Past exam of the physics B course of the University of Cambridge / 2015 / Paper 2 / Section B / B6

At an interface with unit normal $\hat{\mathbf{n}}$ pointing from medium 1 to medium 2, the magnetic boundary conditions areThe first follows from a pillbox applied to $\nabla \cdot \mathbf{B}=0$. The tangential magnetic field jump from a free surface current follows from a shrinking rectangular Ampère loop; ${\mathbf{K}}_f$ is the free surface current. With no free sheet current, normal $B$ and tangential $H$ are continuous. In linear isotropic media $\mathbf{B}=\mu \mathbf{H}$, so neither all components of $B$ nor all components of $H$ generally remain continuous.

For magnetic field-line refraction at a permeability interface, with angles measured from the normal, these conditions give $\tan {\theta }_2/\tan {\theta }_1={\mu }_2/{\mu }_1$. They determine the bending of magnetic flux at a material interface. In a slender core electromagnet, flux continuity and Ampère's law lead to the magnetic-circuit estimateAn air gap can dominate the reluctance even when the iron core has high magnetic permeability. Boundary conditions also determine demagnetizing fields; fringing, saturation and hysteresis limit the simple linear circuit model. A high-magnetic permeability material redirects flux, whereas a perfect-conductor boundary is a different physical assumption.^[ref]

An inductor stores magnetic energy and opposes changes of current through electromagnetic induction. For a fixed linear device, the flux linkage is $\Psi =N\Phi =LI$, the self-induced electromotive force is $-LdI/dt$, and the passive terminal voltage is $V=LdI/dt$. HenceThe voltage leads a sinusoidal current by a quarter cycle. A long solenoid has $B\simeq \mu NI/l$ and $L\simeq \mu N^{2}A/l$; its energy agrees with the volume integral of $B^2/(2\mu )$.

For two coupled coils, ${\Psi }_1=L_1{I}_1+\mathcal{M}{I}_2$, ${\Psi }_2=\mathcal{M}{I}_1+L_2{I}_2$, with signs set by coil orientations. Reciprocity gives the same mutual inductance in both equations. For the energy stored in a nonlinear inductor, $U=\int Id\Psi$ replaces the simple quadratic formula, and hysteresis can dissipate energy. Real coils also have resistance, core loss and parasitic capacitance, so an ideal inductance model has a finite useful frequency range.^[ref]

In a linear, nondispersive, lossless medium, the field energy density and energy flux areThe Poynting theorem follows by dotting the two curl Maxwell equations with the other field and subtracting:The right-hand side is power transferred from the field to free charges. In vacuum the energy density is ${ϵ}_0{E}^2/2+B^2/(2{\mu }_0)$.

Static electric or magnetic fields store energy even though their time derivatives vanish. Static Poynting flow occurs because static crossed fields can also have a nonzero Poynting flux: for example, a steady current-carrying coaxial line has radial electric and azimuthal magnetic fields, carrying energy along the line toward the load. Time independence therefore does not imply zero energy flow.

For phasors whose real parts represent the physical fields,For a travelling plane wave in vacuum the electric and magnetic contributions are equal and $⟨S⟩=c⟨u_{\mathrm{em}}⟩$. In dispersive matter the simple macroscopic energy formula needs modification to include the material response; it is not a universal formula for plasma energy.^[ref]

Reflected and transmitted waves arise because the fields must satisfy the electromagnetic boundary conditions. With no free surface charge or current, tangential $E,H$ and normal $D,B$ are continuous. At normal incidence between lossless media, define the real wave impedance $Z_j=\sqrt{{\mu }_j/{ϵ}_j}$. Writing $r=E_r/E_i$ and $t=E_t/E_i$ in a common transverse electric-field direction givesFor the power normalization of Fresnel amplitudes, power fractions are $R=|r{|}^2$ and $T=(Z_1/Z_2)|t{|}^2$, so $R+T=1$. An electric-field amplitude coefficient alone is not a transmitted power fraction.

At oblique incidence, phase matching along the interface gives Snell's law, $n_1\sin {\theta }_i=n_2\sin {\theta }_t$. Applying the same boundary conditions separately to the two linear polarizations gives the Fresnel equations. For nonmagnetic dielectrics the parallel-polarized reflection vanishes at $\tan {\theta }_B=n_2/n_1$. Above the critical angle into a lower-index lossless medium, the transmitted field is evanescent and reflection is total. A perfect conductor instead has zero tangential electric field and unit reflected power; finite conductivity introduces penetration and absorption.^[ref]