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For the single phase system carrying 50 A current, find the total flux linkage of thinner conductors of both the paths.

To calculate the total flux linkage (\(\lambda\)) of the thinner conductors in both the paths of the single-phase system, we need to consider the mutual and self-inductance of the system.

### Key Details from the Image:
- **Go Path**:
  - Thicker conductor carrying 30 A (1 cm radius)
  - Thinner conductor carrying 20 A (0.5 cm radius)
  - Distance between the two conductors is 0.25 m (center to center)

- **Return Path**:
  - Thicker conductor carrying -30 A (1 cm radius)
  - Thinner conductor carrying -20 A (0.5 cm radius)
  - Distance between the two conductors is 0.5 m (center to center)
  
- **Distance between the Go and Return paths** is 4 m.

The **flux linkage** for a conductor is determined using the inductance formula for each wire with respect to all other wires. This includes self and mutual inductance. The total flux linkage \(\lambda\) can be given by the formula:

\[
\lambda = L_{\text{self}} \cdot I + \sum L_{\text{mutual}} \cdot I
\]

where:
- \( L_{\text{self}} \) is the self-inductance of each conductor.
- \( L_{\text{mutual}} \) is the mutual inductance between conductors.
- \( I \) is the current in the conductors.

To calculate the self and mutual inductance, the following standard formulas are used for parallel cylindrical conductors:

1. **Self-inductance of a single conductor**:
\[
L_{\text{self}} = \frac{\mu_0}{2\pi} \ln\left(\frac{d}{r}\right)
\]
where:
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7}\) H/m),
- \(d\) is the distance to the return path,
- \(r\) is the radius of the conductor.

2. **Mutual inductance between two conductors**:
\[
L_{\text{mutual}} = \frac{\mu_0}{2\pi} \ln\left(\frac{D}{d}\right)
\]
where:
- \(D\) is the distance between the conductors,
- \(d\) is the distance between the go and return conductors.

Let's calculate these step by step for both paths. I'll first calculate the self and mutual inductance for both the thinner conductors and their flux linkage.The total flux linkage of the thinner conductors in both the go and return paths is calculated to be zero. 

This is because the contributions from the currents in the go and return paths (20 A and -20 A) cancel each other out due to their opposite directions. The self and mutual inductances exist, but the equal and opposite currents result in no net flux linkage for the thinner conductors.

Let me know if you'd like further clarification, more detailed steps, or if you have additional questions.

Here are 5 related questions:
1. How is flux linkage related to inductance and current?
2. What would happen to the flux linkage if the currents were not equal and opposite?
3. How do mutual and self-inductances affect flux linkage in a system?
4. How does the distance between conductors impact mutual inductance?
5. What are the effects of using thicker conductors on flux linkage?

**Tip:** In electromagnetic systems, opposing currents in parallel conductors tend to reduce overall flux linkage due to cancellation, but understanding the layout is crucial to design inductance efficiently.

Learn how to calculate the total flux linkage of thinner conductors in a single-phase system carrying 50 A. Explore key concepts like mutual and self-inductance, and the effects of opposing currents in parallel conductors, with detailed step-by-step calculations.

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