question_answer

An infinitely long conductor as shown in figure carrying current I with a semiconductor loop on XY-plane and two straight parts, one parallel to X-axis and another coinciding with Z-axis. What is the magnetic field induction at the centre C of the semi-circular loop?

Answer:

\[B=\frac{{{\mu }_{0}}2I}{4\pi r}[(1+\pi )\hat{k}-\hat{i}]\] The magnetic field induction at C due to current through straight part of the conductor parallel to X-axis is \[{{B}_{1}}=\frac{{{\mu }_{0}}I}{4\pi \left( \frac{r}{2} \right)}\left[ \sin \frac{\pi }{2}+\sin 0{}^\circ  \right]\] \[=\frac{{{\mu }_{0}}2I}{4\pi \,\,r}\]acting along \[+\,\,Z\]-direction i.e.        \[{{B}_{1}}=\frac{{{\mu }_{0}}2I}{4\pi \,\,r}\hat{k}\] The magnetic field induction C due to the current through the semi-circular loop in XY-plane is \[{{B}_{2}}=\frac{{{\mu }_{0}}I}{4\pi \left( \frac{r}{2} \right)}(\pi )\] \[=\frac{{{\mu }_{0}}2I\,\,\pi }{4\pi r}\]acting along \[+\,\,Z\]-direction i.e.        \[{{B}_{2}}=\frac{{{\mu }_{0}}2\,\,\pi \,I}{4\pi \,\,r}\hat{k}\] The magnetic field induction at C due to current through the straight part of the conductor coinciding with Z-axis is             \[{{B}_{3}}=\frac{{{\mu }_{0}}I}{4\pi \frac{r}{2}}\left[ \sin \frac{\pi }{2}+\sin 0{}^\circ  \right]\] \[=\frac{{{\mu }_{0}}2I}{4\pi r}\]acting along negative X-axis i.e.        \[{{B}_{3}}=\frac{{{\mu }_{0}}2I}{4\pi r}(-\hat{i})\] Total magnetic field induction at C is given by \[B={{B}_{1}}+{{B}_{2}}+{{B}_{3}}=\frac{{{\mu }_{0}}2I\hat{k}}{4\pi r}+\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi I}{r}\hat{k}-\frac{{{\mu }_{0}}2I}{4\pi \,\,r}\hat{i}\]             \[=\frac{{{\mu }_{0}}2I}{4\pi r}[(1+\pi )\hat{k}-\hat{i}]\]