A) \[{{\mu }_{0}}/17\]
B) \[\sqrt{3}{{\mu }_{0}}/2\pi \]
C) \[{{\mu }_{0}}/2\pi \]
D) \[3{{\mu }_{0}}/2\pi \]
Correct Answer: C
Solution :
[c] \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{i}_{2}}}{(r/2)}-\frac{{{\mu }_{0}}}{4\mu }\frac{2{{i}_{1}}}{(r/2)}=\frac{{{\mu }_{0}}}{4\pi }\frac{4}{r}({{i}_{2}}-{{i}_{1}})\] \[=\frac{{{\mu }_{0}}}{4\pi }\frac{4}{5}(5-2.5)=\frac{{{\mu }_{0}}}{2\pi }.\]
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