JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 14)
As shown in the figure, a current of $2 \mathrm{~A}$ flowing in an equilateral triangle of side $4 \sqrt{3} \mathrm{~cm}$. The magnetic field at the centroid $\mathrm{O}$ of the triangle is
_30th_January_Evening_Shift_en_14_1.png)
(Neglect the effect of earth's magnetic field)
$4 \sqrt{3} \times 10^{-4} \mathrm{~T}$
$4 \sqrt{3} \times 10^{-5} \mathrm{~T}$
$3 \sqrt{3} \times 10^{-5} \mathrm{~T} $
$\sqrt{3} \times 10^{-4} \mathrm{~T}$
Explanation
$${B_{net}} = {{{\mu _0}i} \over {4\pi r}}(\sin \alpha + \sin \beta ) \times 3$$
$$ = {{{\mu _0} \times 2} \over {4\pi \times (2 \times {{10}^{ - 2}})}} \times \left( {{{\sqrt 3 } \over 2} + {{\sqrt 3 } \over 2}} \right) \times 3$$
$$ = {10^{ - 7}} \times {10^2}(3\sqrt 3 )$$
$$ = 3\sqrt 3 \times {10^{ - 5}}$$ T
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