A wire carrying a current of 140 ampere is bent into the form of a circle of radius `6 cm`. The flux density at a distance of `8 cm` on the axis passing through the centre of the coil and perpendicular to its plane is `( I n Wb//m^(2)(` approximately `) )`

Options

A. `pixx10^(-4)`
B. `2pi xx 10^(-4)`
C. `(pi)/(2)xx10^(-4)`
D. `(1)/(pi)xx10^(-4)`

Explanation

To solve the problem, we need to calculate the magnetic flux density (B) at a point located 8 cm away from the center of a circular wire carrying a current. The wire is bent into a circle with a radius of 6 cm, and the current flowing through it is 140 A.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Current (I) = 140 A
- Radius of the circular wire (r) = 6 cm = 0.06 m
- Distance from the center of the coil to the point on the axis (x) = 8 cm = 0.08 m
- Permeability of free space (μ₀) = 4π × 10⁻⁷ T·m/A

2. **Use the Formula for Magnetic Field (B) at a Point on the Axis of a Circular Loop:**
The magnetic field (B) at a distance x from the center of a circular loop of radius r carrying current I is given by the formula:
\[ B = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}} \]

3. **Substitute the Values into the Formula:**
- Substitute I = 140 A, r = 0.06 m, and x = 0.08 m into the formula:
\[ B = \frac{(4\pi \times 10^{-7}) \times 140 \times (0.06)^2}{2((0.06)^2 + (0.08)^2)^{3/2}} \]

4. **Calculate the Denominator:**
- Calculate \( (0.06)^2 + (0.08)^2 \):
\[ (0.06)^2 = 0.0036 \quad \text{and} \quad (0.08)^2 = 0.0064 \]
\[ (0.06)^2 + (0.08)^2 = 0.0036 + 0.0064 = 0.0100 \]
- Now calculate \( (0.0100)^{3/2} \):
\[ (0.0100)^{3/2} = 0.000316227766 \quad \text{(approximately)} \]

5. **Calculate the Numerator:**
- Calculate \( (4\pi \times 10^{-7}) \times 140 \times (0.06)^2 \):
\[ (0.06)^2 = 0.0036 \]
\[ 4\pi \times 10^{-7} \times 140 \times 0.0036 \approx 2.35619449 \times 10^{-9} \quad \text{(approximately)} \]

6. **Combine the Results:**
- Now substitute back into the formula for B:
\[ B = \frac{2.35619449 \times 10^{-9}}{2 \times 0.000316227766} \approx \frac{2.35619449 \times 10^{-9}}{0.000632455532} \approx 3.72 \times 10^{-6} \, \text{T} \]

7. **Convert to Wb/m²:**
- Since 1 T = 1 Wb/m², we can express the result as:
\[ B \approx 3.72 \times 10^{-6} \, \text{Wb/m}^2 \]

8. **Final Result:**
- The approximate value of the flux density at a distance of 8 cm on the axis is:
\[ B \approx 3.72 \, \text{μT} \quad \text{(or)} \quad 3.72 \times 10^{-6} \, \text{Wb/m}^2 \]