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 :
Find B due to each arc at 0 and add them using vector addition. Since magnetic field at the centre of an arc is equal to \[B=\frac{{{\mu }_{0}}1}{4\pi r}\,\theta \] Hence, net B = \[B=\frac{{{\mu }_{0}}I}{4\pi }\,\,\left[ \frac{1}{r}-\frac{1}{2r}+\frac{1}{3r} \right]\,\,\theta \,\,=\frac{5{{\mu }_{0}}I\theta }{24\pi r}\]
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