A) 0
B) \[\frac{2{{\mu }_{0}}}{\pi }\frac{I}{r}\]
C) \[-\frac{2{{\mu }_{0}}}{\pi }\frac{I}{r}\]
D) \[\frac{2{{\mu }_{0}}}{\pi }\frac{Ia}{{{r}^{2}}}\]
Correct Answer: A
Solution :
[a] From ampere circuital law \[\oint{B.dl={{\mu }_{0}}i}\] Where, i= Net current passing through the loop Now \[i=I+I-2I=0\] \[\therefore \,\,\,B.2\pi r=0\Rightarrow B=0\]You need to login to perform this action.
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