Current $I$ is flowing in conductor shaped as shown in the figure. The radius of the curved part is $r$ and the length of straight portion is very large. The value of the magnetic field at the centre $O$ will be

• Correct answer
  A

  $\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}\left(\frac{3 \pi}{2}+1\right)$
• B

  $\frac{\mu_{0} I}{4 \pi r}\left(\frac{3 \pi}{2}-1\right)$
• C

  $\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}\left(\frac{\pi}{2}+1\right)$
• D

  $\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}\left(\frac{\pi}{2}-1\right)$

Solution

$\mathrm{B}_{\mathrm{A}}=0$

$$ \begin{aligned} \mathrm{B}_{\mathrm{B}} &=\frac{\mu_{0}}{4 \pi} \frac{(2 \pi}{} \\ &=\frac{\mu_{0}}{4 \pi} \frac{3 \pi \mathrm{I}}{2 \mathrm{r}} \\ \mathrm{B}_{\mathrm{C}} &=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}} \otimes \end{aligned} $$ So, net magnetic field at the centre $$ \begin{aligned} &=\mathrm{B}_{\mathrm{A}}+\mathrm{B}_{\mathrm{B}}+\mathrm{B}_{\mathrm{C}} \\ &=0+\frac{\mu_{0}}{4 \pi} \frac{3 \pi \mathrm{I}}{2 \mathrm{r}}+\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}}\left(\frac{3 \pi}{2}+1\right) \end{aligned} $$

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