Past exam of the physics B course of the University of Cambridge / 2012 / Paper 1 (Physics B, Part IB, Paper 1, 2 June 2012)

Original paper: Public original, physical PDF pages 128–135.

Use absolute thermodynamic temperatures and assume a constant-parameter Arrhenius equation, with composition held fixed so that reaction rate is proportional to the rate constant. Write $R(T)=A{e}^{-\beta /T}$, where $\beta =E_a/\mathcal{R}$. two-temperature Arrhenius extrapolation eliminates the unknown Arrhenius prefactor and givesConsequentlyA tenfold increase for one $20\mathrm{K}$ interval does not imply another tenfold increase for the next interval: the Arrhenius plot is linear in inverse temperature. Two measured rates alone do not specify a unique extrapolation without a rate law for its temperature dependence; the result above states that modelling assumption explicitly.

Choose the plate polarity so that the gap electric field points along $+z$ when $V>0$. Away from the plate edges, the quasistatic electromagnetic approximation gives $E_z=V_0\sin \omega t/d$. A circular Ampere-Maxwell law contour of radius $r<a$ then givesIt is displacement current, rather than a conduction current across the air, that supplies the enclosed current term. The direction follows the right-hand rule for increasing positive electric flux; it reverses every half cycle. In the ideal extension that sets the gap electric field to zero outside the plate footprint, $B_{\varphi }={\mu }_0{ϵ}_0\omega V_0{a}^2\cos \omega t/(2dr)$ for $r>a$. That extension is an edge approximation, not an exact exterior solution for finite plates. Neglect fringing field of a capacitor and retardation, requiring $d\ll a$ and $\omega a/c\ll 1$.

For a hollow, air-filled rectangular electromagnetic waveguide with conducting walls, transverse confinement givesA transverse electric mode of a rectangular waveguide permits one zero index, but not two; a transverse magnetic mode of a rectangular waveguide requires both indices positive. Take $w=20\mathrm{mm}$ and $h=10\mathrm{mm}$. The dominant mode is ${\mathrm{TE}}_{10}$, withThe waveguide cutoff frequency is the limiting value: positive axial group velocity and propagating power require $f>f_c$. There is no transverse electromagnetic mode in this single hollow conducting pipe; otherwise a zero cutoff would incorrectly follow.

The Stefan–Boltzmann law gives solar luminosity $L_{\odot }=4\pi R_{\odot }^2\sigma T_{\odot }^4$. At orbital distance $r$, the incident irradiance is $\sigma T_{\odot }^4{R}_{\odot }^2/r^2$. A spherical absorber intercepts a disc of area $\pi R_M^2$ but radiates over $4\pi R_M^2$. Thus the uniform-temperature radiative equilibrium estimate isThe planetary radius cancels. This is a black-body radiation estimate with zero reflected fraction and complete redistribution of absorbed heat. More generally the balance fixes $⟨T^4{⟩}^{1/4}$, rather than the arithmetic mean of a nonuniform surface temperature. It excludes atmospheric heating and thermal lag; no measured surface temperature is inferred.

For the inertia tensor of a uniform cube, reflection symmetry makes every off-diagonal element of the central inertia tensor vanish. Permutation of the three axes makes its diagonal elements equal. Since $M=8\rho a^3$ and the mean of each squared coordinate in the cube is $a^2/3$,Every central axis is therefore a principal axis, including the body diagonal. The angular momentum is parallel to the angular velocity, and its magnitude is

In a charge-free region of homogeneous permittivity, Gauss's law and $\mathbf{E}=-\nabla V$ give Laplace's equation, ${\nabla }^{2}V=0$. Prescribed surface electric potentials specify a Dirichlet problem. Its Uniqueness of the Dirichlet problem permits replacing conductors by fictitious charges outside the physical solution region: a harmonic candidate with the same boundary conditions is the physical solution. For an exterior problem its behaviour at infinity must also be prescribed. For a proof, subtract two candidate potentials to obtain $u$ with zero boundary values. Green's first identity gives $\int |\nabla u{|}^2{d}^{3}x=0$, so $u$ is constant and its boundary value fixes it to zero. Conditions on a normal electric field instead determine a potential only up to an additive constant when the usual compatibility condition holds.

A cylindrical Gaussian surface around an infinite line charge gives $E_r=\lambda /(2\pi {ϵ}_{0}r)$. The electric potential difference is thereforeAn isolated infinite line charge has no finite potential zero at infinity. For the opposite pair, however, the logarithmic growth terms cancel and one can choose $V\to 0$ there. Define $r_b^2=(x-b{)}^2+y^2$ and $r_c^2=(x-c{)}^2+y^2$. superposition givesOn $x^2+y^2=bc$,Thus the circle, and its cylindrical extension, is an equipotential. This is an Apollonius circle: a constant ratio of distances gives a circle rather than a constant distance from either charge.

For the cylinder above the plane, call its centre height $H>a$ and set $s=\sqrt{H^2-a^2}$. The cylinder-plane image charges are $+\lambda$ at height $s$ and $-\lambda$ at height $-s$. Their distances are equal on the grounded plane, while their distance ratio is $(H+s)/a$ on the circular surface. The positive fictitious line lies inside the metal, so the exterior is charge-free and Uniqueness of the Dirichlet problem applies. A Gaussian surface enclosing the physical cylinder confirms that its charge per unit length is indeed $\lambda$. Its voltage isThe physical field energy per unit length at fixed cylinder charge is $U^{\prime }=\lambda V_c/2$. Differentiate with respect to height, keeping $a$ and $\lambda$ fixed:The electrostatic attraction per unit length of a cylinder above a plane is downward for either sign of charge. Placing the fictitious line at the cylinder centre would give the wrong answer at this separation. The long-cylinder approximation neglects the ends.

For fixed, magnetically linear geometry, self-inductance relates a coil's own magnetic flux linkage to its electric current, ${\Lambda }_i=L_i{I}_i$. mutual inductance relates the linkage in one coil to the current in the other, ${\Lambda }_{21}=M{I}_1$. These coefficients enter the induced electromotive forces through ${\mathcal{E}}_i=-d{\Lambda }_i/dt$; reversing a chosen winding orientation reverses the sign assigned to $M$.

In the ideal confined-field toroid, a circular magnetostatic Ampère law contour encloses $N_{1}I$, soWrite $w=b-a$ for the height of the square cross-section. The inductance of a rectangular-section toroid follows by integrating the magnetic flux density over that cross-section; it cannot be replaced by its value at an arbitrary mean radius:Consequently, for winding normals chosen consistently,The equality means complete shared magnetic flux linkage; with leakage one instead has $|M|<\sqrt{L_1{L}_2}$.

For the crack, use the radial-path approximation for a cracked toroid, with an air length $d$ on each path and core length $2\pi r-d$. Continuity of the normal magnetic flux density at a crack face gives $H_g={\mu }_r{H}_c$. HenceIntegrating the radial paths yieldsHere $q=1.5915\mathrm{cm}$ and $M_d/M_0=0.47079$: about a 53% reduction. The small ratio $d/a$ does not justify expanding in ${\mu }_{r}d/a$, which is about ten. The crack's low magnetic permeability gives appreciable magnetic reluctance. This is an estimate within a confined-flux model; actual redistribution near the slit is outside its accuracy. Narrow-gap circuit models assume small gap width relative to transverse dimensions and approximately directed flux.^[ref]

For the intact ring, ideal poloidal winding currents and the no-leakage boundary condition admit $\mathbf{H}=N_{1}I\hat{\mathbf{\varphi }}/(2\pi r)$ with zero radial and vertical components. It is curl-free inside the current-free core, divergence-free for uniform magnetic permeability, tangent to the core boundary, and has the required circulation; uniqueness of the Neumann problem removes an additional homogeneous scalar-potential field. This explains zero non-azimuthal components in the stipulated intact-ring idealization. A literal slit breaks that rotational symmetry, so zero transverse components everywhere is not exact for the cracked ring: magnetic fringing and flux redistribution occur. In fact, extending the approximate $B(r)\propto 1/(r+q)$ unchanged around a current-free core would give $[\nabla \times \mathbf{H}{]}_z\ne 0$. Thus that radial-path calculation and a claim of exact purely azimuthal field in the cracked geometry cannot both be exact. The numerical answer above retains the intended circuit approximation while making its limitation explicit.

For a steady thin-wire electric current, the Biot-Savart law iswhere the source element follows positive current. For a circular loop, source points are $(R\cos \varphi ,R\sin \varphi ,0)$ and the observer is $(0,0,z)$. Their cross product with $d{\mathbf{\ell }}^{\prime }$ has axial component $R^{2}d\varphi$; transverse components integrate to zero. Thus the on-axis magnetic field of a circular current loop, multiplied by the number of coincident turns, isTake positive current counterclockwise as seen from $+z$. magnetic field lines pass upward through the loop and return around its exterior, closing rather than terminating. They fan outward above its centre and converge below it. The original numerical Biot-Savart law integration below illustrates these directions; the wire singularities are masked.

The axisymmetric magnetic field near a regular axis follows from Gauss's law for magnetism. In cylindrical coordinates, regularity gives $B_{\rho }(0,z)=0$, while symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$. Integrate ${\partial }_{\rho }(\rho B_{\rho })=-\rho B_0^{\prime }(z)$ from the axis:A formal $C(z)/\rho$ term is excluded by regularity. For $z>0$ the axial field decreases upward, so the radial component is outward, consistent with the sketch.

On the small receiving coil, $d\mathbf{\ell }=ad\varphi \hat{\mathbf{\varphi }}$. The Lorentz force $I_{2}d\mathbf{\ell }\times \mathbf{B}$ has axial component $-I_{2}a{B}_{\rho }(a,d)d\varphi$. Contributions from the axial field cancel radially around the loop. ThereforeWith both currents giving magnetic dipole moments along $+z$ and $d>0$, this is attraction. Reversing one current reverses the force. The small coil has $m_2=N_2{I}_2\pi a^2$, so the result is exactly the leading magnetic dipole force $F_z=m_2{B}_0^{\prime }(d)$. Equivalently $U=-{\mathbf{m}}_2\cdot {\mathbf{B}}_0$. The approximation requires $a\ll R$ and negligible variation over the receiving coil; corrections begin at relative order $a^2/(R^2+d^2)$.

Use peak-amplitude phasors with time factor $e^{-i\omega t}$ and $\mu ={\mu }_0$. Faraday's law of induction and the Ampere-Maxwell law giveIn the good-conductor limit $\sigma \gg \omega ϵ$, take the root with positive attenuation into the slab:The real physical field is $\mathrm{Re}[{\tilde{E}}_y{e}^{-i\omega t}]$. A thick slab excludes a returning wave from its rear face. The skin depth is the electric-field amplitude decay length; the intensity decay length is $\delta /2$.

Faraday's law of induction fixes ${\tilde{H}}_z=k{\tilde{E}}_y/({\mu }_0\omega )$. The surface impedance of a good conductor and the complete transmitted magnetic field are consequentlyThe sign of the imaginary part belongs to the stated time convention. Average the Poynting vector using complex conjugation and the factor one half appropriate to peak phasors:As an independent energy check, $-{\partial }_x⟨N_x⟩=\sigma |E_0{|}^2{e}^{-2x/\delta }/2$, the mean Joule heating per unit volume.

Ohm's law gives ${\tilde{J}}_y=\sigma {\tilde{E}}_y$ and ${\tilde{B}}_z={\mu }_0{\tilde{H}}_z$. Hence the averaged Lorentz force per unit volume isIntegrating through the illuminated half-space gives the positive inward radiation pressureFinally let $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$ be the vacuum wave impedance. Tangential-field boundary conditions at normal incidence give the electric-amplitude reflection coefficient $r=(Z-Z_0)/(Z+Z_0)$. Since $|Z|\ll Z_0$,Here $N_0=(1-R)I_{\mathrm{inc}}$ by surface energy balance. Combining the leading formulas gives $p\simeq 2I_{\mathrm{inc}}/c$, the nearly perfect-reflector limit. The displayed reflection expansion also assumes a good conductor relative to the incident vacuum wave, $\omega {ϵ}_0/\sigma \ll 1$.

A normal mode is a motion in which all displacement coordinates oscillate at one angular frequency with fixed amplitude ratios. For small conservative oscillations, the symmetric stiffness and mass matrices give the generalized eigenvalue problem for small oscillations, $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. For positive mass and a stable equilibrium, its mass-orthogonal eigenvectors span the motion. A superposition of these modes, with phases and amplitudes fixed by initial displacements and velocities, gives the general solution; different modes need not share a phase.

Use upward end displacements. The spring compressions are $x_0-x_1$ and $x_0-x_2$, while the beam's centre rises by $q=(x_1+x_2)/2$. The potential energy isStatic force balance gives $2k{x}_0=mg$, cancelling the linear terms and leavingFor a uniform beam, its central moment of inertia in the vertical plane is $m{l}^2/12$. To first order in the tilt, $\theta =(x_2-x_1)/l$. Its translational and rotational kinetic energies areThus $\mathsf{M}=(m/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $\mathsf{K}=k1$. The normal modes of a rigid beam on two end springs have eigenvectorsThe first translates without tilt; the second rocks about a stationary centre. Ordinary frequencies are ${\omega }_s/(2\pi )$ and ${\omega }_a/(2\pi )$.

Assume the system had settled before the end load leaves, and put that load at end 1. Its extra static spring load is $Mg$ at that end and zero at the other. Therefore $x_1(0)=-Mg/k$, $x_2(0)=0$, and both initial velocities vanish because no impulse is imparted. Decomposing into the two normal modes givesOnly the symmetric mode contributes to the centre position:The datum $z_{\mathrm{eq}}$ is the unloaded beam's equilibrium height; no absolute height is supplied. The rocking mode is present even though it cancels out of this coordinate. These results use small tilt, $Mg/(kl)\ll 1$, ideal light vertical springs and negligible damping.

For an attractive inverse-square force, write $V(r)=-\kappa /r$ with $\kappa >0$, and take nonzero conserved angular momentum $J$. The conic section orbit hasThe physically allowed cases are ellipse for $-m{\kappa }^2/(2J^2)\le E<0$, parabola for $E=0$, and hyperbola for $E>0$. The minimum allowed energy gives the circular special case $e=0$. Here $E$ is measured relative to infinity. Zero angular momentum instead gives radial collision or escape trajectories, interpreted as degenerate limits rather than regular closed ellipses.

For a repulsive inverse-square force, $V(r)=+\kappa /r$. The effective potential is positive and decreases from infinity to zero, so a trajectory reaching a finite radius has $E>0$. For nonzero angular momentum, it is the accessible branch of a hyperbola, with $e^2=1+2E{J}^2/(m{\kappa }^2)>1$. There are no bound ellipses or finite-distance zero-energy parabolas. At zero angular momentum, the motion is straight inward followed by straight outward after reflection at a radial turning point.

For the scattering calculation, take the heavy target as fixed and define $\kappa =2Z{e}^2/(4\pi {ϵ}_0)>0$. The incoming kinetic energy and angular momentum are $E=m{v}^2/2$ and $J=mvb$. In the original sketch, the horizontal dashed incoming asymptote has perpendicular distance $b$ from the nucleus, while $d$ is the radius from the nucleus to the point where $\dot{r}=0$.

At closest approach, the velocity is tangential and $v_t=J/(md)=vb/d$. conservation of energy givesChoose the positive root and put $s=\kappa /(m{v}^2)$:Head-on incidence gives $d_{\mathrm{rep}}=2s$, and repulsion always gives $d_{\mathrm{rep}}>b$. With an attractive Coulomb force, change the sign of the potential term:For $b>0$, the attractive and repulsive inverse-square closest approaches obey $d_{\mathrm{att}}{d}_{\mathrm{rep}}=b^2$. For $b=0$ attraction causes a collision with the point centre, so $d=0$ is a singular endpoint, not a regular scattering turning point. Finite nuclear size is excluded by the stated point-force model.

For the repulsive inverse-cube central force $\mathbf{F}={\kappa }_3\hat{\mathbf{r}}/r^3$, the torque vanishes, $\mathbf{r}\times \mathbf{F}=0$, and angular momentum is still conserved. Its potential energy is $+{\kappa }_3/(2r^2)$, since $-dV/dr=+{\kappa }_3/r^3$. ThusThe new coefficient is denoted ${\kappa }_3$ because its units are $\mathrm{N}{\mathrm{m}}^3$, unlike the $\mathrm{N}{\mathrm{m}}^2$ units of the inverse-square coefficient. Conserving angular momentum depends on centrality, not on a particular radial power law.

Measure $\theta$ from the upward vertical, let ${\mathbf{e}}_3$ point from the support along the symmetry axis, and use the support as the reference point for all moments of inertia. Choose ${\mathbf{e}}_2=(\hat{\mathbf{z}}-\cos \theta {\mathbf{e}}_3)/\sin \theta$ and ${\mathbf{e}}_1={\mathbf{e}}_2\times {\mathbf{e}}_3$. These form an instantaneous principal-axis frame; its transverse axes need not spin with marks painted on the body. The three infinitesimal Euler angles rotations add as vectors:Resolving the vertical direction givesA convention matters here: in genuinely material-fixed transverse axes, the first two components are rotated by the spin angle $\chi$. For one conventional choice they are $\dot{\theta }\cos \chi +\dot{\varphi }\sin \theta \sin \chi$ and $-\dot{\theta }\sin \chi +\dot{\varphi }\sin \theta \cos \chi$. Since $I_1=I_2$, that rotation changes neither kinetic energy nor the axial and vertical conserved momenta. The simplified displayed components use the instantaneous transverse frame just defined.^[ref]

For a frictionless fixed support, the support does no work and gravity is conservative, so the total energy is constant. The heavy symmetric top Lagrangian isThe two cyclic coordinates $\chi ,\varphi$ giveEquivalently, gravitational torque has no component along the vertical or symmetry axis. Eliminate $\dot{\varphi }=(J_z-J_3\cos \theta )/(I_1{\sin }^2\theta )$ and the axial spin from $E=T+V$:This is the heavy symmetric top reduction at fixed $J_3,J_z$. A constant inclination must satisfy the reduced equation of motion $I_1\ddot{\theta }=-U_{\mathrm{eff}}^{\prime }$, not merely have zero instantaneous velocity. Differentiation at fixed conserved momenta gives, for $0<\theta <\pi$ and $\Omega =\dot{\varphi }$,Therefore uniform precession of a heavy symmetric top requiresFor a prescribed upward inclination $0<\theta <\pi /2$, real precession rates exist only if the quadratic discriminant is nonnegative:For positive spin the absolute-value sign can be dropped. The two roots are the slow and fast branches and coalesce at the threshold. The familiar upright-limit critical magnitude is $2\sqrt{I_{1}mgh}$; strict inequality gives gyroscopic stabilization of an inverted symmetric top. This limiting stability condition should not be confused with a positive minimum over every possible inclination. As $\theta$ approaches $\pi /2$ from below, the fixed-inclination threshold approaches zero. At the horizontal angle $\Omega =mgh/J_3$ for $J_3\ne 0$; below the horizontal the quadratic admits real solutions even at $J_3=0$. The stated minimum therefore needs the inclination convention made explicit.

Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.