Past exam of the physics B course of the University of Cambridge / 2011

Original paper: Public original, physical PDF pages 279–282.

For an outward radial electric field, Gauss's law gives the enclosed chargeTaking the enclosing sphere to infinity gives $Q_{\mathrm{total}}=0$. A nonzero local field is compatible with zero net charge. Indeed, for $r>0$,Positive inner charge is balanced by negative outer charge. There is no point charge at the origin, since ${\lim }_{r\to 0}4\pi {ϵ}_0{r}^2{E}_r=0$. The field's magnitude does not specify its inward/outward sign, but reversing a globally radial direction leaves the zero total unchanged. This is net charge from the asymptotic radial electric field.

In steady state the incoming volumetric flow rate equals the discharge. Bernoulli's principle between a nearly stationary free surface and the atmospheric outlet gives the ideal Torricelli's law speed $\sqrt{2gh}$. The specified discharge coefficient multiplies the opening-flow rate, soThus, with $A=10^{-4}{\mathrm{m}}^2$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,Neglect surface-speed and capillary corrections. The coefficient includes contraction and loss effects; it should not be applied a second time to the calculated height.

Take $f$ as the tensile force exerted on the band, so reversible stretching supplies work $fdx$. At fixed other mechanical variables,Equality of mixed derivatives of the Helmholtz free energy yields the thermoelastic Maxwell relation for a stretched bandFor entropy-dominated rubber elasticity, increasing extension reduces configurational entropy, consistently giving a tensile force that increases with temperature at fixed extension. A force defined as the band's restoring force on its support has the opposite sign.

For a nonabsorbing interface at normal incidence, the Fresnel equations give the intensity reflectanceThe two air–glass interfaces each transmit a fraction $1-\mathcal{R}$ of their incident power. Omitting all internally reflected paths,The power-transmission factors, rather than electric-field amplitudes alone, are multiplied. Absorption, coatings and interference are excluded by this approximation.

Use a local horizontal plane, constant gravity and first order in Earth's angular speed. Let $x$ point along the initial horizontal shot, $y$ to the gunner's left and $z$ upward. At the North pole, $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. Maximum unperturbed range at equal launch and landing heights requires a $45^{\circ }$ elevation, givingThe leading Coriolis acceleration is $\ddot{y}=-2\Omega v_x$, while vertical motion has no first-order Coriolis correction. With zero initial lateral displacement and velocity,Here $u=500\mathrm{m}{\mathrm{s}}^{-1}$, $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\Omega \simeq 7.2921\times 10^{-5}{\mathrm{s}}^{-1}$. The miss is to the right of the aiming direction, independent of its azimuth at the pole. The flight time is about $72.1\mathrm{s}$, so $\Omega t_f\simeq 0.0053\ll 1$. This polar projectile Coriolis deflection uses the usual flat, uniform-gravity approximation; planetary curvature and variation of gravity are beyond its accuracy. Using the polar gravity rather than $9.81$ changes the estimate by less than one percent. Earth's sidereal rotation period, rather than the solar day, sets $\Omega$.^[ref]

Gauss's law connects electric flux through a closed surface to its enclosed charge:The differential form follows from the divergence theorem. The law is exact; extracting a field magnitude by simple division additionally needs symmetry. For spherical symmetry, $E_r(r)=Q(r)/(4\pi {ϵ}_0{r}^2)$; a uniformly charged ball therefore has a field linear in $r$ inside and an inverse-square field outside. An infinite line of linear charge density $\Lambda$ gives $E_r=\Lambda /(2\pi {ϵ}_{0}r)$ by a cylindrical surface. A uniformly charged infinite sheet gives $E=\sigma /(2{ϵ}_0)$ on either side by a pillbox.

An electrostatic conductor has zero interior field and constant potential; excess charge occupies its surface. A pillbox gives $E_{n,\mathrm{out}}=\sigma /{ϵ}_0$ when the inside is conducting. More generally, $\hat{\mathbf{n}}\cdot ({\mathbf{E}}_2-{\mathbf{E}}_1)=\sigma /{ϵ}_0$. For dielectric matter, $\nabla \cdot \mathbf{D}={\rho }_{\mathrm{free}}$ separates bound polarization charge from free charge. Gauss's law alone does not establish a uniform field for an arbitrary finite plate or recover a detailed nonsymmetric charge distribution from one flux measurement.

An electric dipole has equal and opposite charges separated by a small vector from negative to positive, with electric dipole moment $\mathbf{p}=q\mathbf{d}$. More generally, a localized neutral distribution has $\mathbf{p}=\int {\mathbf{r}}^{\prime }\rho ({\mathbf{r}}^{\prime })d^3{r}^{\prime }$, independent of origin. At distances large compared with its size, the leading nonzero term in the multipole expansion isThe field points along $\mathbf{p}$ on either axial side and opposite $\mathbf{p}$ in the equatorial plane; its axial magnitude is twice the equatorial magnitude at equal radius. Near the charges the finite-charge expression must replace the far-field approximation.

A permanent dipole in an externally imposed slowly varying electrostatic field has interaction energy $-\mathbf{p}\cdot \mathbf{E}$, torque $\mathbf{\tau }=\mathbf{p}\times \mathbf{E}$ and force $\nabla (\mathbf{p}\cdot \mathbf{E})$ for fixed $\mathbf{p}$. A uniform field aligns it but gives no net force. For a linearly induced dipole $\mathbf{p}=\alpha \mathbf{E}$, the formation-inclusive energy is $-\alpha E^2/2$, not $-\alpha E^2$: the moment itself grows while the field is applied. Dipole density gives electric polarization, whose bound charges are ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\sigma }_b=\mathbf{P}\cdot \hat{\mathbf{n}}$.

For a fixed circuit, Faraday's law of induction readsSurface orientation and contour direction obey the right-hand rule. The negative sign is Lenz's law: a passive induced current acts against the flux change, consistently with energy conservation. The induced electric field generally has nonzero curl and cannot be represented by one single-valued electrostatic potential.

For a material circuit moving with local velocity $\mathbf{u}$, the measured electromotive force includes the magnetic Lorentz force on its carriers:This distinguishes transformer induction in a stationary coil from motional induction in a changing circuit. A straight sliding rod of length $l$ moving perpendicular to itself and to uniform $B$ has $|\mathcal{E}|=Blv$. With circuit resistance $R$, the electrical power $(Blv{)}^2/R$ equals the external mechanical power needed to oppose magnetic drag. An ideal transformer gives $V_2/V_1=N_2/N_1$ in magnitude from common flux linkage; real leakage, resistance and core losses limit that simple result. pickup coils measure flux change rather than an absolute static magnetic field, so integration and a reference state are needed to recover a field history.

In charge- and current-free vacuum, Maxwell equations reduce toTaking a curl and using $\nabla \times \nabla \times \mathbf{E}=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$ givesand the same wave equation for $\mathbf{B}$. For a plane wave $\mathbf{E}=\mathrm{Re}[{\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$, $\omega =ck$, $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$ and ${\mathbf{B}}_0=\hat{\mathbf{k}}\times {\mathbf{E}}_0/c$. Thus the fields are transverse, mutually perpendicular and in phase; polarization describes the electric-field trajectory at a fixed point.

The energy density and Poynting vector areFor a linearly polarized sinusoid with peak electric amplitude $E_0$, the mean intensity is ${ϵ}_{0}c{E}_0^2/2$. Electromagnetic momentum density is $\mathbf{S}/c^2$; normal absorption produces radiation pressure $⟨S⟩/c$, while ideal retroreflection doubles it. Linear superposition permits interference and standing waves, whose local electric and magnetic fields need not have the traveling-plane-wave equality point by point.

The microscopic magnetic response becomes a macroscopic field problem through the magnetization $\mathbf{M}$, the magnetic dipole moment per unit volume. Definein magnetostatics. Bound currents are represented by ${\mathbf{J}}_b=\nabla \times \mathbf{M}$ and surface current ${\mathbf{K}}_b=\mathbf{M}\times \hat{\mathbf{n}}$. Normal $B$ is continuous across an interface, while $\hat{\mathbf{n}}\times ({\mathbf{H}}_2-{\mathbf{H}}_1)={\mathbf{K}}_{\mathrm{free}}$. It is generally incorrect to impose continuity of every component of both fields. A linear isotropic material has $\mathbf{M}=\chi \mathbf{H}$ and ${\mu }_r=1+\chi$, but a ferromagnet usually requires a nonlinear, history-dependent constitutive relation.

Internal field differs from externally applied field. For uniform magnetization along a principal axis of an ellipsoid, the demagnetizing field is $-N_{d}M$ and $H_{\mathrm{int}}=H_{\mathrm{applied}}-N_{d}M$. A linear specimen therefore has apparent susceptibility $\chi /(1+N_d\chi )$ relative to the applied field. Shape is especially important for large susceptibility. A long closed magnetic circuit reduces stray fields; a bar magnet still has demagnetizing fields despite the absence of magnetic monopoles. Equivalent magnetic-pole descriptions are bookkeeping for magnetization divergence, not free magnetic charge.^[ref]

In paramagnetism, permanent localized moments are preferentially aligned by the field, against thermal disorder. For independent spin-one-half moments with projections $\pm \mu$, the two Zeeman energies givein the dilute weak-field limit with $B\simeq {\mu }_{0}H$. This is a Curie law; the response saturates at $n\mu$ when alignment dominates. More generally, spin $J$ gives $\chi ={\mu }_{0}n(g{\mu }_B{)}^{2}J(J+1)/(3k_{B}T)$. Interactions can produce a Curie–Weiss response $C/(T-\Theta )$ over a suitable range. Conduction-electron Pauli paramagnetism need not show the localized-moment $1/T$ law.

In diamagnetism, the field induces a magnetic moment opposing it, so the linear susceptibility is negative. The microscopic cause is the change in orbital electronic motion, and a closed-shell insulating material has no permanent atomic moment to orient. This weak response is often only mildly temperature dependent away from other transitions. It also exists as a background in materials with a stronger positive response. Perfect flux expulsion in a superconductor is a distinct collective phenomenon and should not be identified with ordinary weak atomic diamagnetism.

In ferromagnetism, exchange interactions stabilize spontaneous magnetic order below the Curie temperature. A specimen can have zero net moment while its magnetic domains remain locally magnetized in different directions. Domain-wall motion and moment rotation produce a large response; anisotropy and pinning lead to remanence, coercivity and hysteresis. Increasing field eventually approaches saturation. Heating above the ordering transition removes spontaneous order, although an applied field still induces a response. Soft magnetic cores favor easy reversal and small cyclic loss; hard permanent-magnet materials favor retained magnetization and resistance to reversal. These properties concern collective order and barriers, not merely independent moments all following a weak field.^[ref]

A practical distinction requires a controlled field sweep and a separate magnetic readout. A toroidal core with drive winding $N_1$, path length $l$ and current $I$ has approximately $H=N_{1}I/l$. A pickup winding measures ${\mathcal{E}}_2=-N_{2}AdB/dt$; integrating after calibration gives $B$, and $M=B/{\mu }_0-H$. Reversing the sweep reveals a loop whose positive loss area per volume is $\oint HdB$. An air-core/empty-holder reference, known geometry and turn count, offset correction, temperature control and slower sweeps test background, heating and rate-dependent eddy-current losses. A single force or susceptibility reading cannot establish spontaneous order or hysteresis. The original curves below are illustrative responses, not measured material data.

A galilean transformation relates inertial frames in uniform relative translation. For $\mathbf{V}$ the velocity of the primed origin,Absolute time and invariant acceleration make Newton's second law retain its form for the usual nonrelativistic interactions. For a particle of mass $m$, ${\mathbf{p}}^{\prime }=\mathbf{p}-m\mathbf{V}$ and $K^{\prime }=K-\mathbf{V}\cdot \mathbf{p}+m{V}^2/2$. Momentum and kinetic energy are frame dependent, although an isolated system's conservation laws hold in either inertial frame. Relative velocities and collision predictions are consistent between frames. A rotating or accelerating frame instead requires inertial-force terms. Maxwell equations with a fixed light speed are not Galilean invariant; Lorentz transformations replace these relations when relativistic effects matter.

Kepler's laws of planetary motion state that, in the relative two-body description, bound trajectories are ellipses with the attracting body at a focus, the radius vector sweeps equal areas in equal times, and the square of the period is proportional to the cube of the semimajor axis for a fixed total mass. Their Newtonian derivation is especially informative. With relative coordinate $\mathbf{r}$ and ${\mu }_g=G(M+m)$, the inverse-square acceleration is $-{\mu }_g\mathbf{r}/r^3$. Its zero torque conserves $\mathbf{h}=\mathbf{r}\times \dot{\mathbf{r}}$, fixing a plane and the areal velocity $h/2$.

For $u=1/r$, the orbit equation is $u^{\prime \prime }+u={\mu }_g/h^2$ as a function of polar angle. ThereforeBound negative-energy motion has $0\le e<1$ and semilatus rectum $h^2/{\mu }_g=a(1-e^2)$. Dividing ellipse area by areal velocity givesFor a planet much lighter than its star, replace $M+m$ by $M$. Zero-energy and positive-energy encounters are parabolic and hyperbolic rather than periodic ellipses. Perturbations from other bodies and departures from point-mass Newtonian gravity can precess or otherwise modify the orbit; Kepler's laws are a two-body approximation, not an exact rule for every observed planetary path.

A tidal force measures the difference of gravitational acceleration across an extended object. A freely falling reference frame removes a spatially uniform acceleration, but not its gradient. For a distant point mass $M$ at distance $R$ and direction $\hat{\mathbf{n}}$ from the reference centre, a small internal displacement $\mathbf{x}$ givesThe tidal tensor has eigenvalues $2GM/R^3$ radially and $-GM/R^3$ in each transverse direction: radial stretching and transverse squeezing. Its trace vanishes in the source-free region, consistent with the gravitational Gauss law. An associated relative potential is ${\Phi }_{\mathrm{tide}}=-GM[3(\hat{\mathbf{n}}\cdot \mathbf{x}{)}^2-x^2]/(2R^3)$.

The scale is $GMl/R^3$ across size $l$, not the centre's absolute gravity $GM/R^2$. This distinction explains how a more distant massive object can exert a large force but a modest tide. Opposite-side stretching supports the elementary two-bulge picture; real ocean tides also involve rotation, basin geometry and fluid dynamics, so the static gradient alone does not predict coastal heights or timing. The same gradient stresses satellites and can disrupt a self-gravitating body when it competes with the body's own gravitational restoring acceleration.

A rapidly spinning gyroscope approximately has angular momentum $\mathbf{L}=I_3{\omega }_3\hat{\mathbf{e}}$ along its axle. With no torque in an inertial frame, this vector retains its direction. A transverse torque changes its direction according to $d\mathbf{L}/dt=\mathbf{\tau }$; it does not initially push the axle in the direction of the applied force.

For a fixed pivot, centre of mass $l\hat{\mathbf{e}}$ from it and gravity $-Mg\hat{\mathbf{z}}$, the torque is $Mgl\hat{\mathbf{z}}\times \hat{\mathbf{e}}$. In slow steady precession about the vertical, $d\mathbf{L}/dt=\Omega \hat{\mathbf{z}}\times \mathbf{L}$, givingThe factor $\sin \theta$ appears both in torque and in the radius traced by the angular-momentum tip, so it cancels for a nonvertical axle. Reversing spin reverses this precession. The full symmetric-top steady condition is $I_1\cos \theta {\Omega }^2-I_3{\omega }_3\Omega +Mgl=0$; the approximation neglects its quadratic term. Sudden release can also produce nutation, so large spin alone does not guarantee a steady sweep.

A wheel-on-pivot experiment can compare precession periods at different measured spin speeds and reverse the spin. Keep the pivot and centre of mass fixed; bearing friction, spin decay, an initial push and support motion can mimic departures from the predicted inverse-spin relation. A freely gimballed direction reference and a gravity-loaded precessing wheel test different torque conditions.^[ref]

A fluid is described on a continuum scale by density, pressure and velocity fields. An Eulerian field is measured at fixed position; a material element changes at rate $D/Dt={\partial }_t+\mathbf{v}\cdot \nabla$. mass conservation requiresFor incompressible flow this becomes $\nabla \cdot \mathbf{v}=0$. A streamline is instantaneously tangent to velocity, whereas a pathline follows a particle through time; they coincide in a steady flow but need not in an unsteady one.

For an ideal inviscid fluid, pressure and body forces give the Euler equation. In the constant-density gravitational case,A stationary impermeable boundary requires zero normal velocity, but inviscid theory does not impose no slip. Taking its steady projection along a streamline gives $p/\rho +v^2/2+gz=\mathrm{constant}$. If the flow is also irrotational in the connected region, that constant is shared between streamlines. Inviscid does not mean irrotational: rotating ideal-fluid flows are possible. A velocity potential exists locally for an irrotational field, and in incompressible flow it obeys Laplace's equation. The Bernoulli assumptions must be checked before converting a pressure difference into a speed.^[ref]

Real fluids transmit shear stress. An incompressible Newtonian fluid with constant dynamic viscosity $\eta$ obeysThe constitutive shear stress is proportional to strain rate, and stationary solid boundaries usually impose no slip. viscosity diffuses vorticity and converts organized mechanical energy into heat; its local dissipation $2\eta e_{ij}{e}_{ij}$ is nonnegative. It cannot generally be included merely by changing the constant in a lossless Bernoulli equation.

In a long straight circular tube, steady fully developed laminar flow hasThis Hagen-Poiseuille equation follows by balancing the axial pressure gradient with viscous shear and setting $v_z(a)=0$. The parabolic profile, the fourth-power radius sensitivity and the pumping power $\Delta pQ$ are experimentally distinct consequences. Entrance regions, rough walls, non-Newtonian response or a changing temperature require additional modelling.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscosity. Small values favor creeping-flow balances; high values permit thin boundary layers, separation and turbulence, although transition depends on geometry and perturbations, not one universal critical number. At high Reynolds number an outer flow can be approximately inviscid while near-wall viscosity still determines drag. This resolves why a zero-drag ideal potential-flow model cannot predict the resistance of an ordinary moving body. In turbulent flow, fluctuating Reynolds stresses redistribute momentum and increase the required pressure drop; averaging the equations introduces closure requirements.

A controlled tube experiment measures pressure difference and collected volume per time while changing forcing. Use tracers or a velocity probe to test profile shape, allow sufficient entrance length, and monitor radius and temperature independently. Repeat with reversed flow and at several rates to test offsets and linearity. Dye mixing alone does not measure viscosity or establish a fully developed velocity profile. The original figure below separates a specified laminar velocity profile from an independent pressure/discharge measurement.

Use $c_p,c_V$ for the heat capacities of the entire $n$-mole sample. For a stable simple compressible gas, a Maxwell relation giveswhere ${\alpha }_V=V^{-1}(\partial V/\partial T{)}_p$ and ${\kappa }_T=-V^{-1}(\partial V/\partial p{)}_T>0$. This heat-capacity difference and mechanical stability explains the sign without assuming the internal energy depends only on temperature. It is strictly positive for an ordinary expanding gas; the unrestricted word “always” needs the qualification that equality is possible when thermal expansion vanishes.

For an ideal gas, thermodynamic consistency givesConsequently $U=U(T)$, while $H=U+pV=U+nRT$. Differentiating gives $c_p-c_V=nR$: at fixed pressure extra supplied heat permits expansion work in addition to the internal-energy increase.

Differentiate the supplied entropy at fixed $n$:Thus $U=5nRT/2$, consistently recovering $pV=nRT$, andThe coefficient in this entropy corresponds to five active quadratic thermal degrees of freedom per molecule. It should not be replaced by the monatomic value. Reference scales needed to make logarithms dimensionless are absorbed into the entropy constant and do not affect these derivatives.

With open windows the house equilibrates mechanically at approximately the same atmospheric pressure before and after heating. It is an open system: its mole number is not fixed. The isobaric ideal-gas internal-energy density isAt a fixed house volume, $n_{300}/n_{273}=273/300=0.91$: fewer molecules occupy the house, each with greater mean energy. Heating energy can enter while enthalpy leaves with escaping air. Treating the room as a sealed fixed-$n$ gas would give the wrong energy-density comparison.

Keep the problem's one-mole convention throughout. The excluded-volume parameter $b$ represents the loss of available translational volume due to short-range repulsion; it is an effective covolume, not simply the literal molecular material volume. The attraction parameter $a$ reduces the pressure below its purely repulsive value. In a mean-field picture, interaction energy is proportional to pair density, so the attractive free-energy contribution scales as $-a/V$ for one mole and the pressure correction as $-a/V^2$. Equivalently, molecular cohesive forces and the number striking a wall each supply a density factor. These parameters approximate microscopic correlations rather than measuring every pair potential exactly.

At fixed temperature, $dF=-pdV$. Substituting $p=RT/(V-b)-a/V^2$ and integrating gives the isothermal free-energy change of a Van der Waals gasThe arbitrary temperature-only term cancels. The internal-energy derivative is insteadExpansion raises the attractive internal energy toward zero. As a check, the entropy change is $R\ln [(V_2-b)/(V_1-b)]$, so $\Delta F=\Delta U-T\Delta S$. Constant $a,b$ and a homogeneous branch with $V>b$ are assumed. If a subcritical equilibrium path crosses coexistence, a Maxwell construction replaces the unstable homogeneous continuation; the displayed integral is the direct equation-of-state result, not a claim that such a continuation is physically stable.

At the Van der Waals critical point, the critical isotherm has a stationary inflection:Equating the resulting expressions for $R{T}_c$ gives $2V_c=3(V_c-b)$, henceIn reduced variables $x=V/V_c$, the critical curve is $p/p_c=8/(3x-1)-3/x^2$: it decreases through $(1,1)$ with zero first and second derivative, rather than having separate liquid/gas turning points at the critical temperature.

In the Carnot cycle, let $1\to 2$ be reversible isothermal expansion at $T_h$, $2\to 3$ reversible adiabatic expansion, $3\to 4$ isothermal compression at $T_c$, and $4\to 1$ adiabatic compression. The pressure–volume curve is clockwise. The entropy–temperature figure below uses temperature on the horizontal axis and entropy on the vertical axis, literally an $S$–$T$ plot: isotherms are vertical and adiabats horizontal. The hot isotherm increases entropy by $\Delta S$ and absorbs $Q_h=T_h\Delta S$; the cold one rejects $Q_c=T_c\Delta S$. Thus $W=(T_h-T_c)\Delta S$ and ${\eta }_C=1-T_c/T_h$.

For the second engine, choose the working direction with heat addition during $A\to B$ at $p_1$ and rejection during $C\to D$ at $p_2$. A reversible ideal-gas adiabat obeys $T\propto p^{(\gamma -1)/\gamma }$. DefineThe two constant-pressure heat transfers are $Q_{\mathrm{in}}=c_p(T_B-T_A)$ and $Q_{\mathrm{out}}=c_p(T_C-T_D)=Q_{\mathrm{in}}/r$, where rejected heat is quoted positive. Therefore the Brayton cycle efficiency isfor $\gamma =5/3$. All heat-capacity amount conventions cancel in the ratio.

Since $T_B>T_A$ and $T_C=T_B/r$, $T_D=T_A/r$, the maximum is at B and the minimum at D. The Brayton-Carnot efficiency comparison isThe inequality is strict for a nondegenerate cycle with positive heat input. Although each process can be reversible, the isobaric heat transfer spans a range of temperatures; making the whole cycle reversible requires a succession of matched-temperature reservoirs, rather than just the hottest and coldest reservoirs.

For an ideal diatomic gas with active rotation and frozen vibration, $\gamma =7/5$ andHere $0<p_2/p_1<1$. This comparison holds at the same pressure ratio and assumes temperature-independent heat capacities in the stated mode regime. Frozen rotation or excited vibration would require their actual heat capacities.

A microstate specifies the microscopic state: classically the positions and momenta of all particles, or quantum mechanically an allowed many-body state. A macrostate specifies a few coarse variables such as energy, volume and particle number. Many microstates can give the same macrostate. A statistical ensemble is a probability assignment over hypothetical copies prepared with the same macroscopic constraints; it does not require a sample to be literally divided into many physical copies. An observable's ensemble mean is $⟨A⟩={\sum }_i{p}_i{A}_i$, with an integral replacing the sum in classical phase space.

For an isolated system at fixed $U,V,N$, the microcanonical ensemble assigns equal probability to accessible microstates in a narrow energy shell. If their number is $\Omega (U,V,N)$, the Boltzmann entropy is $S=k_B\ln \Omega$. Thermal contact between two initially isolated systems allows energy to be shared while their sum stays fixed. The most probable division maximizes $S_1(U_1)+S_2(U-U_1)$, so equilibrium requires equal $\partial S/\partial U$. This gives the statistical temperature definitionFor positive temperature, energy tends to flow from lower entropy slope to higher entropy slope until those slopes agree. The narrow energy-shell width represents macroscopic resolution; it must not exclude the physically accessible states.

For a small subsystem exchanging energy with a much larger bath, the bath has energy $U_{\mathrm{tot}}-E_i$ when subsystem microstate $i$ has energy $E_i$. Expanding its entropy gives $S_b(U_{\mathrm{tot}}-E_i)=S_b(U_{\mathrm{tot}})-E_i/T+\cdots$. Counting compatible bath states therefore yields the Boltzmann distribution and canonical ensemble:The exponential weights individual microstates. An energy level of degeneracy $g_j$ has total probability $g_j{e}^{-\beta E_j}/Z$; a more degenerate excited level can outweigh one ground-state level even though each of its microstates is less probable. The Gibbs entropy $-k_B{\sum }_i{p}_i\ln p_i$ gives $S=k_B\ln Z+U/T$. The equilibrium energy variance is $⟨(\Delta E{)}^2⟩=k_B{T}^2{c}_V$, linking measurable fluctuations to a macroscopic response.

If particle exchange is also allowed, a reservoir fixes $T$ and chemical potential $\mu$. The grand canonical ensemble weights a state by $e^{-\beta (E_i-\mu N_i)}/\Xi$, with $\Xi ={\sum }_{N,i}{e}^{-\beta (E_i-\mu N)}$. Mean particle number is $k_{B}T\partial \ln \Xi /\partial \mu$, while volume remains a constraint in this version. Thus the ensemble follows the physical exchange conditions: fixed energy, thermal contact, or thermal and particle contact.

In ordinary large short-range systems, relative fluctuations often decrease as $N^{-1/2}$ and the ensembles give the same bulk thermodynamics away from singular limits. Finite systems, critical fluctuations and long-range interactions require care. Interpreting a time average as an ensemble average additionally requires appropriate sampling or mixing dynamics; it is not guaranteed solely by naming an ensemble. thermal equilibrium probability assignments do not describe every driven or relaxing state.

Original examination: Physical PDF pages 283–291; Physics B Paper 2, 8 June 2011. All questions and both field forms in B9 are covered. Approximations not fixed by the wording are stated explicitly.

For the entropy balance of a thermostatted resistor, electrical work becomes heat $Q=I^{2}Rt=2.0\times 10^7\mathrm{J}$. At fixed temperature the resistor has no net change of thermodynamic state; it exports this heat to its thermostat. Treating the electrical supply as a source of work,Entropy is produced by Joule heating even though it does not accumulate in the resistor. Additional entropy production inside a nonideal power supply is outside the specified model.

For a uniformly magnetized ring with a narrow gap, take the local direction of magnetization as positive around the magnetic circuit. Neglect leakage and fringing in the narrow gap, and take the cross section as uniform. Continuity of magnetic flux gives $H_g=H_r+M$. With no free current, Ampère's law gives $l{H}_r+a{H}_g=0$. HenceThe rod's field is a demagnetizing field opposite its magnetization; its magnitude is $aM/(l+a)$. The gap field points with the magnetization and $B_g={\mu }_0{H}_g$.

Put the positive plate at $z=d/2$ and the negative plate at $z=-d/2$. Reflection through $z=0$, combined with reversal of the two plate charges, makes the tangential fields cancel at every point of the mid-plane. Thus $E_x=E_y=0$ at the corner as well as inside the capacitor.

For the uniform-charge estimate at a capacitor corner, assemble four identical square pairs around their common corner. Their normal fields there are equal by rotation symmetry and add to the central field $\sigma /{ϵ}_0$ of the larger plate pair. With the bulk parallel-plate relation $\sigma \simeq {ϵ}_{0}V/d$, the conventional corner estimate isThis quarter-field result uses prescribed uniform surface charge, not symmetry alone for finite equipotential conductors. Real conducting plates redistribute charge near their edges; $d\ll a$ justifies the bulk relation away from the edges but does not remove the local corner fringe. For that exact conductor problem only the zero tangential components follow from the stated reflection symmetry; its normal coefficient requires a fringing-field boundary-value solution. The standard parallel-plate formula explicitly neglects edge effects.^[ref]

For a monatomic ideal gas, $\gamma =C_P/C_V=5/3$. If the expansion is reversible adiabatic, $T{V}^{\gamma -1}$ is constant, soAdiabatic isolation alone does not imply this temperature law. For example, free expansion into vacuum has $Q=W=0$, so an ideal gas remains at $300\mathrm{K}$. Thus $75\mathrm{K}$ is the usual intended answer with the additional isentropic assumption; the literal adjective “adiabatic” does not uniquely select it.^[ref]

Choose $z$ normal to the sheet and $x$ along the in-plane projection of the rotation axis. A uniform square has $I_x=I_y=m{a}^2/12$ and $I_z=m{a}^2/6$. Since ${\omega }_x={\omega }_z=\omega /\sqrt{2}$, the two nonzero angular-momentum components are $J_x=I_x\omega /\sqrt{2}$ and $J_z=I_z\omega /\sqrt{2}$. ThereforeAngular momentum is not parallel to the prescribed rotation axis. Multiplying the moment of inertia about that axis by $\omega$ would give the projection of angular momentum on the axis, not its magnitude.

Apply Faraday's law to a shrinking rectangular loop crossing the interface. For bounded fields the magnetic-flux term vanishes with its area, giving ${\mathbf{E}}_{t,+}={\mathbf{E}}_{t,-}$. A shrinking Gaussian pillbox similarly gives $D_{n,+}-D_{n,-}={\sigma }_{\mathrm{free}}$. Thus normal $D$ is continuous when no free surface charge is present, even though a bound polarization charge may remain. These are the assumptions needed for the stated dielectric-interface boundary conditions.

Write $ϵ={ϵ}_r{ϵ}_0$, $r_{+}=\sqrt{(x-a{)}^2+y^2+z^2}$ and $r_{-}=\sqrt{(x+a{)}^2+y^2+z^2}$. Use each region's own permittivity in its Coulomb potential:At $x=0$, continuity of potential and normal displacement givesSolving,This permittivity convention for transmitted image charges matters: if the vacuum Coulomb prefactor is used even for ${\varphi }_{-}$, its combined effective source is $2Q/(1+{ϵ}_r)$ instead. The physical potential is the same. The two numbered leaves identify how these sources satisfy the equations in their respective regions.

The real charge feels only the image contribution after removing its own singular self-field. The image at $(-a,0,0)$ is separated from it by $2a$, soFor the bound charge, the outward normal of the dielectric is $+\hat{x}$ and $\mathbf{P}={ϵ}_0({ϵ}_r-1){\mathbf{E}}_{-}$. Differentiating ${\varphi }_{-}$ at $x=0$ gives $E_{-,x}=-a(Q+Q^{\prime })/[4\pi ϵ(a^2+y^2+z^2{)}^{3/2}]$. ThereforeIts integral over the plane is $Q^{\prime \prime }$, consistent with the external image representation. For ${ϵ}_r=1$ both the image force and induced surface charge vanish.

For the dielectric half-space the two formal sources coincide at $(a,0,0)$, outside the region being represented. Their sum is $Q+Q^{\prime }=2{ϵ}_{r}Q/(1+{ϵ}_r)$ in the dielectric Coulomb kernel, henceThere is no charge singularity inside this half-space. The expression satisfies Laplace's equation there, decays at infinity, and matches the vacuum solution's tangential electric field and normal electric displacement. The coincident image is a mathematical representation, not an extra physical charge in the vacuum region.