A current of 1.5 A is flowing through a triangle of side 9 cm each .The magnetic field at the centroid of the triangle is : (Assume that the current is flowing in the clockwise direction .)

Options

A. `2sqrt(3)xx10^(-5)T`, inside the plane of triangle
B. `2sqrt(3)xx10^(-7)T` , outside the plane of triangle
C. `3xx10^(-7)T` ,outside the plane of triangle
D. `3xx10^(-5)T`, inside the plane of triangle

Answer

To find the magnetic field at the centroid of an equilateral triangle carrying a current, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Triangle and Current**:
We have an equilateral triangle with each side measuring 9 cm. The current of 1.5 A flows in a clockwise direction around the triangle.

2. **Determine the Centroid**:
The centroid of an equilateral triangle is located at a distance of \( \frac{h}{3} \) from the base, where \( h \) is the height of the triangle. The height \( h \) can be calculated using the formula:
\[ h = \frac{\sqrt{3}}{2} \times a \]
where \( a \) is the side length. For our triangle:
\[ h = \frac{\sqrt{3}}{2} \times 9 \text{ cm} = \frac{9\sqrt{3}}{2} \text{ cm} \]

3. **Calculate the Distance from Each Side to the Centroid**:
The distance from each side of the triangle to the centroid is \( \frac{h}{3} \):
\[ d = \frac{h}{3} = \frac{9\sqrt{3}}{6} = \frac{3\sqrt{3}}{2} \text{ cm} \]

4. **Use the Formula for Magnetic Field**:
The magnetic field \( B \) at the centroid due to a straight current-carrying wire is given by:
\[ B = \frac{\mu_0 I}{4\pi d} \]
However, since we have three sides contributing to the magnetic field, we multiply this by 3:
\[ B = 3 \times \frac{\mu_0 I}{4\pi d} \]

5. **Substitute Values**:
Here, \( I = 1.5 \, A \) and \( d = \frac{3\sqrt{3}}{2} \text{ cm} = \frac{3\sqrt{3}}{200} \text{ m} \) (since we need to convert cm to m). The permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \).
\[ B = 3 \times \frac{4\pi \times 10^{-7} \times 1.5}{4\pi \times \frac{3\sqrt{3}}{200}} \]

6. **Simplify the Equation**:
The \( 4\pi \) cancels out:
\[ B = 3 \times \frac{10^{-7} \times 1.5}{\frac{3\sqrt{3}}{200}} = 3 \times \frac{10^{-7} \times 1.5 \times 200}{3\sqrt{3}} = \frac{10^{-7} \times 300}{\sqrt{3}} \]

7. **Calculate the Final Value**:
\[ B = \frac{300 \times 10^{-7}}{\sqrt{3}} \approx 173.21 \times 10^{-7} \, T = 1.732 \times 10^{-5} \, T \]

8. **Determine the Direction of the Magnetic Field**:
Using the right-hand rule, since the current is flowing in a clockwise direction, the magnetic field at the centroid will be directed into the plane of the paper.

### Final Answer:
The magnetic field at the centroid of the triangle is approximately \( 1.732 \times 10^{-5} \, T \) directed into the plane of the paper.