Past exam of the physics B course of the University of Cambridge / 2011 / Paper 2 / Section A

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.