Question 22 of 276

An infinitely long straight wire carrying current $I$ is bent in a planer shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :

A$-\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi-1) \hat{i}$B$\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi-1) \hat{i}$C$-\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi+1) \hat{i}$D$\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi+1) \hat{i}$

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Answer: (A) $-\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi-1) \hat{i}$

$\overrightarrow{\mathrm{B}}_0=\overrightarrow{\mathrm{B}}_{\mathrm{AB}}+\overrightarrow{\mathrm{B}}_{\mathrm{DE}}+\overrightarrow{\mathrm{B}}_{\mathrm{BCD}}$
$=\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}} \hat{\mathrm{i}}+\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}} \hat{\mathrm{i}}-\frac{\mu_0 \mathrm{i}}{2 \mathrm{r}} \hat{\mathrm{i}}$
$=\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}} \hat{\mathrm{i}}-\frac{\mu_0 \mathrm{i}}{2 \mathrm{r}} \hat{\mathrm{i}}$
$=\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}(1-\pi) \hat{\mathrm{i}}$
$=-\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}(\pi-1) \hat{\mathrm{i}}$