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Let $K=e{U}_a$, where $U_a=100\mathrm{V}$, and let $v_z$ be the incident electron speed. Nonrelativistically, $m{v}_z^2=2e{U}_a$. Take the plate length along the beam as $\ell =0.01\mathrm{m}$, their separation as $d=0.01\mathrm{m}$, and their centre-to-screen distance as $D=0.30\mathrm{m}$. A signed deflecting voltage $V_d$ gives the transverse acceleration $a_x=e{V}_d/(md)$, choosing its sign so that positive $V_d$ deflects the electron towards positive $x$. The transit time is $t_p=\ell /v_z$. The displacement inside the plates and in the subsequent field-free drift isThus the two screen edges requireThe transverse deflection at plate exit is $x_p=x\ell /(2D)=4.17\mathrm{mm}$ at either extreme, smaller than the half-gap $5\mathrm{mm}$, so the beam clears the plates. The outgoing slope is $v_x/v_z=x/D\simeq 0.833$; no small-angle replacement for the trajectory was used. The kinetic energy remains far below $m{c}^2$, so the nonrelativistic approximation is adequate. If “30 cm behind” is instead measured from the downstream plate edge, replace $D$ by $0.305\mathrm{m}$, giving approximately $\pm 164\mathrm{V}$; the intended thin-deflector estimate is about $\pm 170\mathrm{V}$.

Choose $\mathbf{B}=B\hat{\mathbf{z}}$ and a signed tangential velocity $\mathbf{v}=u\hat{\mathbf{\varphi }}$, with $u>0$ counterclockwise viewed from $+z$. The magnetic part of the Lorentz force is $euB\hat{\mathbf{r}}$, while the electric part is $-eV\hat{\mathbf{r}}/(\eta r)$. The radial acceleration for a circular orbit is $-u^2/r$, regardless of the sign of $u$. ConsequentlyOne valid signed-velocity convention therefore has $a=m$, $b=eBr$, $c=-eV/\ln (r_2/r_1)$. If $v$ is required to be a nonnegative speed, the two senses of rotation instead satisfywith the plus sign for counterclockwise motion. For positive $V$, the electric force is inward, so it can supply circular motion in either direction; in one direction the magnetic force opposes it, while in the other it helps. A signed negative root must not be discarded as an unphysical speed; it denotes the opposite orbit direction.

With $V=+20\mathrm{V}$ in the outer-minus-inner convention, $\eta =\ln 1.5=0.405465$ and the radial electric field at the orbit is $E_r=-986.52\mathrm{V}{\mathrm{m}}^{-1}$. The signed roots areUsing the proton mass and charge givesThe speeds are therefore $48.8$ and $96.7\mathrm{km}{\mathrm{s}}^{-1}$, in opposite senses of rotation. Both are nonrelativistic. The slower counterclockwise orbit has an outward magnetic force; the faster clockwise orbit has an inward magnetic force. Reversing $\mathbf{B}$ interchanges the senses, while reversing the electrode polarity changes the quadratic and may eliminate real orbits; its sign must therefore be specified.

Let $U>0$ be the drop in electric potential along the positive electric charge's path. By conservation of energy, $qU=m{v}^2/2$, so the nonrelativistic exit speed is $v=\sqrt{2qU/m}$. Apply a uniform magnetic field perpendicular to the incident velocity and to the desired orbital plane. The magnetic Lorentz force is perpendicular to the velocity, does no work, and supplies the centripetal force:For example, $\mathbf{v}=v\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{z}}$ give initial deflection along $-y$ for positive $q$. Reversing the magnetic field reverses the bending. A uniform electric field alone would give constant acceleration and a parabolic path, not uniform circular motion. The displayed result assumes $qU\ll m{c}^2$; the exact relativistic momentum relation would give $R=\sqrt{q^2{U}^2/c^2+2mqU}/(qB)$.

Add a uniform electric field satisfying $\mathbf{E}=-{\mathbf{v}}_0\times \mathbf{B}$. Then the Lorentz force $q(\mathbf{E}+{\mathbf{v}}_0\times \mathbf{B})$ vanishes for the desired velocity. With the axes used above, $\mathbf{E}=E\hat{\mathbf{y}}$, perpendicular to both ${\mathbf{v}}_0$ and $\mathbf{B}$, withThe crossed fields form a velocity selector. All particles must first have traversed the same electric potential difference from rest: that is what converts selection of speed into selection of charge-to-mass ratio. Equal charge-to-mass ratios yield equal speeds even if the separate masses and charges differ. For other ratios, the initial transverse force is $q(E-vB)$ and is nonzero.

State a time convention before using complex response: take fields proportional to $e^{i(kz-\omega t)}$, with $k=k_r+i\kappa$ and $\kappa \ge 0$ for a passive wave entering the material. In a homogeneous Ohmic medium, Maxwell equations with $\mathbf{J}=\sigma \mathbf{E}$ giveThe ohmic-conductor complex wavenumber describes both phase advance and amplitude decay. Real constant conductivity is only a low-frequency approximation. In the Drude model, $m\dot{\mathbf{v}}+m\gamma \mathbf{v}=-e\mathbf{E}$ giveswith a fixed positive ionic background and no additional bound-electron contribution in this simple model. The plasma frequency marks the inertial electron-response scale. More elaborate band structure, interband transitions and nonlocal transport limit this scalar treatment.^[ref]

In the collisionless nonmagnetic limit, the dispersion relation is ${\omega }^2={\omega }_p^2+c^2{k}^2$. For $\omega >{\omega }_p$,A crest can move faster than $c$ without carrying a new signal faster than light. Below cutoff the ideal wave is evanescent, with penetration length $c/\sqrt{{\omega }_p^2-{\omega }^2}$; it does not propagate as an ordinary real-$k$ bulk wave. Evanescence in the lossless limit is not Joule dissipation. Nonzero collision rate gives a dissipative conductivity and smooths this ideal cutoff picture.

For a good conductor with approximately real $\sigma \gg \omega ϵ$, the spatial branch becomesA field amplitude falls as $e^{-z/\delta }$, while intensity falls as $e^{-2z/\delta }$. The phase velocity is $\omega \delta$ in this limit. Magnetic and electric amplitudes are not related by the vacuum ratio: $Z_s=E/H=(1-i)\sqrt{\mu \omega /(2\sigma )}$ with the chosen time convention. Its positive real part is the surface resistance. The local mean heating is $\frac{1}{2}\mathrm{Re}\sigma |E{|}^2$, consistent with electromagnetic energy flowing into the metal. skin depth is an amplitude length; interpreting it as an intensity length gives a factor-of-two error.

A transmission-line mode propagates chiefly through the dielectric between conductors, even though finite metal conductivity causes loss at its boundaries. With per-length parameters $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$, the telegrapher's equations give, for the same $e^{-i\omega t}$ convention,Writing a forward wave as $e^{-\Gamma z}$, $\Gamma =\alpha -i\beta$, gives ${\Gamma }^2=(R^{\prime }-i\omega L^{\prime })(G^{\prime }-i\omega C^{\prime })$. In the low-loss regime $R^{\prime }\ll \omega L^{\prime }$ and $G^{\prime }\ll \omega C^{\prime }$, set $Z_0\simeq \sqrt{L^{\prime }/C^{\prime }}$ to obtainThis low-loss transmission-line attenuation separates series conductor loss from dielectric leakage loss. Average forward power falls as $e^{-2\alpha z}$. In the normal skin-effect regime, reduced conducting area makes the AC conductor resistance increase roughly as $\sqrt{\omega }$; dielectric loss can supply a separate frequency dependence. Reflection from a mismatched termination changes delivered power without being, by itself, dissipation in the line.^[ref]

To test loss, compare several lengths of otherwise identical line with matched terminations, calibrate connectors and incident/reflected power, and monitor temperature. Separate amplitude attenuation from phase delay and from standing-wave minima. Bulk penetration, line attenuation and dielectric propagation speed are related electromagnetic quantities but describe different experimental geometries.

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.

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]

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.

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.

For a moving loop, Faraday's law of induction is $\mathcal{E}=\oint (\mathbf{E}+\mathbf{u}\times \mathbf{B})\cdot d\mathbf{l}=-d{\Phi }_B/dt$, where $\mathbf{u}$ is the local wire velocity, ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$, and the oriented boundary and surface follow the right-hand rule. For a fixed loop the motional term vanishes. Lenz's law is the minus sign: the induced response opposes the change of linked flux, not necessarily the applied field itself.

Let $s$ increase downhill and choose positive current $I$ to produce an uphill magnetic force. The normal component of the imposed field is $B\cos \theta$, so the changing external flux is $Bl\cos \theta s$. With initially zero current and no resistance, conservation of flux linkage gives $LI=Bl\cos \theta s$. The rod feels gravity vertically downward, a rail reaction normal to the slope, and magnetic force of magnitude $IlB$ horizontally uphill. Its downhill component is $-IlB\cos \theta$; its normal component presses the rod into the rails, giving $N=mg\cos \theta +IlB\sin \theta$.

The inductive oscillation of a sliding rod on inclined rails follows fromThe magnetic force is not itself parallel to the rails. Defining ${\omega }_B=Bl\cos \theta /\sqrt{mL}$ and $s_e=mg\sin \theta /(m{\omega }_B^2)$, release from rest givesThe first stop occurs at $t=\pi /{\omega }_B$; the rod then returns and oscillates between $0$ and $2s_e$. At $s_e$ the force is zero but the speed is maximal. Mechanical potential energy is reversibly exchanged with kinetic energy and $L{I}^2/2$; a perfectly conducting circuit provides no damping.

For the resistive long-time drift of an inductive sliding rod, with loop resistance $R>0$, useJoule heating damps the transient oscillation. On sufficiently long rails the asymptotic motion is steady sliding, not permanent rest:The zero-resistance model has persistent oscillations; its late-time limit cannot be interchanged with taking $R\to 0$. These results assume $B\cos \theta \ne 0$, smooth contacts, negligible rod self-inductance and constant geometry.

Locally, zero curl permits $\mathbf{v}=\nabla \Phi$; for a global single-valued potential the circulation around every noncontractible loop must also vanish. incompressibility gives $\nabla \cdot \mathbf{v}={\nabla }^2\Phi =0$. The exterior of a cylinder is multiply connected, so local irrotationality alone does not exclude a circulating solution.

For inviscid constant-density flow under conservative gravity, Euler's equation is ${\partial }_t\mathbf{v}+(\mathbf{v}\cdot \nabla )\mathbf{v}=-\nabla (p/\rho +{\varphi }_g)$. The identity $(\mathbf{v}\cdot \nabla )\mathbf{v}=\nabla (v^2/2)-\mathbf{v}\times (\nabla \times \mathbf{v})$ gives, for irrotational flow,For steady flow, absorb the potential's time-only gauge term and obtain $p+\rho v^2/2+\rho {\varphi }_g=\mathrm{constant}$ throughout the connected fluid region. Without irrotationality the steady constant need only be common along each streamline.

Use $U$ for the uniform upstream speed and take the flow along $+x$. The boundary conditions are $v_r(a,\theta )=0$ and $\mathbf{v}\to U{\mathbf{e}}_x$ at infinity. A perfect inviscid model imposes no penetration, not no slip. For the intended no-circulation solution, require a globally single-valued potential, so the coefficient of $\theta$ is zero. The logarithmic term would be a net source and violates no penetration, so its coefficient is also zero. Growing terms with $n\ge 2$ violate the uniform far field. For each remaining $n\ge 2$, the surface normal condition then forces its decaying coefficient to vanish. The $n=1$ terms satisfy $C_1=U$ and $D_1=U{a}^2$:An irrelevant additive potential constant is understood. At the same elevation, Bernoulli balance with the upstream stream givesThe force per unit height ispressure symmetry gives zero drag and zero lift in this zero-circulation idealization. D'Alembert's paradox is the failure of this inviscid model to represent real separated viscous wakes and their drag, not a claim that a real pole feels no force.

The zero-circulation assumption matters: $\Phi \to \Phi +\Gamma \theta /(2\pi )$ leaves the stated velocity boundary conditions intact but adds $\Gamma /(2\pi a)$ to the surface tangential speed. It produces $F_y^{\prime }=-\rho U\Gamma$ for positive counterclockwise circulation. Thus the boundary conditions alone do not justify excluding the angular term; an initially noncirculating inviscid start, or an explicitly single-valued potential, supplies the missing condition.^[ref]

In vacuum, displacement current contributes the additional term ${ϵ}_0\partial \mathbf{E}/\partial t$:Taking the divergence and using Gauss law gives $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the local conservation of charge. Without displacement current the curl equation would incorrectly require a divergence-free current even during charge accumulation.

For a charging capacitor, a surface cutting the wire carries conduction current while a surface passing through the gap carries changing electric flux. The displacement current term makes the circulation independent of the spanning surface. It is not transport of free charge across the vacuum gap. In matter the macroscopic equation is $\nabla \times \mathbf{H}={\mathbf{J}}_f+\partial \mathbf{D}/\partial t$, with electric polarization included in $\mathbf{D}$. Together with Faraday's law of induction, the vacuum equations give electromagnetic waves of speed $1/\sqrt{{\mu }_0{ϵ}_0}$.

Viscosity transfers momentum between adjacent fluid layers. A Newtonian fluid has shear stress proportional to strain rate: for simple shear, $\boxed{{\tau }_{xy}=\eta \partial u_x/\partial y}$. Thus parallel plates of area $A$, separation $d$ and relative speed $U$ require force magnitude $\eta AU/d$ in a steady linear profile.

For an incompressible isotropic Newtonian fluid,for constant viscosity. With strain-rate tensor $e_{ij}$, the viscous dissipation rate per unit volume is $2\eta e_{ij}{e}_{ij}\ge 0$. Kinematic viscosity $\nu =\eta /\rho$ is the momentum diffusivity: a disturbance spreads over distance of order $\sqrt{\nu t}$.

The Reynolds number $\rho UL/\eta$ compares inertial and viscous forces. Small Reynolds number supports a creeping-flow approximation, while large values permit thin viscous boundary layers and possible turbulence. It does not mean molecular viscosity has disappeared. Gas viscosity results from molecular momentum transport; a liquid's microscopic transport depends strongly on intermolecular interactions. Non-Newtonian materials require a more general constitutive law.

For a homogeneous isotropic solid with small displacement $\mathbf{u}$, linear elasticity gives ${\sigma }_{ij}={\lambda }_L(\nabla \cdot \mathbf{u}){\delta }_{ij}+2\mu e_{ij}$. Momentum balance becomesA plane wave with displacement parallel to its wavevector is longitudinal, whereas perpendicular displacement is transverse. Writing $K={\lambda }_L+2\mu /3$ for the bulk modulus and $\mu$ for the shear modulus, their bulk speeds areThere are two independent transverse polarizations. The bulk modulus supplies resistance to compression; shear rigidity supplies transverse restoring stress. An ordinary fluid has no static shear rigidity and cannot sustain these propagating bulk shear waves. A thin rod with freely contracting sides instead has longitudinal speed $\sqrt{Y/\rho }$, because its lateral boundary conditions differ from a bulk plane wave. Reflection, refraction and mode conversion occur at interfaces; a free surface also supports Rayleigh waves.

For speed distribution $f(v)$ normalized to one and isotropic directions, the incident number flux isThe positive normal velocity weights the population reaching the wall. Consequently its mean translational energy isThe bulk value $3k_{B}T/2$ would omit the arrival weighting.

For the two vessels, let $r=\sqrt{T_1/T_0}$, with subscripts h and c for hot and cold. Equal opposing effusion rates give $p_h/\sqrt{T_1}=p_c/\sqrt{T_0}$. Number conservation in equal, fixed volumes adds $p_h/T_1+p_c/T_0=2p_0/T_0$. HenceBoth final pressures must be determined; holding the cold temperature fixed does not hold its particle number fixed.

For aperture area $A_h=\pi d^2/4$, the common one-way flux is $J=p_c/\sqrt{2\pi m{k}_B{T}_0}$. The translational heat transfer from hot to cold is thereforeThis is the total heat transfer for a monatomic gas with no active internal modes, neglecting radiation and conduction through vessel walls. For a molecular gas, add $A_{h}J[u_{\mathrm{int}}(T_1)-u_{\mathrm{int}}(T_0)]$ for its internal energy per molecule. The maintained unequal temperatures describe a stationary nonequilibrium state even though the net particle current is zero.

A heat pump delivers $Q_H=Q_S+W$: electrical work moves heat from a source as well as contributing energy itself. Its heating coefficient of performance can therefore exceed one, whereas direct resistance heating supplies at most its electrical input as useful room heat. For an ideal reversible pump,with absolute temperatures and $T_S<T_H$. This is an ideal limit; actual costs also depend on compressor performance and tariffs.

For the annual ground-temperature wave, model the soil as a homogeneous semi-infinite conductor, with depth positive downward and a prescribed surface temperature. Here $\kappa$ is thermal conductivity and $C$ is heat capacity per unit volume, so the thermal diffusivity is $\kappa /C$. Conservation of energy gives $C{\partial }_{t}T=\kappa {\partial }_z^{2}T$. After transients decay, put $T-T_0=\mathrm{Re}[A{e}^{i\omega t-qz}]$. Substitution gives $q^2=i\omega C/\kappa$; the solution bounded at depth uses $q=(1+i)/\delta$. Matching the surface then givesThis approximation identifies the ground surface with air temperature and neglects the pump's perturbation of the ground, as required by the model. The attenuation and phase delay arise from heat diffusion, not propagation at one fixed wave speed.

The graph below covers a full annual cycle. At $z/\delta =\pi /2$, the amplitude is $20e^{-\pi /2}=4.16$ K and the minimum lags January by a quarter-year. At $z/\delta =\pi$, the amplitude is only $0.864$ K and the minimum lags by half a year: the ground is slightly warmer than its annual mean in January.

The seasonal source depth for a reversible heat pump has a subtle limit: a deep source approaches $T_0$ and avoids severe winter cooling, but infinite depth is not the thermal optimum of this single-frequency model. In January, with $x=z/\delta$, maximize $T_S=T_0-T_A{e}^{-x}\cos x$. The first and largest maximum is at $x=3\pi /4$, where $\cos x+\sin x=0$, giving $z=3\pi \delta /4$ and $T_S\simeq 16.34^{\circ }\mathrm{C}$. It is also the optimum of the ideal annually integrated heating work: the heating load is even about January, so its weighted sine term integrates to zero and the depth dependence is again proportional to $e^{-x}\cos x$. Installation cost, finite tube spacing, heat extraction, weather and real pump performance alter this result. The supplied model therefore supports a finite seasonal thermal optimum, while not fixing a practical burial depth in metres.

The Maxwell speed distribution is proportional to $v^2{e}^{-m{v}^2/(2k_{B}T)}$. Setting its logarithmic derivative to zero gives the most probable speed:Use molecular oxygen, with $M\simeq 0.032\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, and $T=300\mathrm{K}$.

Assume uniform charge per unit length and the same angular speed $\Omega$. A charge element at distance $r$ has $\mathbf{v}=\mathbf{\Omega }\times \mathbf{r}$, so the Biot-Savart law gives the centre field of a rigidly rotating charged loop as $dB={\mu }_0\Omega dq/(4\pi r)$. For one square side, with $b=a/2$,Using $dq=Qdx/(4a)$ on each side givesThus $B_{\square }/B_{\circ }=\ln (1+\sqrt{2})\simeq 0.8814$: the circular field is larger by $1/\ln (1+\sqrt{2})\simeq 1.135$. This is a rigidly rotating charge distribution, not a current constrained to the sides of a stationary square.