Solution: Inductance of Coaxial Cable
1. Magnetic Field CalculationConsider a coaxial cable with inner conducting shell radius \(a\) and outer shell radius \(b\). A current \(I\) flows up the inner conductor and returns down the outer conductor. By Ampere’s Law, the magnetic field is non-zero only in the space between the conductors (\(a < r < b\)). $$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I \implies B(2\pi r) = \mu_0 I $$ $$ B = \frac{\mu_0 I}{2\pi r} $$
2. Magnetic Flux CalculationConsider a rectangular cross-section of length \(l=1\) (unit length) and width \((b-a)\). The magnetic flux \(\Phi\) passing through this section is: $$ \Phi = \int B dA = \int_{a}^{b} \frac{\mu_0 I}{2\pi r} (1 \cdot dr) $$ $$ \Phi = \frac{\mu_0 I}{2\pi} \left[ \ln r \right]_a^b = \frac{\mu_0 I}{2\pi} \ln\left(\frac{b}{a}\right) $$
3. Inductance per Unit LengthThe self-inductance \(L\) is defined as the flux per unit current (\(L = \Phi / I\)).