A) \[\frac{5{{\mu }_{0}}I\theta }{24\,\pi r}\]
B) \[\frac{{{\mu }_{0}}I\theta }{24\,\pi r}\]
C) \[\frac{11{{\mu }_{0}}I\theta }{24\,\pi r}\]
D) zero
Correct Answer: A
Solution :
Since magnetic field at the centre of an arc is equal to \[B=\frac{{{\mu }_{0}}I}{4\pi r}\theta \] Hence, net \[B=\frac{{{\mu }_{0}}I}{4\pi }\left( \frac{1}{r}-\frac{1}{2r}+\frac{1}{3r} \right)\theta =\frac{5{{\mu }_{0}}I\theta }{24\pi r}\] (As, field due to straight part is being zero).
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