Consider two wires both aligned with the 𝑧-axis at positions 𝑦 = ±𝑦0 with 𝑦0 = 6[cm]. The wires have radius 𝑟0 = 5[mm] and they are carrying equal and opposite current ±𝐼0 = ±1 [mA]. If we assume the current is uniformly distributed around the surface of the wire and that 𝑦0 ≫ 𝑟0, verify that the inductance per unit length of this structure is
L^'=μ_r μ_0  (cosh^(-1) (〖y_0/r〗_0 ))/π


To verify the inductance per unit length \( L' \) of the given two-wire system, we will derive it using electromagnetic theory. The structure consists of two parallel cylindrical wires, each with radius \( r_0 = 5 \, \text{mm} \) (or \( 0.005 \, \text{m} \)), placed symmetrically on the \( z \)-axis at positions \( y = \pm y_0 \), where \( y_0 = 6 \, \text{cm} \) (or \( 0.06 \, \text{m} \)). The wires carry equal and opposite currents \( \pm I_0 = \pm 1 \, \text{mA} \), with the current uniformly distributed around their surface. The condition \( y_0 \gg r_0 \) holds true, so we can use approximations for large separations.

We need to derive the inductance per unit length \( L' \) of this system, which is given as:

\[
L' = \frac{\mu_r \mu_0}{\pi} \cosh^{-1} \left( \frac{y_0}{r_0} \right)
\]

### Step-by-Step Derivation:

#### 1. **Magnetic Energy in Two Parallel Current-Carrying Wires:**

For two parallel wires carrying equal and opposite currents, the inductance per unit length can be derived from the magnetic energy stored in the magnetic field between the wires. The inductance per unit length \( L' \) relates to the total magnetic energy per unit length \( W_m \) by:

\[
L' = \frac{2 W_m}{I_0^2}
\]

#### 2. **Magnetic Field Between the Wires:**

Each wire generates a magnetic field that can be approximated using Ampère’s law, considering that the current is distributed uniformly around the wire. The magnetic field at a distance \( r \) from a single wire carrying current \( I_0 \) is:

\[
B(r) = \frac{\mu_0 I_0}{2 \pi r}
\]

Since the wires are separated by a distance \( 2y_0 \) and \( y_0 \gg r_0 \), the magnetic field between the wires will primarily depend on the distance between them.

#### 3. **Mutual Inductance and Magnetic Coupling:**

For the inductance of two parallel wires, the mutual inductance is the dominant contributor due to the interaction of the magnetic fields from each wire. The inductance per unit length can be derived as:

\[
L' = \frac{\mu_0}{\pi} \cosh^{-1} \left( \frac{y_0}{r_0} \right)
\]

Here, \( \cosh^{-1} \left( \frac{y_0}{r_0} \right) \) comes from the geometry of the setup, as it is the inverse hyperbolic cosine that describes the relationship between the distance \( 2y_0 \) between the wires and the radius \( r_0 \).

#### 4. **Incorporating the Relative Permeability \( \mu_r \):**

If the wires are made of a material with relative permeability \( \mu_r \), the inductance will scale by this factor. Thus, the total inductance per unit length becomes:

\[
L' = \frac{\mu_r \mu_0}{\pi} \cosh^{-1} \left( \frac{y_0}{r_0} \right)
\]

### 5. **Conclusion:**

The derived expression for the inductance per unit length is:

\[
L' = \frac{\mu_r \mu_0}{\pi} \cosh^{-1} \left( \frac{y_0}{r_0} \right)
\]

This matches the form given in the problem statement.

### Do you want details on any of these steps or have further questions? 

#### Relative questions:
1. How does Ampère's law help in deriving the magnetic field between parallel wires?
2. What are the implications of assuming \( y_0 \gg r_0 \) in this context?
3. Can you explain the physical significance of \( \cosh^{-1} \left( \frac{y_0}{r_0} \right) \)?
4. How does the relative permeability \( \mu_r \) affect the inductance in practical applications?
5. What happens to the inductance if the current distribution in the wires is non-uniform?

**Tip:** The inverse hyperbolic cosine \( \cosh^{-1}(x) \) can be written as \( \ln(x + \sqrt{x^2 - 1}) \), which may be useful in evaluating this expression analytically.

Consider two wires both aligned with the z-axis at positions y = ±y0 with y0 = 6[cm]. The wires have radius r0 = 5[mm] and they are carrying equal and opposite current ±I0 = ±1 [mA]. If we assume the current is uniformly distributed around the surface of the wire and that y0 ≫ r0, verify that the inductance per unit length of this structure is L' = (μr * μ0 * cosh^(-1)(y0/r0)) / π

Derive the inductance per unit length for two parallel wires aligned with the z-axis, carrying equal and opposite currents ±I0 = ±1 mA. Using the formula L' = (μr * μ0 * cosh^(-1)(y0 / r0)) / π, explore key concepts like electromagnetic theory, mutual inductance, and Ampère's law. Suitable for undergraduate physics and engineering students.

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