Past exam of the physics B course of the University of Cambridge / 2014 / Paper 1 / Section B

Specifying potential on the complete boundary gives a Dirichlet boundary condition and a unique harmonic solution. Specifying its normal derivative gives a Neumann boundary condition, with the zero-net-flux compatibility condition and an undetermined additive constant. Mixed conditions on complementary portions also give uniqueness when a potential reference is fixed. In an exterior domain, conditions at infinity are needed. These statements follow by applying Green's first identity to the difference of two solutions. A metal in electrostatic equilibrium is an equipotential: $\Phi =V_0$, tangential electric field zero, and normal field $E_n=\sigma /{ϵ}_0$ just outside a vacuum interface.

For the conducting-wedge potential expansion, put $u=\Phi -V_0$ in the empty sector $0<\theta <\alpha$. For $u=r^{\nu }\Psi (\theta )$, Laplace's equation gives ${\Psi }^{\prime \prime }+{\nu }^2\Psi =0$. Vanishing values on both faces require $\Psi \propto \sin (n\pi \theta /\alpha )$, henceBounded potential at the corner sets $b_n=0$. The coefficients still require outer boundary data; the metal potential alone cannot select them. For a generic nonzero $a_1$, put $p=\pi /\alpha$. Differentiation givesIf $a_1=0$, the first nonzero higher harmonic replaces this leading term. For $\alpha =\pi /2$, $u\simeq 2a_{1}xy$, so $\mathbf{E}=-2a_1(y,x)$ and field lines satisfy $x^2-y^2=\mathrm{constant}$, meeting the metal faces normally and tending to zero field at the corner. For $\alpha =3\pi /2$, the electrostatic corner-field exponent is $-1/3$. A real tip's rounding cuts off this divergence; breakdown or field emission can limit the attainable field.

Applying Gauss's law for magnetism to a pillbox and Ampère's law to an interface-crossing rectangle givesHere $\hat{n}$ points from material 1 to 2 and ${\mathbf{K}}_f$ is free surface current. With no such current, $B_n$ and $H_t$ are continuous. For linear isotropic media, $B_{t2}/{\mu }_2=B_{t1}/{\mu }_1$ and ${\mu }_2{H}_{n2}={\mu }_1{H}_{n1}$. The magnetic fields need not have equal magnitudes on either side.

With negligible leakage, Gauss's law for magnetism makes each cross-section of a magnetic flux tube carry the same flux $\mathcal{F}$. High permeability guides it through the core; a short gap permits neglect of fringing. Flux conservation across these particular geometric sections is an approximation, not a consequence of zero divergence without controlling side leakage.

Let $L$ be the constant-radius arc marked in the figure, and ${\mu }_r=\mu /{\mu }_0$. For each straight taper, the radius changes linearly from $R$ to $r$, giving the reluctance of a conical magnetic poleSumming magnetic reluctances to find the gap field of a tapered electromagnet, and using $B_g=\mathcal{F}/(\pi r^2)$ givesIf $L$ is instead defined as the total magnetic-material length, replace the displayed uniform-arc length by $L-2s$. The comparison untapered core has uniform radius $R$ and material length $L+2s$, soThe original PDF specifies $s/L\ll r/R$; the TeX transcription reverses this inequality. In the printed limit the taper reluctances are small relative to that of the broad arc. The gain is approximately $[h+L/{\mu }_r]/[h+L{r}^2/({\mu }_r{R}^2)]$, approaching $(R/r{)}^2$ if the gap reluctance is also negligible. If the gap dominates, both fields instead approach $NI/h$ and the gain approaches one. magnetic saturation near the small poles, gap fringing, leakage and winding heating limit practical enhancement.

The macroscopic Maxwell equations, with free charge and current, areInside this body $\mathbf{D}={ϵ}_0\mathbf{E}$ and $\mathbf{B}={\mu }_0\mathbf{H}$. The wave magnetic field lies along $z$; its sign follows the local wave component. Faraday's and Ampère's laws giveStatic background fields are excluded from this wave solution. The magnetic force density is along $x$, with $f_x={\mu }_0{J}_y{H}_z$. For a steady periodic state, $\overline{H_z{\partial }_t{E}_y}=-\overline{E_y{\partial }_t{H}_z}$. ThereforeThe result gives radiation-force density without Ohm's law. Integrating through the body gives its pressure as the boundary value of the averaged electromagnetic energy density minus its value deep inside. In both limiting cases below the latter is zero. Take $A$ as the incident electric field's peak amplitude and write $Z_v=\sqrt{{\mu }_0/{ϵ}_0}$.

With no reflected field, $\overline{E_y^2}=A^2/2$ and $\overline{H_z^2}=A^2/(2Z_v^2)$ at the entrance. ThusComplete absorption transfers the incident normal momentum flux to the body.

At a perfect conductor, the reflected electric field cancels the incident tangential field, while their magnetic fields add. Thus $E_y(0)=0$ and $H_z(0)$ has peak amplitude $2A/Z_v$. Taking the zero-penetration limit, including its surface force, givesReversal of the normal momentum doubles the pressure compared with absorption at the same incident Poynting flux.

$Z$ is the terminating load impedance, $Z_0$ the line's characteristic impedance, and $k$ the propagation wave number; $l$ is its length. In the lossless convention printed, an open circuit gives $Z_{\mathrm{in}}=-i{Z}_0\cot kl$, a short circuit gives $i{Z}_0\tan kl$, and a matched load $Z=Z_0$ gives $Z_{\mathrm{in}}=Z_0$ for every length. The divergent or zero limits at isolated electrical lengths are understood continuously.

Put $\theta =kl$, $c_{\theta }=\cos \theta$, $s_{\theta }=\sin \theta$. The voltage/current matrix of one section isFor a pair beginning with $Z_1$, write $M_1{M}_2=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Its entries are $A=c_{\theta }^2-(Z_1/Z_2)s_{\theta }^2$, $D=c_{\theta }^2-(Z_2/Z_1)s_{\theta }^2$, $B=i(Z_1+Z_2)s_{\theta }{c}_{\theta }$, and $C=i(1/Z_1+1/Z_2)s_{\theta }{c}_{\theta }$. The alternating-impedance transmission line has unchanged input $X$, its Bloch impedance, obeying $X=(AX+B)/(CX+D)$. Hence, away from the removable dividing zeros,For real positive $Z_1,Z_2$, select the branch carrying energy into the line or decaying into it. A positive real part selects the pass-band root; infinitesimal passive loss resolves the stop-band branch. At $\theta =m\pi$ the cell matrix is the identity, so the exact fixed-point condition alone does not determine $X$; at odd quarter-wave lengths the appropriate limiting impedance can be zero or infinite.

In a transmission-line stop band, a lossless semi-infinite ladder totally reflects because its physical input is reactive. Equivalently its cell hasThus, for nonzero contrast, the stop-band interiors satisfyWith ${\theta }_c=\arcsin [2\sqrt{Z_1{Z}_2}/(Z_1+Z_2)]$, these are $m\pi +{\theta }_c<kl<(m+1)\pi -{\theta }_c$; band-edge reflection follows by a limit. If $k=\omega /u$, multiply the endpoints by $u/l$ for angular frequencies. Identical impedances have no stop band. A finite ladder can transmit by tunnelling, unlike the semi-infinite idealization.