Past exam of the physics 2 paper of the University of Oxford / 2016 resit (A10288W1, CP2)

Source paper: www.whatdotheyknow.com/request/physics_past_exam_papers_2015_16/response/973997/attach/3/Cowell%20Attachment.zip?cookie_passthrough=1

The integral Gauss law in vacuum is ${\oint }_{\partial V}\mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enc}}/{ϵ}_0$. A single isolated point charge has spherical symmetry, so its electric field is radial and constant in magnitude on a centred sphere. The flux is $4\pi r^2{E}_r$, giving Coulomb's lawIn electrostatic equilibrium, a conductor is an equipotential: a tangential electric field would move its mobile charges. The electric field inside the conducting material is zero. A thin pillbox straddling its surface therefore gives $E_{n}A=\sigma A/{ϵ}_0$, henceHere $\sigma$ is the charge density on the particular face and $\hat{\mathbf{n}}$ points into vacuum. For an isolated thin plate with total charge divided equally between two faces, each face instead has half that total areal density. The local boundary result must not be confused with the $\sigma /(2{ϵ}_0)$ field of an isolated nonconducting infinite charged sheet.

Introduce cylinder length $\ell$ and line charge density $\lambda =Q/\ell$; only charge per length matters in the long-cylinder approximation. A coaxial cylindrical Gaussian surface gives $E_{r}2\pi r\ell =Q/{ϵ}_0$. Choosing the outer cylinder's electric potential to be zero,The capacitance per unit length follows from $C=Q/[V(a)-V(b)]$:If $Q$ is intended as a line charge rather than a finite total charge, replace $\lambda$ by that $Q$. Neglecting end effects, equal opposite cylinder charges make the exterior electric field vanish; the potential difference between the conductors remains the quantity defining their capacitance.

Define $\mathbf{d}$ to point from the negative charge to the positive charge. The electric dipole moment is $\mathbf{p}=q\mathbf{d}$. In a uniform electric field, the two forces cancel, so the net force is zero. They nevertheless form a couple with torqueFor a positive angular displacement increasing $\theta$, the restoring torque component is $-pE\sin \theta$. Thus $dU/d\theta =pE\sin \theta$ andwith the additive constant chosen to make $U=0$ at $\theta =\pi /2$. Alignment is a stable minimum and antialignment an unstable maximum. The zero-force result uses a uniform field; a field gradient can exert a net force on a dipole.

Faraday's law of induction gives the electromotive force $\mathcal{E}=-d{\Phi }_B/dt$. For a fixed circuit, $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{l}$; for moving material wire, include the motional term $\mathbf{u}\times \mathbf{B}$ in that line integral. Lenz's law fixes the sign: the induced response opposes the change of linked flux, rather than necessarily opposing the field itself.

For the long solenoid, Ampère's law gives $B\simeq {\mu }_{0}nI$ inside and negligible exterior field away from the ends. With $N=nD$ turns and area $\pi a^2$, the linked flux is $N{\Phi }_B={\mu }_0{n}^{2}D\pi a^{2}I$. Hence the self-inductance isThis assumes $D\gg a$, closely spaced turns, negligible end fringing and an air core with relative magnetic permeability close to one. A finite core permeability or significant fringing changes the answer.

For a steady uniform current density in a nonmagnetic wire, the enclosed electric current at radius $r<a$ is $I_{\mathrm{enc}}=I{r}^2/a^2$. A circular Amperian loop and Ampère's law then giveThe magnetic field circles the wire according to the right-hand rule relative to the current direction. Its magnitude increases linearly from zero on the axis. The uniform-current assumption excludes skin effect.

Outside the wire, the Amperian loop encloses the entire electric current, soIts direction is again azimuthal, fixed by the right-hand rule. The inside and outside formulas join continuously at the maximum $B(a)={\mu }_{0}I/(2\pi a)$. The outer branch decays as $1/r$.

The adjacent-corner resistance of a resistor cube follows from Kirchhoff's current law and cube symmetries. Hold adjacent corners $A=(0,0,0)$ and $B=(1,0,0)$ at potentials $V$ and zero. The two corners $(0,1,0),(0,0,1)$ have a common potential $x$; their neighbours $(1,1,0),(1,0,1)$ have $V-x$. Write the remaining two potentials as $z,V-z$.

At a corner of the first pair and at $(0,1,1)$, Kirchhoff's current law givesThus $4x-z=2V$ and $4z-2x=V$, yielding $x=9V/14$, $z=4V/7$. The supplied current at $A$ isThe direct edge between the terminals is one of three current paths leaving $A$; treating the other eleven resistors as a single obvious series chain would miss the branching.

Huygens' principle treats every point of an advancing wavefront as a source of secondary wavelets; their common envelope forms the next wavefront. Consider two points of an incident plane wave reaching the interface at different times. During the same interval $\Delta t$, the remaining incident portion travels $v_1\Delta t$ while the first transmitted wavelet expands a distance $v_2\Delta t$. The two wavefront right triangles share their interface segment, soThis is Snell's law, with both angles measured from the surface normal. The speeds are phase velocities at the common frequency in stationary isotropic media. If the equation requires $|\sin {\theta }_2|>1$, there is no propagating refracted ray and total internal reflection applies instead.

In vacuum, with possible sources $\rho$ and $\mathbf{J}$, Maxwell's equations areTheir integral forms, with compatible orientations and fixed surfaces, areCharge- and current-free space additionally sets $\rho =0$, $\mathbf{J}=0$. Vacuum constitutive constants alone do not imply that no charges exist.

Take the curl of the source-free Ampere-Maxwell law, use Faraday's law of induction and $\nabla \cdot \mathbf{B}=0$:ConsequentlyFor the specified complex plane wave, the wavevector is $k\hat{\mathbf{z}}$ and $\omega =ck$. Faraday's law of induction fixesThe physical fields are the real parts. They are in phase and their Poynting vector points along $+z$.

The instantaneous electric and magnetic energy densities are $u_E={ϵ}_0{E}^2/2$ and $u_B=B^2/(2{\mu }_0)$. Since $E=cB$, they are equal for this ideal travelling plane wave, and so are their cycle averages: $⟨u_E⟩/⟨u_B⟩=1$. For real peak amplitude $E_0$, each average is ${ϵ}_0{E}_0^2/4$. The Poynting vector and mean intensity satisfyIt represents transported energy per area per time, in $\mathrm{W}{\mathrm{m}}^{-2}$.

For the numerical moonlight estimate, adopt $L_{\odot }=3.828\times 10^{26}\mathrm{W}$ and lunar mean radius $R_M=1.7374\times 10^6\mathrm{m}$. These additional astronomical inputs give solar irradiance at the Moon of ${\mathcal{I}}_{\odot }=L_{\odot }/(4\pi d_{\odot }^2)\simeq 1361\mathrm{W}{\mathrm{m}}^{-2}$.^[ref][ref]

Under the simple isotropic point-reflector approximation, the intercepted power is $\pi R_M^2{\mathcal{I}}_{\odot }$, of which fraction $A=0.12$ is redistributed over $4\pi$ steradians. ThusThese are equivalent sinusoidal peak amplitudes; the corresponding root mean square fields are smaller by $\sqrt{2}$. Actual moonlight is broadband and incoherent.

A total reflected fraction alone does not specify the directional brightness of a full Moon. For a Lambertian sphere at full phase, with the same total reflectance $A$, integrating its radiance gives ${\mathcal{I}}_M=(2A/3){\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, instead giving $E_0\simeq 1.30\mathrm{V}{\mathrm{m}}^{-1}$ and $B_0\simeq 4.32\mathrm{nT}$. If 0.12 is instead the geometric albedo, the full-phase formula is ${\mathcal{I}}_M=0.12{\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, giving $E_0\simeq 1.59\mathrm{V}{\mathrm{m}}^{-1}$. NASA's table itself distinguishes Bond albedo from geometric albedo.^[ref] The numerical estimate therefore needs an explicit reflection convention; the first boxed values use the stated isotropic approximation.

The Lorentz force is $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. With no electric field, the exact nonrelativistic component equations for the given quadrupole areThe magnetic force does no work, so the total speed stays constant. In the paraxial approximation, $\dot{z}\simeq v$ and its small change is second order in the transverse motion. For $q>0$, put $\kappa =\sqrt{qA/(mv)}$. Initial travel parallel to the $z$ axis means zero entrance slopes, yieldingThus $x$ is focused while $y$ is defocused. The first focus in the focusing plane occurs atFor negative $q$, interchange the focusing and defocusing planes and use $|q|$ in $\kappa$. A single quadrupole magnet gives a line focus of a general beam, not simultaneous focusing of both transverse coordinates onto one point. The magnet must extend at least to the computed focus, and the slopes must remain small. For a relativistic beam, the corresponding paraxial factor is ${\kappa }^2=|q|A/p$, with $p$ the longitudinal relativistic momentum.

Monochromatic light has one frequency, or a bandwidth sufficiently narrow for the intended measurement. A lamp can illuminate the experiment through a narrow spectral filter or monochromator, followed by a small source aperture at the focus of a collimating lens. The aperture supplies spatial coherence across the slits; the lens supplies approximately parallel incident wavefronts. Two independently illuminated incoherent slits would not provide a stationary fringe pattern.

In the Fraunhofer approximation, the optical path difference at observation angle $\theta$ is $s\sin \theta$. For equal single-slit intensities ${\mathcal{I}}_s$, adding the two wave amplitudes givesThe thin-slit idealization suppresses the individual diffraction envelope. If additionally $|Y|\ll D$, then $\sin \theta \simeq Y/D$ and $\mathcal{I}(Y)\simeq 4{\mathcal{I}}_s{\cos }^2(\pi sY/(\lambda D))$. The far-field condition does not by itself impose this small observation angle.

For order $m=10$ at $Y=0.05\mathrm{m}$,The paraxial estimate is $sY/(10D)=5.88\times 10^{-7}\mathrm{m}$, consistent to the precision of the data.

Let the upper slit contain the glass plate and take positive $Y$ towards that slit. The extra optical path length is $\Delta =(n_g-n_a)t=5.0\times 10^{-5}\mathrm{m}$. Its total upper-minus-lower path difference is $\Delta -s\sin \theta$. The order-zero fringe therefore satisfiesIt moves towards the covered slit. Using $Y_0\simeq D\Delta /s=0.425\mathrm{m}$ here is a poor small-angle approximation: $30^{\circ }$ is not small. The order-zero fringe is being tracked as the optical delay is introduced; because the pattern is periodic in phase, many other bright fringes still lie near the original centre. Neglecting unequal glass transmission changes visibility but not the zero-path-difference condition.

Initially the open switch leaves a series capacitor-resistor-inductor circuit driven by the battery. After a long time the electric current is zero and the capacitor voltage is $V_0$. Closing the switch bypasses and, under the stated idealization, instantaneously discharges the capacitor. The inductor current remains continuous at zero.

The remaining RL circuit obeys $L\dot{I}+RI=V_0$, soThe capacitor's impulsive discharge through the ideal short is distinct from the finite electric current through the coil. A finite switch resistance would resolve that separate fast transient.

Put $t_1=0.005\mathrm{s}$, $\tau =t-t_1$, and $I_1=2(1-e^{-1.875})$. When the switch reopens, the circuit again contains the battery and all three series elements. Inductor current continuity and capacitor voltage continuity giveThus $I^{\prime }(0^{+})=(V_0-R{I}_1)/L\simeq 115.0\mathrm{A}{\mathrm{s}}^{-1}$, not zero. Differentiating the loop equation gives the homogeneous current equationThe battery is still connected; its constant voltage disappears on differentiating, and the eventual capacitor voltage is $V_0$.

With $\alpha =R/(2L)=187.5{\mathrm{s}}^{-1}$ and ${\omega }_0=1/\sqrt{LC}$, the inequality $\alpha <{\omega }_0$ makes the circuit underdamped. The damped angular frequency and ordinary frequency areThe full current satisfying both initial conditions isIts amplitude envelope decays as $e^{-\alpha \tau }$. ThereforeThis is measured after reopening; on the clock starting with the first closure it is $t\simeq 0.02098\mathrm{s}$. An instantaneous current zero is not an amplitude decay measurement.