Two long straight parallel conductors are separated by a distance of 5 cm and carrying current 20 A. what work per unit length of a conductor must be done to increases the separation between conductors to 10 cm, if the current flows in the same direction ?

Answer

To solve the problem of calculating the work done per unit length to increase the separation between two long straight parallel conductors carrying current in the same direction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial separation, \( d_1 = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Final separation, \( d_2 = 10 \, \text{cm} = 0.10 \, \text{m} \)
- Current in each conductor, \( I_1 = I_2 = 20 \, \text{A} \)

2. **Formula for Force Between Conductors:**
The force per unit length \( F \) between two parallel conductors carrying currents in the same direction is given by:
\[ F = \frac{\mu_0 I_1 I_2}{2 \pi d} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) is the permeability of free space, and \( d \) is the separation between the conductors.

3. **Calculate Initial Force per Unit Length:**
Substitute the initial separation \( d_1 \) into the formula:
\[ F_1 = \frac{(4\pi \times 10^{-7}) \cdot (20) \cdot (20)}{2 \pi \cdot 0.05} \]
Simplifying this:
\[ F_1 = \frac{(4 \times 20 \times 20) \times 10^{-7}}{2 \times 0.05} \]
\[ F_1 = \frac{1600 \times 10^{-7}}{0.1} = 16000 \times 10^{-7} = 1.6 \times 10^{-3} \, \text{N/m} \]

4. **Calculate Final Force per Unit Length:**
Now substitute the final separation \( d_2 \):
\[ F_2 = \frac{(4\pi \times 10^{-7}) \cdot (20) \cdot (20)}{2 \pi \cdot 0.10} \]
Simplifying this:
\[ F_2 = \frac{(4 \times 20 \times 20) \times 10^{-7}}{2 \times 0.10} \]
\[ F_2 = \frac{1600 \times 10^{-7}}{0.2} = 8000 \times 10^{-7} = 0.8 \times 10^{-3} \, \text{N/m} \]

5. **Calculate Work Done:**
The work done \( W \) to separate the conductors from \( d_1 \) to \( d_2 \) can be calculated using the integral of force over distance:
\[ W = \int_{d_1}^{d_2} F \, dx \]
Since the force varies with distance, we can express it as:
\[ W = \int_{0.05}^{0.10} \frac{\mu_0 I_1 I_2}{2 \pi x} \, dx \]
Substituting the values:
\[ W = \frac{(4\pi \times 10^{-7}) \cdot (20) \cdot (20)}{2 \pi} \int_{0.05}^{0.10} \frac{1}{x} \, dx \]
The integral of \( \frac{1}{x} \) is \( \ln x \):
\[ W = \frac{800 \times 10^{-7}}{1} [\ln(0.10) - \ln(0.05)] \]
\[ W = 800 \times 10^{-7} \ln(2) \quad (\text{since } \ln(0.10) - \ln(0.05) = \ln(2)) \]

6. **Final Calculation:**
Using \( \ln(2) \approx 0.693 \):
\[ W = 800 \times 10^{-7} \times 0.693 \approx 5.544 \times 10^{-5} \, \text{J/m} \]

### Final Answer:
The work done per unit length to increase the separation between the conductors from 5 cm to 10 cm is approximately:
\[ W \approx 5.544 \times 10^{-5} \, \text{J/m} \]