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Step-by-Step Solution
Step 1: Identify the Relevant Formula
For a straight current-carrying wire, the magnetic field at a point located at perpendicularly shortest distance $r$ from it can be given by:
$$B = \frac{\mu_0\,I}{4\pi\,r} \left(\sin\theta_1 + \sin\theta_2\right).$$
Here, $I$ is the current through the wire, $\mu_0$ is the permeability of free space, and $\theta_1$, $\theta_2$ are the angles subtended by the segment of wire at the point where the field is being calculated.
Step 2: Apply to the Equilateral Triangular Loop
We have an equilateral triangular loop of side $1 \text{ m}$ carrying a current $I = 10 \text{ A}$. The magnetic field at the center is the vector sum of the fields due to each side. Due to the loop's symmetry, each side contributes equally, and the net magnetic field is:
$$B_{\text{net}} = 3 \times \frac{\mu_0\,I}{4\pi\,r}\,(\sin\theta_1 + \sin\theta_2).$$
Step 3: Determine the Angles
In an equilateral triangle, each vertex is $60^\circ$. At the center, each side subtends angles $\theta_1 = 60^\circ$ and $\theta_2 = 60^\circ$. Thus,
$$\sin\theta_1 = \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \sin\theta_2 = \sin 60^\circ = \frac{\sqrt{3}}{2}.$$
Therefore,
$$\sin\theta_1 + \sin\theta_2 = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}.$$
Step 4: Calculate the Distance $r$
The distance $r$ from the center of an equilateral triangle to any of its sides is one-third of the altitude of the triangle. The altitude of an equilateral triangle of side $a$ is $\frac{\sqrt{3}}{2}a$. Hence,
$$r = \frac{1}{3} \times \frac{\sqrt{3}}{2} \times 1 \text{ m} = \frac{1}{2\sqrt{3}} \text{ m}.$$
Step 5: Substitute Numerical Values
We know that
$$
\frac{\mu_0}{4\pi} = 10^{-7} \,\text{N A}^{-2}, \quad I = 10 \,\text{A}, \quad r = \frac{1}{2\sqrt{3}} \,\text{m}.
$$
Substituting into the net magnetic field expression:
$$
\begin{aligned}
B_{\text{net}} &= 3 \times \frac{\mu_0 \, I}{4\pi \, r} \times \bigl(\sin\theta_1 + \sin\theta_2\bigr) \\
&= 3 \times \frac{10^{-7} \times 10}{\frac{1}{2\sqrt{3}}} \times \sqrt{3}.
\end{aligned}
$$
Simplify further:
$$
B_{\text{net}} = 3 \times 10^{-6} \times \frac{10}{\frac{1}{2\sqrt{3}}} \times \sqrt{3}.
$$
Carefully carrying out each step:
$$
B_{\text{net}}
= \frac{3 \times 10^{-7} \times 10 \times 2\sqrt{3} \times \sqrt{3}}{1}
= 3 \times 10^{-7} \times 10 \times 2 \times 3
= 18 \times 10^{-6} \,\text{T}
= 18\,\mu\text{T}.
$$
Step 6: Final Answer
Therefore, the magnitude of the magnetic field at the center of the equilateral triangular current loop is:
$$\boxed{18\,\mu\text{T}}.$$
