A current of $1\,A$ is flowing on the sides of an equilateral triangle of side $4.5\times10^{-2}\,m$ . The magnetic field at the centre of the triangle will be
JEE MAIN 2018, Diffcult
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Here, side of the triangle, $l=4.5 \times 10^{-2} \,\mathrm{m}$ current, $I =1 \,\mathrm{A}$
magnetic field at the centre of the triangle $'O'B = ?$
From figure, $\tan 60^{\circ}=\sqrt{3}=\frac{1}{2 d}$
$\Rightarrow d=\frac{l}{2 \sqrt{3}}=\left(\frac{4.5 \times 10^{-2}}{2 \sqrt{3}}\right)\, \mathrm{m}$
Magnetic field, $B=\frac{\mu_{0} i}{4 \pi d}\left(\cos \theta_{1}+\cos \theta_{2}\right)$
Putting value of $\mu=4 \pi \times 10^{-7}$ and $\theta_{1}$ and $\theta_{2}$
we will get net magenetic field
$=3 \times B=4 \times 10^{-5} \,\mathrm{Wb} / \mathrm{m}^{2}$
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