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 (\( 4\pi \times 10^{-7} \, \text{N/A}^2 \)),
- \( I \) is the current flowing through the loop (in Amperes),
- \( R \) is the radius of the loop (in meters).

### Step 1: Convert the diameter to radius
Given the diameter of the loop is 80 cm, we can find the radius:

\[ R = \frac{\text{Diameter}}{2} = \frac{80 \, \text{cm}}{2} = 40 \, \text{cm} = 0.4 \, \text{m} \]

### Step 2: Substitute the values into the formula
Now, we can substitute the values into the formula. We know:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 \)
- \( I = 2 \, \text{A} \)
- \( R = 0.4 \, \text{m} \)

Substituting these values into the formula gives:

\[ B = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.4} \]

### Step 3: Simplify the expression
Now, simplifying the expression:

\[ B = \frac{4\pi \times 10^{-7} \times 2}{0.8} \]

\[ B = \frac{8\pi \times 10^{-7}}{0.8} \]

\[ B = 10\pi \times 10^{-7} \, \text{T} \]

### Step 4: Substitute the value of \(\pi\)
Now, substituting the value of \(\pi\) (taking \(\pi = 3.14\)):

\[ B = 10 \times 3.14 \times 10^{-7} \, \text{T} \]

\[ B = 31.4 \times 10^{-7} \, \text{T} \]

### Step 5: Convert to scientific notation
To express this in the form of \(xx \times 10^{-6} \, \text{T}\):

\[ B = 3.14 \times 10^{-6} \, \text{T} \]

### Final Answer
Thus, the magnetic field intensity at the center of the loop is:

\[ \text{Answer: } 3.14 \times 10^{-6} \, \text{T} \]

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