The magnitude of the magnetic field at the center of an equilateral triangular loop of side 1 m which is carrying a current of 10 A (in `muT`) is ___________ .
[Take `mu_(0) = 4pi xx 10^(-7) NA^(-2)`]

Explanation

To find the magnitude of the magnetic field at the center of an equilateral triangular loop of side 1 m carrying a current of 10 A, we can follow these steps:

### Step 1: Understand the Geometry
We have an equilateral triangle with each side \( A = 1 \, \text{m} \). The center of the triangle is known as the centroid.

### Step 2: Determine the Distance from the Centroid to a Vertex
The distance \( r \) from the centroid to any vertex of the triangle can be calculated using the formula:
\[ r = \frac{A}{\sqrt{3}} \]
Substituting \( A = 1 \, \text{m} \):
\[ r = \frac{1}{\sqrt{3}} \approx 0.577 \, \text{m} \]

### Step 3: Calculate the Magnetic Field Contribution from One Side
The magnetic field \( B \) at the center due to a straight current-carrying wire is given by:
\[ B = \frac{\mu_0 I}{4 \pi r} \sin \theta \]
For an equilateral triangle, the angle \( \theta \) between the line from the center to the vertex and the perpendicular from the center to the wire is \( 60^\circ \). Thus, \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).

### Step 4: Substitute Values into the Formula
Using \( \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 \) and \( I = 10 \, \text{A} \):
\[ B = \frac{(4\pi \times 10^{-7}) \times 10}{4\pi \times \frac{1}{\sqrt{3}}} \cdot \frac{\sqrt{3}}{2} \]
This simplifies to:
\[ B = \frac{10 \times 10^{-7}}{\frac{1}{\sqrt{3}}} \cdot \frac{\sqrt{3}}{2} \]
\[ B = 10 \times 10^{-7} \cdot \frac{1}{2} = 5 \times 10^{-7} \, \text{T} \]

### Step 5: Total Magnetic Field from All Three Sides
Since there are three sides contributing equally to the magnetic field:
\[ B_{\text{total}} = 3B = 3 \times 5 \times 10^{-7} \, \text{T} = 15 \times 10^{-7} \, \text{T} \]
To convert this to microtesla (\( \mu T \)):
\[ B_{\text{total}} = 15 \times 10^{-1} \, \mu T = 15 \, \mu T \]

### Final Answer
The magnitude of the magnetic field at the center of the equilateral triangular loop is:
\[ \boxed{15 \, \mu T} \]