Past exam of the electromagnetism course of the University of Oxford / 2023 / 2 / b

Work in cylindrical coordinates and use the assumed independence of $\phi$ and $z$. In each material region, the zero curl of the static electric field givesThe tangential electric field is zero at a perfect conductor. The dielectric-interface boundary conditions make its tangential components continuous at each material interface, so $E_z=E_{\phi }=0$ throughout the gap. They also vanish in the empty interior and exterior for the regular solution with no applied external electric field. HenceThis specifies an axis of direction: the sign of $E_{\rho }$ depends on the sign of $V$.

For the static magnetic field, Gauss's law for magnetism impliesThus $\rho B_{\rho }$ is constant in each layer. Continuity of the normal magnetic field across every interface makes that constant global, and regularity at the axis makes it zero. Next use $\nabla \times \mathbf{H}=0$ away from the conducting surfaces. Its $\phi$ component gives $d{H}_z/d\rho =0$. There is no azimuthal surface current density, so $H_z$ is continuous across the conducting surfaces and the material interface. The decay of the exterior magnetic field at infinity therefore forces this global constant to zero. The only remaining component isThe magnetostatic Ampère law fixes its sign: positive inner electric current in the $+z$ direction produces positive $B_{\phi }$ for positive magnetic permeability.

For this radially layered coaxial cable, defineUse the solution regular at the axis with bounded exterior electric potential, so that the empty interior and exterior have the constant electric potentials fixed by their adjacent conducting cylinders.

There is no free charge in either electrical insulator or on their interface. Gauss's law and continuity of normal electric displacement field give $D_{\rho }=C/\rho$ in both layers. The electric potential difference fixes the constant becauseThe electric potential, including the empty regions bounded by the conducting cylinders, is thereforeBoth inner and outer Dirichlet boundary conditions hold, and the electric potential is continuous at the material interface, whereTaking the negative gradient of the electric potential yieldsThe radial electric field generally jumps at $\rho =c$, but $ϵE_{\rho }(c^{-})={ϵ}^{\prime }{E}_{\rho }(c^{+})$, as required by the absence of free surface charge density there. The conducting cylinders carry opposite linear charge densities $\lambda =2\pi C$ and $-\lambda$.

For an Amperian loop of radius $\rho$, the enclosed electric current is zero, $I$, or zero according as $\rho <a$, $a<\rho <b$, or $\rho >b$. The magnetostatic Ampère law therefore gives $H_{\phi }=I/(2\pi \rho )$ in the gap, and zero outside. Using the two magnetic permeabilities,The tangential magnetic field strength is continuous at the material interface, while the tangential magnetic field need not be. At each ideal conducting surface and at the material interface, the displayed electric field and magnetic field formulas specify one-sided limits; a single interface value is unnecessary when the limiting values differ.

Since ${\mathbf{e}}_{\rho }\times {\mathbf{e}}_{\phi }={\mathbf{e}}_z$, the Poynting vector is axial. The magnetic permeability cancels from $\mathbf{E}\times \mathbf{B}/\mu$ in each material layer:The signed Poynting flux through the plane with unit normal $+{\mathbf{e}}_z$ is its surface integral with area element $\rho d\rho d\phi$:ThusThe transmitted electromagnetic energy per unit time is the circuit's terminal power. For $VI>0$, the Poynting flux points in the direction of the inner electric current. The electromagnetic energy travels through the insulating gap and reaches a load at the end of the coaxial cable. The radii and material constants redistribute the electromagnetic energy density and the local Poynting vector, while their integrated Poynting flux depends only on the terminal electric potential difference and electric current.