question_answer

A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point 0 is:             

A) \[\overset{\to }{\mathop{B}}\,=-\frac{{{\mu }_{0}}}{4\pi }\frac{I}{R}\left( \mu \hat{i}\times 2\hat{k} \right)\]

B) \[\overset{\to }{\mathop{B}}\,=-\frac{{{\mu }_{0}}}{4\pi }\frac{I}{R}\left( \pi \hat{i}+2\hat{k} \right)\]

C) \[\overset{\to }{\mathop{B}}\,=\frac{{{\mu }_{0}}}{4\pi }\frac{I}{R}\left( \pi \hat{i}-2\hat{k} \right)\]

D) \[\overset{\to }{\mathop{B}}\,=\frac{{{\mu }_{0}}}{4\pi }\frac{I}{R}\left( \pi \hat{i}+2\hat{k} \right)\]

Correct Answer: B

Solution :

[b] Magnetic field due to segment '1' \[{{\overset{\to }{\mathop{B}}\,}_{1}}=\frac{{{\mu }_{0}}I}{4\pi R}[sin\,{{90}^{o}}+\sin \,{{0}^{o}}](-\hat{k})\] \[=\frac{-{{\mu }_{0}}I}{4\pi R}(\hat{k})={{\overset{\to }{\mathop{B}}\,}_{_{3}}}\] Magnetic field due to segment 2 \[{{B}_{2}}=\frac{{{\mu }_{0}}I}{4R}(-\hat{i})=\frac{-{{\mu }_{0}}I}{4\pi R}(\pi \hat{i})\] \[\therefore \,\,\,\overset{\to }{\mathop{B}}\,\] at centre \[{{\overset{\to }{\mathop{B}}\,}_{c}}={{\overset{\to }{\mathop{B}}\,}_{1}}+\overset{\to }{\mathop{{{B}_{2}}+}}\,\overset{\to }{\mathop{{{B}_{3}}}}\,\] \[=\frac{-{{\mu }_{0}}I}{4\pi R}(\pi \hat{i}+2\hat{k})\]