Past exam of the physics B course of the University of Cambridge / 2013 / Paper 2 / Section A

For the force on a current loop in a bilinear magnetic field, the arrow gives current in the positive $y$ direction on the edge $x=a$, and in the negative $x$ direction on the edge $y=a$. Use $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$. The edges on the coordinate axes contribute zero because the applied field vanishes there. The other two giveThus the force on the square itself, excluding the supply leads, isIt lies in the plane along the diagonal towards increasing $x$ and $y$, at $45^{\circ }$ to either positive axis.

In the ideal semidiurnal tide model, successive high tides occur after the Earth–Moon relative angle advances by $\pi$. Since the motions have the same sense, subtract their angular speeds:The lunar orbital period is sidereal, so the consistent terrestrial rotation period is $T_E=23.9345\mathrm{h}$. With $T_M=27.3\times 24\mathrm{h}$, this gives about 12 h 25 min. The sidereal terrestrial period is distinct from the 24-hour solar day.^[ref]

If the Earth's rotation is approximated as one revolution in 24 hours, the same calculation instead gives 12 h 27 min. That small difference is the day convention, not a change of sign in the relative angular speed. Actual coastal tide times also depend on ocean dynamics and the solar tide.

For the uniform magnetic field in an off-axis cylindrical cavity, let $\mathbf{b}$ point from the conductor's axis to the hole's axis and let the current flow in $+z$. Its uniform current density in the metal isSuperpose a complete cylinder carrying $\mathbf{J}$ and a cylinder occupying the hole carrying $-\mathbf{J}$. Inside a uniformly filled cylinder, Ampère's law gives $\mathbf{B}=({\mu }_0/2)\mathbf{J}\times \mathbf{r}$. For any point in the cavity,There is no remaining position dependence, so the field is uniform. Its magnitude is ${\mu }_{0}Ib/[2\pi (R^2-a^2)]$, perpendicular to the axis-offset vector in the cross-sectional plane. Reversing the current reverses the field.

Use the rotating reference frame equilibrium model, in which the centrifugal potential energy of a particle of mass $m_i$ is $U_i(r)=-m_i{\omega }^2{r}^2/2$. The Boltzmann distribution then givesDefine the concentration ratio as $c=n_1/n_2$. At the axis,Thus the axis favours the lighter species. The expression assumes a dilute classical mixture at one temperature, equilibrium in the rotating reference frame, and sampling sufficiently close to $r=0$.

Define $q=N_{+}-N_{-}$, where $N_{+}$ counts particles in the $+ϵ$ level. Then $N_{\pm }=(N\pm q)/2$, soThe binomial entropy of independent two-level particles follows from counting which of the $N$ independently labelled particles occupy each nondegenerate level. This is the usual distinguishable, or localized, two-level model; occupation constraints of a degenerate quantum gas would require different counting.

For large occupations, Stirling's approximation with $x=q/N$ givesAllowed $q$ runs from $-N$ to $N$ in steps of two. Entropy is largest near equal populations and is zero when all particles occupy one level; use $0\ln 0=0$ at the endpoints.