A circular loop of radius 3 cm is having a current of 12.5 A. The magnitude of magnetic field at a distance of 4 cm on its axis is

Options

A. `5.65 xx 10^(-5)T`
B. `5.27 xx 10^(-5)T`
C. `6.54 xx 10^(-5)T`
D. `9.20 xx 10^(-5)T`

Explanation

To find the magnitude of the magnetic field at a distance of 4 cm on the axis of a circular loop with a radius of 3 cm carrying a current of 12.5 A, we can use the formula for the magnetic field along the axis of a circular loop:

\[ B = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}} \]

Where:
- \( B \) is the magnetic field,
- \( \mu_0 \) 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,
- \( x \) is the distance from the center of the loop along the axis in meters.

### Step-by-Step Solution:

1. **Convert the given measurements to meters:**
- Radius \( r = 3 \, \text{cm} = 3 \times 10^{-2} \, \text{m} \)
- Distance \( x = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m} \)

2. **Substitute the values into the formula:**
\[ B = \frac{4\pi \times 10^{-7} \, \text{T m/A} \times 12.5 \, \text{A} \times (3 \times 10^{-2} \, \text{m})^2}{2((3 \times 10^{-2} \, \text{m})^2 + (4 \times 10^{-2} \, \text{m})^2)^{3/2}} \]

3. **Calculate \( r^2 \) and \( x^2 \):**
\[ r^2 = (3 \times 10^{-2})^2 = 9 \times 10^{-4} \, \text{m}^2 \]
\[ x^2 = (4 \times 10^{-2})^2 = 16 \times 10^{-4} \, \text{m}^2 \]

4. **Calculate \( r^2 + x^2 \):**
\[ r^2 + x^2 = 9 \times 10^{-4} + 16 \times 10^{-4} = 25 \times 10^{-4} \, \text{m}^2 \]

5. **Calculate \( (r^2 + x^2)^{3/2} \):**
\[ (25 \times 10^{-4})^{3/2} = (5 \times 10^{-2})^{3} = 125 \times 10^{-6} = 1.25 \times 10^{-4} \, \text{m}^3 \]

6. **Substitute back into the formula for \( B \):**
\[ B = \frac{4\pi \times 10^{-7} \times 12.5 \times 9 \times 10^{-4}}{2 \times 1.25 \times 10^{-4}} \]

7. **Calculate the numerator:**
\[ 4\pi \times 10^{-7} \times 12.5 \times 9 \approx 4 \times 3.14 \times 10^{-7} \times 12.5 \times 9 = 1.413 \times 10^{-5} \]

8. **Calculate the denominator:**
\[ 2 \times 1.25 \times 10^{-4} = 2.5 \times 10^{-4} \]

9. **Final calculation for \( B \):**
\[ B = \frac{1.413 \times 10^{-5}}{2.5 \times 10^{-4}} \approx 5.652 \times 10^{-5} \, \text{T} \]

### Final Answer:
The magnitude of the magnetic field at a distance of 4 cm on the axis of the circular loop is approximately:
\[ B \approx 5.65 \times 10^{-5} \, \text{T} \]