A circular coil completes 30 rounds each of radius 6m carrying a current of 10 A. Calculate the magnetic flux density at the centre of the coil ?

Options

A. `1.7 xx 10 ^(-10)`
B. `3.187 xx 10 ^(-20)`
C. `3.14 xx 10^(-5) T`
D. `16.7 xx 10 ^(-10)`

Explanation

To find the magnetic flux density at the center of a circular coil, we can use the formula:

\[ B = \frac{\mu_0 \cdot n \cdot I}{2R} \]

where:
- \(B\) is the magnetic flux density,
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T m/A}\)),
- \(n\) is the number of turns of the coil,
- \(I\) is the current in amperes,
- \(R\) is the radius of the coil in meters.

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns, \(n = 30\)
- Radius of the coil, \(R = 6 \, \text{m}\)
- Current, \(I = 10 \, \text{A}\)

2. **Substitute the values into the formula:**
\[ B = \frac{4\pi \times 10^{-7} \cdot 30 \cdot 10}{2 \cdot 6} \]

3. **Calculate the denominator:**
\[ 2 \cdot 6 = 12 \]

4. **Calculate the numerator:**
\[ 4\pi \times 10^{-7} \cdot 30 \cdot 10 = 1200\pi \times 10^{-7} \]

5. **Now substitute back into the equation:**
\[ B = \frac{1200\pi \times 10^{-7}}{12} \]

6. **Simplify the fraction:**
\[ B = 100\pi \times 10^{-7} \]

7. **Calculate the value of \(\pi\):**
\[ \pi \approx 3.14 \]
Thus,
\[ B = 100 \cdot 3.14 \times 10^{-7} = 314 \times 10^{-7} = 3.14 \times 10^{-5} \, \text{T} \]

8. **Final answer:**
\[ B \approx 3.14 \times 10^{-5} \, \text{T} \]

### Conclusion:
The magnetic flux density at the center of the coil is approximately \(3.14 \times 10^{-5} \, \text{T}\).