Explanation

Consider an equilateral triangular loop of side

$$ a = 4\sqrt{3}\,cm = 4\sqrt{3}\times10^{-2}\,m $$

The distance of the centroid from each side of an equilateral triangle is

$$ r = \frac{a\sqrt{3}}{6} $$

Substituting value of $a$,

$$ r = \frac{4\sqrt{3}\times10^{-2}\times\sqrt{3}}{6} = \frac{12\times10^{-2}}{6} = 2\times10^{-2}\,m $$

Magnetic field at a point due to a finite straight current carrying conductor is

$$ B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2) $$

At the centroid of an equilateral triangle,

$$ \theta_1 = \theta_2 = 60^\circ $$

So magnetic field due to one side is

$$ B_1 = \frac{\mu_0 I}{4\pi r}(\sin60^\circ + \sin60^\circ) = \frac{\mu_0 I}{4\pi r}(\sqrt{3}) $$

There are three identical sides, hence total magnetic field at the centroid is

$$ B = 3B_1 = \frac{3\mu_0 I\sqrt{3}}{4\pi r} $$

Substituting $\mu_0 = 4\pi\times10^{-7}$, $I=2\,A$ and $r=2\times10^{-2}\,m$,

$$ B = \frac{3(4\pi\times10^{-7})\times2\sqrt{3}}{4\pi(2\times10^{-2})} $$

$$ B = 3\sqrt{3}\times10^{-5}\,T $$

Comparing with $B = \alpha \times 10^{-5}\,T$,

$$ \alpha = 3\sqrt{3} $$

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