A circular loop of diameter 80 cm carries a steady current 2 A. The magnetic field intensity at the centre of the loop is _________ `xx10^(-6) T`.
(Permeability of vacuum `mu_(0)=4pixx10^(-7) N//A^(2)`, Take `pi=3.14`)

Explanation

To find the magnetic field intensity at the center of a circular loop carrying a steady current, we can use the formula:

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

where:
- \(B\) is the magnetic field intensity at the center of the loop,
- \(\mu_0\) is the permeability of free space,
- \(I\) is the current flowing through the loop,
- \(R\) is the radius of the loop.

### Step-by-Step Solution:

1. **Identify the given values:**
- Diameter of the loop = 80 cm
- Current, \(I = 2 \, \text{A}\)
- Permeability of vacuum, \(\mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2\)

2. **Convert the diameter to radius:**
- Radius \(R = \frac{\text{Diameter}}{2} = \frac{80 \, \text{cm}}{2} = 40 \, \text{cm}\)
- Convert cm to meters: \(R = 40 \, \text{cm} = 0.4 \, \text{m}\)

3. **Substitute the values into the formula:**
\[ B = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.4} \]

4. **Simplify the expression:**
- Calculate the denominator:
\[ 2 \times 0.4 = 0.8 \]
- Now substitute this back into the equation:
\[ B = \frac{4\pi \times 10^{-7} \times 2}{0.8} \]

5. **Calculate the numerator:**
- \(4 \times 2 = 8\)
- Thus, we have:
\[ B = \frac{8\pi \times 10^{-7}}{0.8} \]

6. **Perform the division:**
\[ \frac{8}{0.8} = 10 \]
- Therefore, we can rewrite \(B\):
\[ B = 10\pi \times 10^{-7} \]

7. **Substitute the value of \(\pi\):**
- Using \(\pi \approx 3.14\):
\[ B = 10 \times 3.14 \times 10^{-7} \]
\[ B = 31.4 \times 10^{-7} \, \text{T} \]

8. **Express in scientific notation:**
\[ B = 3.14 \times 10^{-6} \, \text{T} \]

### Final Answer:
The magnetic field intensity at the center of the loop is \(3.14 \times 10^{-6} \, \text{T}\).

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