A circular loop of radius 9.7 cm carries a current 2.3 A. Obtain the magnitude of the magnetic field at the centre of the loop .

Answer

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

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

Where:
- \( B \) is the magnetic field at the center of the loop,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( n \) is the number of turns per unit length (for a single loop, \( n = 1 \)),
- \( I \) is the current in amperes,
- \( r \) is the radius of the loop in meters.

### Step-by-Step Solution:

1. **Identify the given values**:
- Radius of the loop, \( r = 9.7 \, \text{cm} = 9.7 \times 10^{-2} \, \text{m} \)
- Current, \( I = 2.3 \, \text{A} \)
- Number of turns per unit length, \( n = 1 \) (since it's a single loop)

2. **Substitute the values into the formula**:
\[ B = \frac{\mu_0 n I}{2r} \]
\[ B = \frac{4\pi \times 10^{-7} \, \text{T m/A} \cdot 1 \cdot 2.3 \, \text{A}}{2 \cdot (9.7 \times 10^{-2} \, \text{m})} \]

3. **Calculate the denominator**:
\[ 2r = 2 \cdot (9.7 \times 10^{-2}) = 1.94 \times 10^{-1} \, \text{m} \]

4. **Calculate the numerator**:
\[ \text{Numerator} = 4\pi \times 10^{-7} \cdot 2.3 \approx 2.8 \times 10^{-6} \, \text{T m/A} \]

5. **Combine the results**:
\[ B = \frac{2.8 \times 10^{-6}}{1.94 \times 10^{-1}} \approx 1.44 \times 10^{-5} \, \text{T} \]

6. **Final result**:
\[ B \approx 1.49 \times 10^{-5} \, \text{T} \]

### Conclusion:
The magnitude of the magnetic field at the center of the circular loop is approximately \( 1.49 \times 10^{-5} \, \text{T} \).