Question 47 of 276

As shown in the figure, a current of $2A$ flowing in an equilateral triangle of side $4\sqrt{3}cm$ . The magnetic field at the centroid $O$ of the triangle is :

(Neglect the effect of earth’s magnetic field)

A$4\sqrt{3}\times {10}^{-4}T$B$4\sqrt{3}\times {10}^{-5}T$C$\sqrt{3}\times {10}^{-4}T$D$3\sqrt{3}\times {10}^{-5}T$

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Answer: (D) $3\sqrt{3}\times {10}^{-5}T$

Perpendicular distance of centroid from any of the sides will be, $d\tan 60^{\circ}=2\sqrt{3}$
$\Rightarrow d=2cm$
Each side will subtend an angle of $120^{\circ}$ at the centre. Also magnetic field due to all sides will be out of plane of paper, hence total magnetic field can be written as
$B=3\times \dfrac{{\mu }_{0}i}{4\pi d}\left[\sin 60^{\circ}+\sin 60^{\circ}\right]$
$=3\times \dfrac{{10}^{-7}\times 2}{2\times {10}^{-2}}\times \sqrt{3}$
$=3\sqrt{3}\times {10}^{-5}T$