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

Options

A. `3muT`
B. `1muT`
C. `18muT`
D. `9muT`

Answer

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 of the Triangle
An equilateral triangle has all sides equal. Given that the side length is 1 meter, we can denote the vertices of the triangle as A, B, and C.

### Step 2: Determine the Perpendicular Distance (r) from the Center to a Side
The perpendicular distance from the center of the triangle to any side can be calculated using the formula for the height (h) of an equilateral triangle:
\[ h = \frac{\sqrt{3}}{2} \times \text{side length} \]
For our triangle:
\[ h = \frac{\sqrt{3}}{2} \times 1 = \frac{\sqrt{3}}{2} \text{ meters} \]
The distance from the center to the side (r) is:
\[ r = \frac{h}{3} = \frac{\sqrt{3}/2}{3} = \frac{\sqrt{3}}{6} \text{ meters} \]

### Step 3: Calculate the Magnetic Field Due to One Side
The magnetic field (B) at the center due to one side of the triangle is given by the formula:
\[ B_{\text{one side}} = \frac{\mu_0 I}{4 \pi r} \sin(60^\circ) \]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 \)
- \( I = 10 \, \text{A} \)
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)

Substituting the values:
\[ B_{\text{one side}} = \frac{(4\pi \times 10^{-7}) \times 10}{4\pi \times \frac{\sqrt{3}}{6}} \cdot \frac{\sqrt{3}}{2} \]
This simplifies to:
\[ B_{\text{one side}} = \frac{10 \times 6}{\sqrt{3}} \times 10^{-7} \text{ T} \]

### Step 4: Total Magnetic Field at the Center
Since there are three sides contributing to the magnetic field at the center, we multiply the magnetic field due to one side by 3:
\[ B_{\text{total}} = 3 \times B_{\text{one side}} \]
Substituting the expression from above:
\[ B_{\text{total}} = 3 \times \frac{10 \times 6}{\sqrt{3}} \times 10^{-7} \text{ T} \]
This simplifies to:
\[ B_{\text{total}} = \frac{180}{\sqrt{3}} \times 10^{-7} \text{ T} \]

### Step 5: Final Calculation
Calculating the numerical value:
\[ B_{\text{total}} = 60\sqrt{3} \times 10^{-7} \text{ T} \]
Using \( \sqrt{3} \approx 1.732 \):
\[ B_{\text{total}} \approx 60 \times 1.732 \times 10^{-7} \approx 1.0392 \times 10^{-5} \text{ T} \approx 10.39 \mu T \]

### Conclusion
The magnitude of the magnetic field at the center of the equilateral triangular loop is approximately:
\[ B \approx 10.39 \mu T \]