Past exam of the physics B course of the University of Cambridge / 2015 / Paper 1 / Section A

For a smooth isolated conducting sphere, Gauss's law gives the surface electric field $E_s=Q/(4\pi {ϵ}_0{a}^2)$ and electric potential $V=Q/(4\pi {ϵ}_{0}a)=a{E}_s$. At the specified dielectric strength, $V_{\max }=3.0\times 10^3\mathrm{V}$. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}a$, so its capacitor energy isThis is the ideal breakdown-limited energy of a conducting sphere, taking air's permittivity approximately equal to vacuum permittivity and neglecting nearby objects and surface irregularities. The vacuum permittivity used numerically is $8.8542\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$.^[ref]

Take the initial thrower and receiver positions to be $(-a,0)$ and $(a,0)$, with counterclockwise angular velocity $\omega$. Interpret $v$ as the launch speed relative to the thrower. The ball inherits the thrower's tangential velocity, so in an inertial frame its initial horizontal velocity is $(v,-a\omega )$. The flight time across the diameter is $t_f\simeq 2a/v$. During that time the receiver moves approximately $+a\omega t_f$ sideways while the ball moves $-a\omega t_f$. Thus the transverse miss on a rotating platform isThe ball passes behind the receiver in the sense of rotation. Equivalently, in the rotating frame its transverse Coriolis acceleration has magnitude $2\omega v$ to leading order; integrating over $t_f$ gives $d=\frac{1}{2}(2\omega v)t_f^2$. Terms neglected here are higher order in $a\omega /v$. Specifying an initially radial launch velocity in the inertial frame instead would define a different throw and would lose the inherited tangential velocity.

The dynamic viscosity measures the transport of tangential momentum between adjacent layers. The kinetic estimates of gas transport coefficients givewhere $n$ is number density, $m$ the molecular mass, $\lambda$ the mean free path, and ${\sigma }_c$ a collision cross section. At fixed temperature, raising the pressure increases the number of momentum carriers but shortens their mean free path by the same factor. Consequently the dynamic viscosity of a dilute gas is approximately independent of pressure.

For a temperature-independent collision cross section, the same estimate gives $\eta \propto \sqrt{m{k}_{B}T}/{\sigma }_c$, so $\eta \propto T^{1/2}$ in the hard-sphere model. Faster molecules carry momentum across layers more rapidly. Real intermolecular interactions make the effective collision cross section temperature-dependent, changing this simple exponent. The pressure-independence also fails when wall collisions, rather than molecule–molecule collisions, set the transport distance; the continuum estimate assumes a small Knudsen number.

The varying magnetic flux through the ring induces an electromotive force by Faraday's law of induction. Its direction follows Lenz's law. With solenoid current $I_s$, ring current $I_r$, ring self-inductance $L_r$, electrical resistance $R$, and mutual inductance $M(z)$,The ring sits in a nonuniform magnetic field: coupling decreases above the solenoid, so $M>0$ and $M^{\prime }<0$ with a common orientation for the two windings. For $I_s=I_0\cos \omega t$, the in-phase part of the induced ring current is negative:Thus alternating-current induction levitation can balance the ring's weight when $⟨F_z⟩=mg$. A falling mean lifting force with height supplies vertical stability. In the strongly inductive limit the two currents are almost opposite: the ring repels the solenoid during both half cycles. Finite self-inductance matters; an ideal purely resistive loop would carry a current exactly in quadrature with $I_s$ and give zero average force in this model. The source supplies the Joule heating in the real ring, and any mechanical work while it rises.

After the photon has thermalized, its energy $E_{\gamma }=hc/\lambda$ belongs to the macroscopic body's internal energy. The body's heat capacity makes the resulting temperature change negligible. Its equilibrium entropy therefore increases byThe Boltzmann entropy $S=k_B\ln \Omega$ then gives the microstate multiplier from photon absorption:For scale, explicitly choosing a representative visible wavelength $\lambda =500\mathrm{nm}$ and room temperature $T=300\mathrm{K}$ gives $\Delta S=1.32\times 10^{-21}\mathrm{J}{\mathrm{K}}^{-1}$, or $\Delta S/k_B=95.9$, and a multiplier approximately $4.5\times 10^{41}$. These are illustrative choices, not additional supplied data: the exact multiplier depends exponentially on the particular visible wavelength and temperature. The exact Planck constant, speed of light, and Boltzmann constant fix the conversion.^[ref]