An infinitely long straight conductor is bent into the shape as shown below. It carries a current of

Question

An infinitely long straight conductor is bent into the shape as shown below. It carries a current of i ampere and the radius of the circular loop is $r$ metre. Then the magnetic induction at the centre of the circular part is\

Professor PC Thomas Classes · Answer

$\mathrm{B}=\mathrm{B}_{1}+\mathrm{B}_{2}$ where\ $B_{1}=$ field due to straight portion $=\frac{\mu_{0} I}{2 \pi r}$\ (away from the plane of the paper)\ $B_{2}=$ field due to circle $==\frac{\mu_{0} I}{2 r}$\ (into the plane of the paper)\ $\mathrm{B}_{2}>\mathrm{B}_{1}$. So take $\mathrm{B}_{2}+$ ve and $\mathrm{B}_{1}-\mathrm{ve}$\ $	herefore B=\frac{\mu_{0} I}{2 r}-\frac{\mu_{0} I}{2 \pi r}=\frac{\mu_{0} I}{2 r}\left[1-\frac{1}{\pi}ight]$\ (into the plane of paper)\ Rearrange to get the answer\ Look at this figure and note the difference\ $\mathrm{B}=\mathrm{B}_{1}+\mathrm{B}_{2}$\ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ (away from the plane of the paper)\ $B_{2}=\frac{\mu_{0} I}{2 \pi r}+\frac{\mu_{0} I}{2 r}$