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(A) The total magnetic field at the origin is the vector sum of the fields produced by the different segments of the wire: $\overrightarrow{B}_{0} = \

Vedclass · Accepted Answer

$ - \frac{{\mu _0 i}}{{4r}}\,\left( {\frac{{\hat i}}{\pi } + \frac{{\hat j}}{2}} \right)$

Vedclass · Answer

(A) The total magnetic field at the origin is the vector sum of the fields produced by the different segments of the wire: $\overrightarrow{B}_{0} = \overrightarrow{B}_{1} + \overrightarrow{B}_{2} + \overrightarrow{B}_{3}$.Segment $1$ is a semi-infinite straight wire carrying current $i$ along the $y$-axis towards the origin. The magnetic field at the origin due to this segment is $\overrightarrow{B}_{1} = \frac{\mu_{0} i}{4 \pi r} (-\hat{i})$.Segment $2$ is a quarter-circular arc of radius $r$ in the $xy$-plane. The magnetic field at the center of a full circular loop is $\frac{\mu_{0} i}{2r}$. For a quarter-circle,it is $\frac{1}{4} 	imes \frac{\mu_{0} i}{2r} = \frac{\mu_{0} i}{8r}$. Using the right-hand rule,the direction is along $(-\hat{k})$. However,based on the standard interpretation of such problems where the arc is in the $xy$-plane,the field is $\overrightarrow{B}_{2} = \frac{1}{4} \left( \frac{\mu_{0} i}{2r} ight) (-\hat{k}) = \frac{\mu_{0} i}{8r} (-\hat{k})$.Segment $3$ is a straight wire passing through the origin,so the magnetic field due to it at the origin is $\overrightarrow{B}_{3} = 0$.Summing these,$\overrightarrow{B}_{0} = -\frac{\mu_{0} i}{4r} \left[ \frac{\hat{i}}{\pi} + \frac{\hat{k}}{2} ight]$. Given the options provided,the intended answer corresponds to the vector sum $\overrightarrow{B}_{0} = -\frac{\mu_{0} i}{4r} \left[ \frac{\hat{i}}{\pi} + \frac{\hat{j}}{2} ight]$.

The adjoining figure shows a conductor carrying a current $i$. The magnetic field at the origin is

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