Past exam of the physics B course of the University of Cambridge / 2018 / Paper 2 / Section A

For the usual hollow rectangular electromagnetic waveguide, let the propagation axis be $z$ and the interior be $0<x<a$, $0<y<b$. A transverse magnetic polarization has $B_z=0$ and $E_z\ne 0$. At a perfect conductor wall the tangential electric field vanishes. Since $E_z$ is tangential to all four side walls, it satisfies a homogeneous Dirichlet boundary condition around the rectangle.

Separating the transverse Helmholtz equation therefore gives the transverse magnetic mode of a rectangular waveguideA zero value of either index makes $E_z$ identically zero, rather than producing a nonzero field uniform along that direction. The transverse electric and magnetic fields of this TM family are obtained from transverse derivatives of $E_z$, so they also vanish. Thus both TM indices must be nonzero: neither ${\mathrm{TM}}_{n0}$ nor ${\mathrm{TM}}_{0m}$ exists in this rectangular hollow-guide model. In contrast, transverse electric modes obey a Neumann boundary condition on $B_z$ and may have one zero index. The claim is about this waveguide geometry, not all possible guides.

Use absolute temperatures $T_c=278.15\mathrm{K}$ and $T_h=298.15\mathrm{K}$. Let ${\dot{Q}}_c$ be the heat extracted from the ground, ${\dot{Q}}_h$ the heat delivered indoors, and $P$ the electrical input power. For a reversible heat pump, the entropy transferred per unit time balances:Consequently the coefficient of performance isThe requested heat delivery is a rate, since the duration is unspecified:Of this, $13.9\mathrm{kW}$ comes from the ground and $1.0\mathrm{kW}$ from the electricity. A coefficient of performance greater than one is consistent with conservation of energy because the heat pump transports heat as well as converting electrical energy.

Let the particle have mass $m$, and assume its force is derived from a time-independent potential energy $V(x)$ during the stated interval. Differentiation givesFor $n\ne 0$, choose a branch on which $t=(x/A{)}^{1/n}$ is real. Conservation of energy is the quickest integration of Newton's second law:The additive constant $E$ is arbitrary. For instance, the displayed expression applies directly for $t>0$, $A\ne 0$ and $x/A>0$, without raising a possibly negative $A$ to a fractional power. Its derivative gives $-V^{\prime }(x)=mn(n-1)A{t}^{n-2}$ along the trajectory, as required. The second term has units of kinetic energy because $A$ has units $\mathrm{m}{\mathrm{s}}^{-\mathrm{n}}$.

For $n=1$, the velocity is constant, so the potential energy is constant on the visited interval; the expression above reduces to such a constant. For $n=0$, or $A=0$, the particle is stationary: the trajectory only fixes $V^{\prime }$ to vanish at its position and does not determine the potential energy elsewhere. These exceptional cases cannot determine a unique nonconstant function.

Model air as a dilute ideal gas of molecules described by the hard sphere model. The number density and mean free path areThe factor $\sqrt{2}$ accounts for the relative motion of the colliding molecules. A molecular collision diameter has to be estimated because none is supplied. Taking $d\simeq 0.37\mathrm{nm}$, appropriate as an order-of-magnitude size for the main constituents of air, givesThus the mean free path is about $7\times 10^{-8}\mathrm{m}$, or $70\mathrm{nm}$. A slightly smaller assumed diameter $0.35\mathrm{nm}$ gives about $76\mathrm{nm}$, illustrating the intended precision of the estimate.

The appropriate quantity to minimize is the energy including exchange of particles with the Electron reservoir, $\mathcal{E}=E-\mu N$. The occupation of an interacting localized electron level follows fromFor repulsion $U>0$, the first Electron costs $ϵ$ and the second costs $ϵ+U$. Comparing all three grand energies givesFor attraction $U<0$, the singly occupied state never minimizes the grand energy. At the crossing of the empty and doubly occupied states, ${\mu }_{\ast }=ϵ+U/2$, its grand energy is $-U/2>0$, while both competing states have zero grand energy. HenceFor $U=0$ the two thresholds coincide at $\mu =ϵ$, giving a direct $0\to 2$ step. At a threshold, distinct occupations are degenerate; zero-temperature minimization alone does not select a unique occupation. A grand canonical ensemble at small positive temperature smooths the steps, withIn particular, the equal-weight empty/pair limit for $U<0$ gives mean occupation one exactly at its crossing; this is not a finite-width singly occupied plateau.