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Step-by-Step Solution
Step 1: Identify the geometry and parameters
• The triangle is equilateral, with side length
$4\sqrt{3}\,\mathrm{cm}$ (i.e., $4\sqrt{3}\times10^{-2}\,\mathrm{m}$).
• A current of $2\,\mathrm{A}$ flows through each side of the triangle.
• We need the magnetic field at the centroid (O) of the triangle.
Step 2: Determine the distance from centroid to each side
In an equilateral triangle of side $a$, the altitude is $h = \frac{\sqrt{3}}{2}a$.
For side $a = 4\sqrt{3}\,\mathrm{cm}$, the altitude becomes
$h = \frac{\sqrt{3}}{2} \times 4\sqrt{3} = 6\,\mathrm{cm}.$
The centroid divides the altitude in a 2:1 ratio. Hence, the distance from the centroid to any side (often called the perpendicular distance) is
$ \frac{h}{3} = \frac{6\,\mathrm{cm}}{3} = 2\,\mathrm{cm} = 2 \times 10^{-2}\,\mathrm{m}.$
Thus, $r = 2 \times 10^{-2}\,\mathrm{m}.$
Step 3: Use the formula for magnetic field due to a finite straight current-carrying wire
For a straight current-carrying segment, the magnetic field at a point located at perpendicular distance $r$ and subtending angles $\alpha$ and $\beta$ is
$ B = \frac{\mu_0\,i}{4\pi\,r}\bigl(\sin\alpha + \sin\beta\bigr). $
In our equilateral triangle, each side subtends identical angles at the centroid. Because of symmetry, the net field is the vector sum of contributions from the three sides. Each side creates the same magnitude of field, and the directions add up symmetrically.
Step 4: Calculate the contribution from one side and multiply by 3
Let $B_1$ be the magnetic field at the centroid due to one side of the triangle. Then the total magnetic field $B_{\text{net}}$ from all three sides is
$ B_{\text{net}} = 3 \times B_1. $
Using the angles for an equilateral triangle geometry, one can show that
$ \sin\alpha + \sin\beta = \sqrt{3}, $
so the expression often simplifies neatly in such symmetric problems.
Step 5: Substitute numerical values
Using $ \mu_0 = 4\pi\times 10^{-7}\,\mathrm{H\,m^{-1}} $, $ i = 2\,\mathrm{A} $, and $ r = 2 \times 10^{-2}\,\mathrm{m} $:
$
B_{\text{net}}
= 3 \times \frac{\mu_0\,i}{4\pi\,r} \bigl(\sin\alpha + \sin\beta\bigr)
= 3 \times \frac{4\pi \times 10^{-7} \times 2}{4\pi \times 2 \times 10^{-2}} \times \sqrt{3}.
$
First, notice $ 4\pi $ in numerator and denominator cancels out:
$
= 3 \times \frac{10^{-7} \times 2}{2 \times 10^{-2}} \times \sqrt{3}
= 3 \times \frac{2 \times 10^{-7}}{2 \times 10^{-2}} \times \sqrt{3}.
$
Simplify inside:
$
= 3 \times \left(\frac{2 \times 10^{-7}}{2 \times 10^{-2}}\right) \times \sqrt{3}
= 3 \times \left(\frac{10^{-7}}{10^{-2}}\right) \times \sqrt{3}
= 3 \times 10^{-5} \times \sqrt{3}.
$
Thus,
$
B_{\text{net}} = 3\,\sqrt{3} \times 10^{-5}\,\mathrm{T}.
$
Step 6: State the final answer
Therefore, the magnetic field at the centroid of the equilateral triangle is
$ 3\,\sqrt{3} \times 10^{-5}\,\mathrm{T}. $
