A wire carrying current $\text{I}$ has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to $\text{X}$ -axis while semicircular portion of radius $\text{R}$ is lying in $\text{Y}$ -$\text{Z}$ plane. Magnetic field at point $\text{O}$ is :

• A

  $\overrightarrow{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{I}}{\mathrm{R}}\mathrm{ }\left(\mathrm{\pi }\widehat{\mathrm{i}}+2\widehat{\mathrm{k}}\right)$
• B

  $\overrightarrow{\mathrm{B}}=-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{I}}{\mathrm{R}}\mathrm{ }\left(\mathrm{\pi }\widehat{\mathrm{i}}-2\widehat{\mathrm{k}}\right)$
• Correct answer
  C

  $\overrightarrow{\mathrm{B}}=-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{I}}{\mathrm{R}}\mathrm{ }\left(\mathrm{\pi }\widehat{\mathrm{i}}+2\widehat{\mathrm{k}}\right)$
• D

  $\overrightarrow{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{I}}{\mathrm{R}}\mathrm{ }\left(\mathrm{\pi }\widehat{\mathrm{i}}-2\widehat{\mathrm{k}}\right)$

Solution

${\overrightarrow{B}}_c = {\overrightarrow{B}}_1+{\overrightarrow{B}}_2+{\overrightarrow{B}}_3.$

$B_1 = \frac{{\mu }_0}{4\pi R} \left(\sin 90+sin\theta \right)-\widehat{k}$

$B_1 = \frac{{\mu }_0}{4\pi R} \left(-\widehat{k}\right)=B_3$

B due to segment 2

$B_2 = \frac{{\mu }_0 I}{4\pi }\times \pi (-\hat{i})=-\frac{{\mu }_{0}I}{4R}\hat{i}$

So B at center ${\overrightarrow{B}}_c= {\overrightarrow{B}}_1+ {\overrightarrow{B}}_2+{\overrightarrow{B}}_3$

$B_c= \frac{-{\mu }_{0}I}{4R} \left(i+\frac{2\widehat{k}}{\pi }\right)= \frac{-{\mu }_{0}I}{4\pi R} (\pi \widehat{i}+2\widehat{k})$

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