Past exam of the physics B course of the University of Cambridge / 2010 / Paper 1 / Section A

For a fixed linear circuit, self-inductance relates the magnetic flux linkage produced by its own current to that current: $\Lambda =LI$, with $\Lambda ={\sum }_{\mathrm{turns}}{\Phi }_B$. The self-induced electromotive force is $\mathcal{E}=-d\Lambda /dt=-LdI/dt$ when $L$ is constant. Its unit is the henry, $1\mathrm{H}=1Wb/A$.

mutual inductance relates the linkage in one circuit to current in another: ${\Lambda }_2=M_{21}{I}_1$ and ${\Lambda }_1=M_{12}{I}_2$. For stationary reciprocal linear media, $M_{12}=M_{21}=M$ with consistently oriented turns, and ${\mathcal{E}}_2=-Md{I}_1/dt$. Orientation fixes the sign. The coupled magnetic energy is $W=\frac{1}{2}{L}_1{I}_1^2+\frac{1}{2}{L}_2{I}_2^2+M{I}_1{I}_2$, so positivity requires $M^2\le L_1{L}_2$. Moving coils or nonlinear cores require differentiation of the full linkage rather than treating the inductance as a constant.

Use local east, north and up axes. At latitude $\lambda$, $\mathbf{\Omega }=(0,\Omega \cos \lambda ,\Omega \sin \lambda )$ and eastward velocity is $(v,0,0)$. The Coriolis acceleration isIts horizontal part is southward, so the lift must acquire a northward component: bank left, with the north wing lower. For the requested local Coriolis correction, let the vertical lift balance $mg$. The Coriolis bank angle for eastward flight isUse $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\Omega =7.2921\times 10^{-5}{\mathrm{s}}^{-1}$. Including the upward Coriolis term replaces $g$ by $g-2\Omega v\cos \lambda$ and changes the answer by only about $0.1\%$. This is the horizontal Coriolis compensation, not a full calculation of a curved constant-latitude flight path. The rotation rate follows Earth's sidereal period.^[ref]

Let $Q_c$ be heat extracted from the cold reservoir, $Q_h$ the heat delivered to the surroundings and $W$ the supplied work. The entropy inequality for a cyclic refrigerator isEquality is the reversible Carnot refrigerator limit. Thereforeenergy conservation gives $W_{\min }=Q_h-Q_c=74\mathrm{J}$. The requested rejected heat is 75 J, while 74 J is the minimum work; both have to be distinguished. Any entropy production increases the required rejection at these fixed reservoir temperatures.

For a plane wave propagating along $\hat{\mathbf{k}}$ in vacuum, Maxwell equations give $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$, with $\mathbf{E}\perp \hat{\mathbf{k}}$. HenceTaking $E=E_0\cos (\mathbf{k}\cdot \mathbf{r}-\omega t)$, the period average of the squared cosine is $1/2$. ThusThe fields are in phase, so the travelling-wave energy flux never reverses. The peak amplitude convention is essential; using a root-mean-square electric field would remove the factor $1/2$.

Use a circular Kepler orbit to eliminate the terrestrial gravitational parameter: $T^2/r^3$ is constant for both satellites. The reference orbital radius is $r_1=(6400+150)\mathrm{km}=6550\mathrm{km}$, measured from Earth's centre. A geostationary orbit is circular, equatorial and prograde, with the sidereal rotation period, approximately $86164\mathrm{s}$. ThusIts altitude is about $3.61\times 10^4\mathrm{km}$ above the stated surface. The approximate low-orbit input sets the accuracy; replacing it by a modern measured gravitational parameter would answer a different numerical estimate. A $24$-hour approximation changes the radius by less than $0.2\%$. Matching the period alone would not make an inclined or eccentric orbit stationary above one point.