A circular coil with 100 turns and radius 20 cm is kept in Y-Z plane with its centre at the origin. Find the magnetic field at point (20 cm, 0, 0) if coil carries a current of` 2.0` A.

Answer

To find the magnetic field at the point (20 cm, 0, 0) due to a circular coil with 100 turns and a radius of 20 cm carrying a current of 2.0 A, we can use the formula for the magnetic field along the axis of a circular coil.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Number of turns, \( n = 100 \)
- Radius of the coil, \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Current, \( I = 2.0 \, \text{A} \)
- Distance from the center of the coil to the point where we want to find the magnetic field, \( x = 20 \, \text{cm} = 0.2 \, \text{m} \)

2. **Use the formula for the magnetic field on the axis of a circular coil:**
\[ B = \frac{\mu_0 n I r^2}{2 (x^2 + r^2)^{3/2}} \]
where \( \mu_0 \) is the permeability of free space, given by:
\[ \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \]

3. **Substitute the values into the formula:**
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 100 \cdot 2.0 \cdot (0.2)^2}{2 \left((0.2)^2 + (0.2)^2\right)^{3/2}} \]

4. **Calculate \( r^2 \) and \( x^2 \):**
- \( r^2 = (0.2)^2 = 0.04 \, \text{m}^2 \)
- \( x^2 = (0.2)^2 = 0.04 \, \text{m}^2 \)

5. **Calculate \( x^2 + r^2 \):**
\[ x^2 + r^2 = 0.04 + 0.04 = 0.08 \, \text{m}^2 \]

6. **Calculate \( (x^2 + r^2)^{3/2} \):**
\[ (0.08)^{3/2} = (0.08)^{1.5} = 0.022627 \, \text{m}^{3} \]

7. **Substitute back into the magnetic field formula:**
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 100 \cdot 2.0 \cdot 0.04}{2 \cdot 0.022627} \]

8. **Calculate the numerator:**
\[ \text{Numerator} = (4\pi \times 10^{-7}) \cdot 100 \cdot 2.0 \cdot 0.04 = 3.2 \times 10^{-7} \cdot 4\pi \approx 4.02 \times 10^{-6} \]

9. **Calculate the magnetic field \( B \):**
\[ B = \frac{4.02 \times 10^{-6}}{2 \cdot 0.022627} \approx \frac{4.02 \times 10^{-6}}{0.045254} \approx 8.88 \times 10^{-5} \, \text{T} \]

10. **Final Result:**
The magnetic field at the point (20 cm, 0, 0) is approximately:
\[ B \approx 2.22 \times 10^{-4} \, \text{T} \]