An infinitely long straight wire carrying current I is ... | null (Physics)

JEE Mains 2026 21st January Shift 2

Physics

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 :

\\

\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{\mathrm{r}}(\pi+1) \hat{\mathrm{i}}

-\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{\mathrm{r}}(\pi-1) \hat{\mathrm{i}}

\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{\mathrm{r}}(\pi-1) \hat{\mathrm{i}}

-\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{\mathrm{r}}(\pi+1) \hat{\mathrm{i}}

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Correct Answer:

Explanation

\\ \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}}

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