Past exam of the physics 2 paper of the University of Oxford / 2019 (A10288W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1. The cover identifies the ordinary sitting of 11 June 2019; the annual bundle's filename does not determine the year. Letters below distinguish successive unlettered blocks of the source.

Take an electric current $I_1$ in the long solenoid. Away from its ends, Ampère's circuital law gives the approximately uniform magnetic fieldHere ${\mu }_0$ is the vacuum permeability; an air core is understood. Each turn of the short surrounding coil encloses the magnetic flux $B_1\pi R^2$. Thus its magnetic flux linkage is ${\Lambda }_2=N_2{B}_1\pi R^2$, and the mutual inductance isThe short coil lies in the nearly uniform central magnetic field; its own axial length therefore does not enter. With consistent winding orientations, Faraday's law of induction gives ${\mathcal{E}}_2=-Md{I}_1/dt$. Reversing a winding reverses this signed coupling, while the stated magnitude is unchanged.

The method of images replaces electrical conductors by fictitious image charges outside the physical region. Choose the image charges so that the physical electric charges and their images produce the prescribed constant electric potential on every conducting boundary. The uniqueness theorem for the electrostatic potential then identifies this constructed electric potential with the physical solution in the charge-containing region. Several conducting planes can require several, or infinitely many, image charges; a single reflected image is sufficient for one infinite plane.

Put the plane at $z=0$ and the point charge at $(0,0,d)$. For $z>0$, use an image charge $-q$ at $(0,0,-d)$:This vanishes on the plane. An arbitrary constant can be added if a different reference electric potential is chosen. Excluding the point charge's own electric field, Coulomb's law givesThe separation from the image charge is $2d$, and the force is attractive for either sign of $q$. No imposed external electric field is included.

At electrostatic equilibrium the electric field in the metal is zero. The empty cavity also has zero electric field, by the uniqueness theorem for the electrostatic potential. The electric charge lies on the outer surface. For the isolated conducting sphere, spherical symmetry and Gauss's law giveTaking the electric potential to vanish at infinity and using $E_r=-d\varphi /dr$ givesThe electric potential is continuous at the surface, while the normal electric field jumps by $\sigma /{ϵ}_0$, with surface charge density $\sigma =Q/(4\pi R^2)$.

Electrical contact equalizes electric potential and preserves total electric charge. The usual isolated-sphere approximation assigns the two conducting spheres the capacitances $4\pi {ϵ}_{0}R$ and $8\pi {ϵ}_{0}R$. It then givesThese are accurate when the conducting spheres are far apart and joined by a thin wire. Literal touching changes the answer: each conducting sphere polarizes the other, so its electric potential is not its isolated value. The source does not specify an approximation which would remove this effect.

For completeness, the exact result for ideal externally tangent conducting spheres can be obtained by the method of images. Let their radii be $a,b$, their centre separation $D=a+b$, and their common electric potential $V$. Begin with central image charges $4\pi {ϵ}_{0}Va$ and $4\pi {ϵ}_{0}Vb$. Reflect each external image charge $q^{\prime }$ at centre-distance $s$ into the other conducting sphere of radius $a$: the new image charge is $-a{q}^{\prime }/s$ at centre-distance $a^2/s$ toward the original. Repeat, alternately, in the two conducting spheres. Pairing successive opposite-sign terms gives the convergent net electric chargesEach reflection cancels a nonconstant surface electric potential; the remaining central term fixes $V$. The total internal image charge equals the exterior electric flux divided by $1/{ϵ}_0$, so these sums give the physical net electric charges.

For $a=R,b=2R$, multiply both dimensionless sums by the common factor $4\pi {ϵ}_{0}V(2R/3)$. Their values follow by writing reciprocals as integrals and setting $t=u^3$:Dividing by $S_a+S_b$ enforces conservation of electric charge and yieldsThis concrete touching-sphere result disproves applying the isolated $1:2$ sharing rule to literal contact.

Let $\theta =\omega t+{\theta }_0$ be the angle between the coil's area normal and the magnetic field. Its magnetic flux linkage is $\Lambda =NB\pi r^2\cos \theta$. Consequently Faraday's law of induction givesThe sign refers to the winding orientation used to define the normal. The initial phase ${\theta }_0$ depends on the chosen time origin; one can set it to zero when the normal first points along $+x$. The minimum absolute value is zero, attained at $\theta =m\pi$: the coil's normal is parallel or antiparallel to the $x$-axis, and the plane of the coil is perpendicular to that axis. At this orientation the magnetic flux is extremal and its derivative with respect to time vanishes.

This is a plane electromagnetic wave with linear polarization. $\mathbf{E}$ is the electric field; $E_0$ its peak amplitude; $x$ the position along propagation; $t$ the time; $k=2\pi /\lambda$ the wavenumber; and $\omega =2\pi f$ the angular frequency. Here $\lambda$ is the wavelength, $f$ the frequency, and $\hat{\mathbf{y}}$ the unit vector in the direction of the electric field. Constant phase $kx-\omega t$ moves along $+x$ with phase velocity $\omega /k$.

Put $\theta =kx-\omega t$. The differential Faraday's law of induction givesThe accompanying oscillating magnetic field, excluding any independent static background, is therefore $\mathbf{B}=(k/\omega )E_0\cos \theta \hat{\mathbf{z}}$. To determine the propagation speed, also use the source-free Ampere-Maxwell law:ThusThe electric field and magnetic field oscillate in phase, perpendicular to one another and to propagation; $\mathbf{E}\times \mathbf{B}$ points along $+x$.

A wirewound resistor uses a metal alloy chosen for stable electrical resistivity. Compared with a typical carbon composition resistor, it can offer tighter resistor tolerance, a smaller temperature coefficient of resistance, lower excess electrical noise, and greater continuous electrical dissipation when its construction provides an adequate heat path. These are design-dependent advantages, rather than guarantees for every alloy or package.

The winding has parasitic inductance and interturn capacitance. Its electrical impedance therefore ceases to be a nearly real electrical resistance at sufficiently high frequencies; an approximate model is a series circuit $R+i\omega L$ shunted by a small parasitic capacitance. A resonance can occur, making a wirewound resistor unsuitable where a nearly frequency-independent electrical resistance is required. Noninductive winding patterns reduce this problem. For comparable small-signal duties it is usually larger and more expensive. A carbon composition resistor is compact and has relatively small parasitic inductance, but commonly has poorer stability, larger resistor tolerance, and more excess electrical noise.

In the paraxial approximation, apply refraction at each surface of the thin lens in air. With signed radii $R_1$ and $R_2$, the first surface obeys $n/v_1-1/u=(n-1)/R_1$, and the second obeys $1/v-n/v_1=(1-n)/R_2$. The separation of the surfaces is neglected, so the intermediate image coordinate $v_1$ cancels:This is Lensmaker's equation, with Cartesian coordinate system object coordinate $u<0$ for a real object. For an equiconvex thin lens and light travelling left to right, $R_1=R$ and $R_2=-R$. Hence

Set the red image at coordinate zero, the blue image at $\delta =0.010\mathrm{m}$, and the screen at $D=1.000\mathrm{m}$. Let the correcting thin lens be at $x$, with $\delta <x<D$. Both original images are then real objects for the correcting thin lens, with positive object distances $u_r=x$, $u_b=x-\delta$, and common positive image distance $v=D-x$. The thin lens equation givesThe refractive indices and Lensmaker's equation requireSubstitute the two focal lengths and cancel $D-x$:Thus the correcting thin lens is $100\mathrm{mm}$ beyond the red image and $900\mathrm{mm}$ before the screen. The required focal lengths and surface radius areThe blue focal length is shorter because the blue refractive index is larger; these two differing object distances are mapped to the same screen, removing the specified longitudinal chromatic aberration.

For a circular coil, use the Biot-Savart lawAt an axial position $x$, every wire element is at distance $r=\sqrt{a^2+x^2}$. The transverse magnetic field contributions cancel by rotational symmetry. The axial part of $d\mathbf{\ell }\times \mathbf{r}$ is $ad\ell$, and $\oint d\ell =2\pi a$. Multiplying by the $N$ turns givesThe electric current's sense fixes the axial direction by the right-hand rule.

Put the two coil centres at $z=\pm a/2$. By superposition, their axial magnetic field isBoth magnetic field contributions point the same way. Evaluation givesThe corresponding coefficients are $0.715542$ and $0.676777$. Each individual magnetic field falls as one moves away from its own coil; their sum is much flatter near the midpoint.

For a single coil, differentiating $(a^2+x^2{)}^{-3/2}$ twice produces $3(4x^2-a^2)(a^2+x^2{)}^{-7/2}$. At $x=a/2$ this vanishes. Thus the pair has $B^{\prime }(0)=0$ by reflection symmetry and $B^{\prime \prime }(0)=0$ by its chosen separation; all odd derivatives vanish there. In fact,This Helmholtz coil arrangement produces a nearly uniform central magnetic field, useful for controlled charged particle in a uniform magnetic field and magnetic field calibration. The small orbit used below lies within that central region.

The electron follows a semicircle under the transverse Lorentz force. Parallel, oppositely directed incoming and outgoing paths are separated by a diameter, so the orbit radius is $r=1.00\mathrm{cm}$. Acceleration from rest through the electric potential difference $V$ gives kinetic energy $m_e{v}^2/2=eV$, where $e>0$ is the elementary electric charge. Equating the Lorentz force magnitude to the required centripetal force givesThe magnetic field must be perpendicular to the plane of the turn, with its sign chosen for the electron's negative electric charge. In the illustration the electron enters rightwards at the upper endpoint, so a magnetic field into the page bends it downwards.

The nonrelativistic estimate has $v\simeq 0.198c$. If the small special relativity correction is retained, use $pc=\sqrt{K(K+2m_e{c}^2)}$ with $K=eV$ and $B=p/(er)$. This gives $B\simeq 3.389\times 10^{-2}\mathrm{T}$, about $0.49\%$ larger. The leading answer is adequate at the precision implicit in the classical treatment.

The fixed point charge produces electric potential $\varphi (r)=k_{e}q/r$, where $k_e=1/(4\pi {ϵ}_0)$. The external quasistatic work is the increase in Coulomb potential energy:The endpoints have radii $2d$ and $d$, irrespective of their opposite coordinate signs. One must use a path avoiding the singular point charge at the origin. The electric field is conservative on such paths, so the endpoint calculation suffices; a literal straight path through the origin is inadmissible. The work done by the electric field has the opposite sign.

For the negative test electric charge, use the same change in the fixed point charge's electric potential:The external agent extracts energy, since the final Coulomb potential energy is lower. Again the path must avoid the fixed point charge, and the work done by the electric field is $-W_{-}$.

The electric dipole moment is $\mathbf{p}=qa\hat{\mathbf{y}}$. Write $\theta$ for the angle measured from $+y$, with $\hat{\mathbf{\theta }}$ pointing toward increasing $\theta$. The exact electric potential isFor $r\gg a$, expand the reciprocals to first order in $a/r$. Their common $1/r$ terms cancel, giving the electric dipole approximationThe radial electric field follows from $E_r=-\partial \varphi /\partial r$, and the tangential electric field from $E_{\theta }=-(1/r)\partial \varphi /\partial \theta$. The omitted terms are of relative order $a^2/r^2$ away from zeros of a component. Multiplying by the test electric charge gives the force componentsThe electric field lines emerge from the positive electric charge and terminate on the negative electric charge. They are mirror-symmetric about the electric dipole axis and bow out across its equatorial plane. In the point-electric dipole limit their meridional curves obey $dr/(rd\theta )=E_r/E_{\theta }=2\cot \theta$, hence $r=C{\sin }^2\theta$.

In a Cartesian coordinate system the leading electric dipole field givesThereforeThe force is downwards on the equator and upwards on the negative axial side. A free positive electric charge released from rest at the left-hand point initially accelerates downwards. As it moves into $x<0,y<0$, the force also acquires a positive $x$ component. Its path bends toward the right; the force continually changes as its position changes. An initial velocity was not specified, so the following more precise trajectory assumes release from rest and neglects radiation reaction.

There is a useful exact result within the point-electric dipole approximation. Put $A=k_{e}pQ>0$; the potential energy is $U=Ay/r^3$, a homogeneous function of degree $-2$. Thus $\mathbf{r}\cdot \mathbf{F}=2U$. For a particle of mass $m$, Newton's second law and conservation of energy implyAt release, both kinetic energy and potential energy vanish, so $E=0$; also $d(r^2)/dt=0$. Consequently $r=d$ throughout the motion. Parametrize the lower semicircle by $x=-d\cos \alpha$, $y=-d\sin \alpha$. ThenThe electric charge gains speed until the bottom point, where the force is inward and the speed is $\sqrt{2A/(m{d}^2)}$. It then slows, reaches $(d,0)$ at rest, and reverses. Thus, under these assumptions,Its trajectory is not an electric field line: the force determines acceleration, rather than velocity direction. For the finite two-charge electric dipole, corrections of order $a^2/d^2$ perturb the circular trajectory; an arbitrary nonzero initial velocity also changes it.

Use complex phasors with time dependence $e^{i\omega t}$ and assume an unloaded output. The capacitor has electrical impedance $Z_C=1/(i\omega C)$. The voltage divider formed with the resistor gives the transfer functionFor input $V_i(t)=V_{i0}\cos \omega t$, the output amplitude and phase shift areso $V_o(t)=|{\tilde{V}}_o|\cos (\omega t-{\tan }^{-1}(\omega RC))$.

Here $\tau =RC=16.0\mu \mathrm{s}$, and the cutoff frequency is $f_c=1/(2\pi RC)=9.947\mathrm{kHz}$. This is an RC low-pass filter: at small frequency the capacitor has large electrical impedance and the output nearly equals the input; at large frequency it increasingly shunts the signal to ground. The amplitude then falls as $1/f$, and the phase shift approaches $-90^{\circ }$.

No numerical input amplitude is supplied, so the output voltages are specified as fractions of $V_{i0}$. An application is suppression of high-frequency electrical noise on a slowly varying signal. A finite load changes the transfer function and must be included if its electrical impedance is comparable with the circuit's output electrical impedance.

For an ideal operational amplifier under negative feedback, the grounded non-inverting terminal makes the inverting terminal a virtual ground, and no electric current enters either input. The feedback resistor and capacitor have parallel electrical impedanceKirchhoff's current law at the inverting node gives ${\tilde{V}}_i/R_1+{\tilde{V}}_o/Z_f=0$, and henceWith the stated components, $R_{2}C=16.0\mu \mathrm{s}$ and $R_2/R_1=5$. Thus this inverting low-pass amplifier has exactly the same ideal cutoff frequency and normalized frequency response as the passive RC low-pass filter, with output multiplied by $-5$:Its amplitude ratios at the three stated frequencies are $4.99975$, $3.52616$, and $0.97560$. One principal choice of phase shift is $180^{\circ }-{\tan }^{-1}(\omega \tau )$: respectively $179.424^{\circ }$, $134.848^{\circ }$, and $101.252^{\circ }$. Adding or subtracting $360^{\circ }$ describes the same output; the extra inversion must be retained.

The passive RC low-pass filter is simple, inexpensive, requires no power supply, and has no operational amplifier saturation or finite bandwidth limitation. It offers no voltage gain; its output impedance is finite, and both source electrical resistance and loading can change the observed frequency response. Its resistor also produces thermal noise.

The active inverting low-pass amplifier provides a low-frequency voltage gain of magnitude five and, within the usable bandwidth, a small output impedance, so it can drive a load with much less change in frequency response. The input electrical resistance is approximately $R_1=1\mathrm{k}\Omega$ and may itself load the source. It needs a power supply and adds electrical noise and offset; finite operational amplifier gain, bandwidth, slew rate, available output swing, and output electric current limit the ideal formula. Its inversion must also suit the application. The best choice depends on required voltage gain, source and load electrical impedances, and the signal's frequency and amplitude.

A mercury-vapor lamp is an extended, spontaneously emitting source, so practical interference requires spatial filtering and spectral selection. A narrow illuminated source slit gives sufficient spatial coherence; selecting one strong spectral line improves temporal coherence and avoids overlapping differently spaced interference fringes.

In the classical double-slit experiment, both apertures are illuminated by the same narrow source. The arrangement is symmetric, permits direct measurement and adjustment of slit separation, and has equal optical materials in the two paths. Its main drawbacks are low transmitted intensity through narrow apertures, the need for a well-made closely spaced slit pair and alignment, and the finite single-slit diffraction envelope. A broad source washes out interference fringes.

Lloyd's mirror uses the direct wave from a narrow source and the reflected wave, which appears to come from its virtual image behind a plane mirror. It requires only one physical source slit; the mirror can be adjusted to make a small effective source separation without manufacturing a double aperture. Only the direct-reflected overlap region shows interference fringes. Finite mirror size, alignment, surface quality, unequal intensities and the polarization-dependent reflection coefficient can reduce visibility. For the usual grazing reflection from glass, the reflected electric wave acquires approximately a $\pi$ phase shift, so the zero path-difference position is dark rather than the bright central fringe of the symmetric classical double-slit experiment. Both arrangements still require adequate spatial coherence and temporal coherence.

Illuminate a narrow entrance slit with the mercury-vapor lamp, use a collimating lens to illuminate two parallel narrow slits coherently, and insert a bandpass filter selecting the green line. Place a screen or scanning detector a measured distance $D$ beyond the double slit. The source slit and the two apertures should be parallel. The aperture widths should be appreciably smaller than their separation to provide an extended diffraction envelope; the green filter should reject the other strong mercury lines. The layout is drawn below.

For slit separation $d$ and observation angle $\theta$, the optical path difference in air is $\Delta =d\sin \theta$. At a bright interference fringe, $\Delta =m\lambda$. For a distant screen and small angles, $\sin \theta \simeq \tan \theta \simeq y/D$, givingMeasuring several interference fringe spacings and dividing by their number reduces relative position error. The derivation assumes the slits have equal phase on illumination and that the screen region lies within their common diffraction envelope.

The two wavelengths have independent interference patterns. Their interference fringe spacings areDifferent optical frequencies do not maintain a fixed mutual phase over the detector's integration time, so their intensities add. For narrow equal slits, suppressing the slowly varying diffraction envelopes, one may writeThe relative line strengths are not supplied; the figure illustrates equal peak intensities rather than inferring a quantitative brightness ratio. At $y=0$ both colours are bright, producing a blue-green mixture. Away from the centre the green and blue interference fringes progressively separate, creating alternating colour and brightness variations; places where one line is dark can still be bright in the other. Neither two-line mixing nor its central maximum is generally white.

Across $-15\mathrm{mm}\le y\le 15\mathrm{mm}$, the green maxima occur at $y=2.730m\mathrm{mm}$ for $m=-5,\dots ,5$, and the blue maxima at $y=2.185n\mathrm{mm}$ for $n=-6,\dots ,6$. Near $y=\pm 10.92\mathrm{mm}$, the fourth green and fifth blue maxima nearly coincide: $4{\beta }_g=10.920\mathrm{mm}$ and $5{\beta }_b=10.925\mathrm{mm}$. The $5\mu \mathrm{m}$ mismatch shows that these are near coincidences, not exact ones for the given wavelengths. The plotted central interval includes all these maxima.