The dimensions of the outer conductor of a coaxial cable are \(b\) and \(c\), where \(c>b\). Assuming \(\mu=\mu_{0}\), find the magnetic energy stored per unit length in the region \(b<\rho

Ampere's law states that the closed line integral of the magnetic field B around a closed loop equals to the total current I passing through the loop times the permeability constant μ. In the case of a coaxial cable, we can consider a cylindrical loop of radius ρ in the region b<ρ<c, with height equal to 1 (since we're looking for energy per unit length). Then, the line integral of B around the loop can be simplified as the product of B and the circumference of the loop (since B is always tangent to the loop). Let's denote the current density J, which is uniform between the conductors, as: \(/space J=I/(\pi(c^2 - b^2))/space\) Now, we can apply Ampere's law to find the magnetic field B: \(/space B \cdot 2 \pi \rho = \mu I \cdot (ρ^2 - b^2)/(\pi(c^2 - b^2))/space\)

By solving the equation from Step 1, we can find the expression for the magnetic field B as a function of ρ: \(/space B(\rho) = \mu I /(2 \pi) \cdot (ρ^2 - b^2)/(ρ(c^2 - b^2))/space\)

Now that we have determined the magnetic field B, we can find the magnetic energy density u, which is given by the following expression: \(/space u = B^{2}/(2 \mu)/space\) Substituting the expression for B from Step 2: \(/space u(\rho) = \mu I^2 /(8 \mu^2 \pi^2) \cdot [(ρ^2 - b^2)/(ρ(c^2 - b^2))]^2/space\)

Finally, in order to find the magnetic energy stored per unit length in the region b<ρ<c, we need to integrate the energy density expression from Step 3, over the given region. The differential volume element for cylindrical coordinates is dV=ρdρdφdz, but since we are looking for the energy per unit length, the integration will be carried out only with respect to ρ and φ: \(/space U = \int_{\phi=0}^{2\pi} \int_{\rho=b}^c u(\rho) \rho d\rho d\phi /space\) Substituting the expression for u(ρ) and carrying out the integration: \(/space U = \mu I^2 /(8 \mu^2 \pi^2) \int_{\phi=0}^{2\pi} \int_{\rho=b}^c [ (ρ^2 - b^2)/(ρ(c^2 - b^2))]^2 \rho d\rho d\phi/space\) After evaluating the integral, we find the total magnetic energy stored per unit length: \(/space U = I^{2}/(8\pi^2) (\ln c - \ln b) /space\) So, the magnetic energy stored per unit length in the region b<ρ<c for a uniformly distributed total current I flowing in opposite directions in the inner and outer conductors is: \(/space U = I^{2}/(8\pi^2) (\ln \frac{c}{b}) /space\)