A wire carrying 100A is bent into a circular loop of 5cm radius. The flux density at its centre in tesla is

Options

A. `22.56 xx 10^(-4)`
B. `23.08 xx 10^(-4)`
C. `12.56 xx 10^(-4)`
D. `40.06 xx 10^(-4)`

Answer

To find the magnetic flux density at the center of a circular loop formed by a wire carrying a current, we can use the formula for the magnetic field (B) at the center of a circular loop:

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

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

### Step-by-Step Solution:

1. **Identify the given values**:
- Current, \( I = 100 \, \text{A} \)
- Radius, \( R = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)

2. **Convert the radius to meters**:
- \( R = 5 \, \text{cm} = 0.05 \, \text{m} = 5 \times 10^{-2} \, \text{m} \)

3. **Substitute the values into the formula**:
- We know \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).
- Substitute \( I \) and \( R \) into the formula:
\[ B = \frac{4\pi \times 10^{-7} \times 100}{2 \times (5 \times 10^{-2})} \]

4. **Calculate the denominator**:
- \( 2R = 2 \times (5 \times 10^{-2}) = 1 \times 10^{-1} = 0.1 \, \text{m} \)

5. **Calculate the magnetic flux density**:
\[ B = \frac{4\pi \times 10^{-7} \times 100}{0.1} \]
\[ B = \frac{4\pi \times 10^{-5}}{0.1} = 4\pi \times 10^{-6} \, \text{T} \]

6. **Use the value of \(\pi\)**:
- Approximating \(\pi \approx 3.14\):
\[ B \approx 4 \times 3.14 \times 10^{-6} \, \text{T} = 12.56 \times 10^{-6} \, \text{T} \]

7. **Final Result**:
- The magnetic flux density at the center of the loop is:
\[ B \approx 12.56 \times 10^{-6} \, \text{T} = 12.56 \, \mu T \]

### Conclusion:
The flux density at the center of the loop is approximately \( 12.56 \times 10^{-6} \, \text{T} \) or \( 12.56 \, \mu T \).