The diameter of a circular coil is 0.16m and it has 100 turns. If a current of 5 ampere is passed through the coil, then the intensity of magnetic field at a point on the axis at a distance 0.06 m from its centre will be -

Options

A. `2xx10^(-3)Wb//m^(2)`
B. `2xx10^(-2)Wb//m^(2)`
C. `2xx10^(3)Wb//m^(2)`
D. `2xx10^(2)Wb//m^(2)`

Answer

To find the intensity of the magnetic field at a point on the axis of a circular coil, we can use the formula for the magnetic field due to a circular coil at a distance from its center along its axis. The formula is given by:

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

Where:
- \( B \) = magnetic field intensity
- \( \mu_0 \) = permeability of free space = \( 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( n \) = number of turns per unit length (in this case, total turns since we are considering the entire coil)
- \( I \) = current in amperes
- \( a \) = radius of the coil
- \( x \) = distance from the center of the coil along the axis

### Step-by-Step Solution:

1. **Determine the radius of the coil:**
The diameter of the coil is given as 0.16 m, so the radius \( a \) is:
\[ a = \frac{0.16}{2} = 0.08 \, \text{m} \]

2. **Identify the number of turns and current:**
The number of turns \( n \) is given as 100, and the current \( I \) is given as 5 A.

3. **Identify the distance from the center:**
The distance \( x \) from the center of the coil to the point where we want to find the magnetic field is given as 0.06 m.

4. **Calculate \( a^2 \) and \( x^2 \):**
\[ a^2 = (0.08)^2 = 0.0064 \, \text{m}^2 \]
\[ x^2 = (0.06)^2 = 0.0036 \, \text{m}^2 \]

5. **Calculate \( a^2 + x^2 \):**
\[ a^2 + x^2 = 0.0064 + 0.0036 = 0.0100 \, \text{m}^2 \]

6. **Calculate \( (a^2 + x^2)^{3/2} \):**
\[ (a^2 + x^2)^{3/2} = (0.0100)^{3/2} = 0.000316227766 \, \text{m}^3 \]

7. **Substitute all values into the magnetic field formula:**
\[ B = \frac{4\pi \times 10^{-7}}{2} \cdot \frac{100 \cdot 5 \cdot (0.08)^2}{(0.0100)^{3/2}} \]
\[ B = 2\pi \times 10^{-7} \cdot \frac{100 \cdot 5 \cdot 0.0064}{0.000316227766} \]
\[ B = 2\pi \times 10^{-7} \cdot \frac{3.2}{0.000316227766} \]
\[ B = 2\pi \times 10^{-7} \cdot 10118.03399 \]
\[ B \approx 6.36 \times 10^{-3} \, \text{T} \]

### Final Answer:
The intensity of the magnetic field at the point on the axis at a distance of 0.06 m from the center of the coil is approximately \( 6.36 \times 10^{-3} \, \text{T} \).