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 carrying a current, we can follow these steps:

### Step 1: Understand the Geometry
We have an equilateral triangle with each side of length \( A = 1 \, \text{m} \). The center of the triangle is where we want to calculate the magnetic field.

### Step 2: Calculate the Distance from the Center to a Side
For an equilateral triangle, the distance from the center to any side (perpendicular distance) can be calculated using the formula:
\[ d = \frac{A}{\sqrt{3}} \]
Substituting \( A = 1 \, \text{m} \):
\[ d = \frac{1}{\sqrt{3}} \approx 0.577 \, \text{m} \]

### Step 3: Calculate the Magnetic Field Contribution from One Side
The magnetic field \( B \) at a distance \( d \) from a long straight wire carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2\pi d} \]
Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{NA}^{-2} \) and \( I = 10 \, \text{A} \).

Substituting the values:
\[ B = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times \frac{1}{\sqrt{3}}} \]
This simplifies to:
\[ B = \frac{4 \times 10^{-6}}{2 \times \frac{1}{\sqrt{3}}} = \frac{4 \times 10^{-6} \sqrt{3}}{2} = 2 \sqrt{3} \times 10^{-6} \, \text{T} \]

### Step 4: Calculate the Total Magnetic Field
Since there are three sides to the triangle, and the magnetic fields due to each side at the center will be equal in magnitude and will add up vectorially. The angle between the magnetic fields from two adjacent sides is \( 120^\circ \).

Using the formula for the resultant of three equal vectors \( B \) at \( 120^\circ \):
\[ B_{\text{total}} = 3B \cdot \sin\left(\frac{120^\circ}{2}\right) = 3B \cdot \sin(60^\circ) = 3B \cdot \frac{\sqrt{3}}{2} \]

Substituting \( B = 2 \sqrt{3} \times 10^{-6} \):
\[ B_{\text{total}} = 3 \cdot 2 \sqrt{3} \cdot \frac{\sqrt{3}}{2} \times 10^{-6} = 3 \cdot 3 \times 10^{-6} = 9 \times 10^{-6} \, \text{T} \]

### Step 5: Convert to Microtesla
Since \( 1 \, \text{T} = 10^6 \, \mu\text{T} \):
\[ B_{\text{total}} = 9 \, \mu\text{T} \]

### Final Answer
The magnitude of the magnetic field at the center of the equilateral triangular loop is \( \boxed{9 \, \mu\text{T}} \).

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