In the figure shown, the magnetic field induction as the point $\mathrm{O}$ will be

• Correct answer
  A

  $\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$
• B

  $\left(\frac{\mu_{0}}{4 \pi}\right)\left(\frac{i}{r}\right)(\pi+2)$
• C

  $\left(\frac{\mu_{0}}{4 \pi}\right)\left(\frac{\mathrm{i}}{\mathrm{r}}\right)(\pi+1)$
• D

  $\frac{\mu_{0}}{4 \pi r} \frac{i}{r}(\pi-2)$

Solution

Field due to a straight wire of infinite length is $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$ ifthe point is on a line perpendicular to its length while at the centre of a semicurcular coil is $\frac{\mu_{0} \pi i}{4 \pi r}$
$$ \begin{array}{l} \therefore \mathrm{B}=\mathrm{B}_{\mathrm{a}}+\mathrm{B}_{\mathrm{b}}+\mathrm{B}_{\mathrm{c}} \\ =\frac{\mu_{0}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}+\frac{\mu_{0}}{4 \pi} \frac{\pi \mathrm{i}}{\mathrm{r}}+\frac{\mu_{0}}{4 \pi \mathrm{r}} \frac{\mathrm{i}}{\mathrm{r}} \\ =\frac{\mu_{0}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}(\pi+2) \text { out of the phase } \end{array} $$

Hence, the correct answer is option: 'A'.