# Inductance per Unit Length of Two Parallel Conducting Cylinders ## Problem Summary - Two infinitely long, parallel conducting cylinders. - Radius: \( a \) - Center-to-center distance: \( b \gg a \) - Positioned alongside the \( z \)-axis in the \( xz \)-plane, symmetric about the \( z \)-axis. - Currents: \( +I \) in one cylinder and \(-I\) in the other. - Medium has relative permeability: \( \mu_r = \pi \). --- ## Assumptions - Cylinders are long enough to neglect edge effects. - The distance \( b \) is much larger than \( a \), simplifying mutual inductance calculations. - Uniform current distribution over the cross-section. --- ## Goal Calculate the **inductance per unit length**, \( L' \), of the two cylinders. --- ## Step 1: Magnetic Field Due to a Single Cylinder ### Magnetic field of a single long wire with current \( I \): \[ B(r) = \frac{\mu_ \mu_r I}{2\pi r} \] - \( \mu_ \): permeability of free space. - \( \mu_r = \pi \): relative permeability of the medium. - \( r \): distance from the wire. ### Magnetic field at a distance \( r \) from the cylinder's axis: \[ B(r) = \frac{\mu_ \mu_r I}{2\pi r} \] --- ## Step 2: Magnetic Flux Linkage ### For a single cylinder: - Magnetic flux through a cross-section (radius from \( a \) to some large radius \( R \)): \[ \Phi = \int_{a}^{R} B(r) \cdot 2\pi r \, dr \] \[ \Phi = \int_{a}^{R} \frac{\mu_ \mu_r I}{2\pi r} \times 2\pi r \, dr = \mu_ \mu_r I \int_{a}^{R} dr = \mu_ \mu_r I (R - a) \] - For inductance per unit length, consider the flux per unit length as \( R \to \infty \): \[ \Phi = \mu_ \mu_r I \ln \left(\frac{R}{a}\right) \] --- ## Step 3: Mutual and Self-Inductance ### Self-inductance per unit length: \[ L_{\text{self}}' \sim \frac{\mu_ \mu_r}{2\pi} \ln \left(\frac{R}{a}\right) \] ### Mutual inductance between the two cylinders: - The magnetic flux linkage in one cylinder due to the current in the other: \[ \Phi_{12} = \frac{\mu_ \mu_r}{2\pi} \ln \left(\frac{b}{a}\right) \] (assuming the field extends from \( a \) to \( b \)). - Since currents are opposite, the flux linkage contributes negatively to the total inductance. --- ## Step 4: Total Inductance per Unit Length ### For two conductors carrying currents \( +I \) and \( -I \): \[ L' = \frac{\mu_ \mu_r}{\pi} \left[ \ln \left(\frac{b}{a}\right) - 1 \right] \] *The term "-1" accounts for the self-inductance correction, considering the finite radius \( a \).* --- ## **Final Expression:** \[ \boxed{ L' = \frac{\mu_ \pi}{\pi} \left[ \ln \left(\frac{b}{a}\right) - 1 \right] = \mu_ \pi \left[ \ln \left(\frac{b}{a}\right) - 1 \right] } \] where: - \( \mu_ \) is the permeability of free space. - \( \pi \) is the relative permeability of the medium. - \( a \) is the radius of the cylinders. - \( b \) is the center-to-center distance between the cylinders. --- ## **Key Points** - The inductance depends logarithmically on the ratio \( \frac{b}{a} \). - The permeability of the medium enhances the inductance proportionally. - The large separation \( b \gg a \) simplifies the calculation, focusing on the mutual inductance term.