor PC Thomas Classes · Answer

$\mathrm{B}=\mathrm{B}_{1}+\mathrm{B}_{2}+\mathrm{B}_{3}$\ $\mathrm{B}_{1}=$ field due to straight portion $=0$\ $B_{2}=$ field due to upper semicircle $=\frac{\mu_{0} I}{4 r}$\ $\mathrm{B}_{3}=$ field due to lower semicircle $=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}}$\ (away from the plane of the paper). If $\mathrm{B}_{2}$ is $+\mathrm{ve}, \mathrm{B}_{3}$ will be -ve\ $	herefore=0+\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}}+\left(\frac{-\mu_{0} \mathrm{I}}{4 \mathrm{r}}ight)=$