question_answer

An equilateral triangular loop is made up of uniform wire as shown in the figure. A current/enters through one of the vertices of triangle and exits from other vertices of the triangle of side 'a' as shown. The magnitude of magnetic field at the centre of the equilateral triangle loop due to itself is

A) \[\frac{3{{\mu }_{0}}\ell }{2\pi a}\]            

B) \[\frac{9{{\mu }_{0}}\ell }{2\pi a}\]

C) \[\frac{3\sqrt{3}{{\mu }_{0}}\ell }{2\pi a}\]            

D)  None of these

Correct Answer: D

Solution :

\[{{i}_{ABC}}=\frac{R}{2R+R}i=\frac{i}{3}\] \[{{i}_{AC}}=\frac{2R}{2R+R}i=\frac{2i}{3}\] Magnetic field at point O (centre) \[\frac{{{\mu }_{0}}(i/3)}{4\pi d}(\sin {{60}^{o}}+\sin {{60}^{o}})+\frac{{{\mu }_{0}}(i/3)}{4\pi d}(\sin {{60}^{o}}+\sin {{60}^{o}})\]\[-\frac{{{\mu }_{0}}(2i/3)}{4\pi d}(\sin {{60}^{o}}+\sin {{60}^{o}})=0\] Hence, the correction option is [d].