2022年IPhO国际物理奥林匹克竞赛真题（理论）

Theory ③22 NLD-S-2 T-0 G-2 G0-2 English (Official) Physical Constants Below is the list of physical constants you might use in your solution in addition to the quantities given in the problem text. Physical constant Notation Numerical value Speed of light in vacuum 299792458m/5 Vacuum permeability(magnetic constant) o 4r.10-1N/A2: [1 N/A2 =1Vs/Am] Vacuum permittivity(electric constant) 8.8541878128(13),10-12F/m [1 F/m=1 As/Vm] Elementary charge 1.602176634.10-18C: [1C=1A-s Electron rest mass me 9.1093837015(28)·10-31kg Proton rest mass mp 1.67262192369(51)·10-27kg Neutron rest mass mn 1.67492749804(95)·10-27kg Atomic mass unit mamu 1.66053906660(50)·10-27kg Gravitational constant G 6.67430(15)-10-1m3/(kgs2) Avogadro constant NA 6.02214076,1023mol-1 Molar gas constant R 8.31446261815324J/(K-mol Boltzmann constant ke 1.380649.10-23J/K Stefan-Boltzmann constant a 5.670374419.·10-sW/(m2.K4) Planck constant h 6.62607015,10-34J·s Reduced Planck constant h=h/(2x) 1.0545718176...10-34」+5 Theory 0022 NLD-S-2 T-1 Q-1 Q1-1 English(Official) Permanent magnets(10 points) Strong permanent magnets are made from NdFeB alloy which obeys a very wide hysteresis loop so that the magnetizationJ can be assumed to be constant over a wide range of applications;in what follows. we assume that J=1.5T/Ho,whereo=4x 10-7 N/A2,and the magnetization of all the permanent magnets is homogeneous.Magnetization is defined as the volume density of the magnetic dipole moment of the matter. Hint 1.The following equality might be useful: 1= 名示0 Hint 2.The magnetic field created by a spherical magnet is identical to that of a point dipole.The magnetic fields created by magnets of other shapes become equivalent to a point dipole fields only at distances much larger than their diameter. Hint 3.Electric and magnetic fields of electric and magnetic point dipoles as functions of coordinates and of the corresponding dipole moment are similar,i.e.one can be obtained from the other by multiplying it by a constant factor. Hint 4.The induced field due to a boundary condition can always be replaced by some configuration of field sources outside the given boundaries. Part A.Interaction of magnets(4.5 points) When the distance to a magnet is much larger than its size,the magnetic field created by it can be approximated with the magnetic field of its dipole momentm, ==2m-m) Herer-,and we have decomposed the dipole moment into components parallel and perpendicular to the radius vector drawn from the dipole to the observation point,=+m. A.1 Find the magnitude of the interaction force between two coaxially placed cylin-0.6pt drical magnets of diameter d =20mm and thickness h =2mm,magnetized parallel to their axis,if the distance between the centres of the magnets is L=20cm.You may assume that L>d,h. A.2 At distances much larger than,the field created by the magnet from task A.1 0.4pt is the same as that created by a circular current I.Find I. Theory NLD-S-2 T-1 Q-2 Q1-2 English(Official) A.3 Find the interaction force between the magnets for the setup of task A.1 if in-1.0pt stead L=5mm.You may assume that dh. A.4 Identical spherical magnets of diameter 6=5mm,bound together by magnetic 1.0pt attraction,form a chain.What is the maximal length i for such a chain which does not break under its own weight when hanging beneath the topmost mag- net?The density of NdFeB magnets p=7500 kg/m3. A.5 Consider the chain from part A.4.Obtain an expression for the magnitude of 1.5pt the magnetic B-field at such a point P which is at distance r from one of the chain's endpoint O,and the angle between the chain and the line Op is (cf. figure below),assuming that >r and rsin6 Part B.Interaction with ferromagnets(3.5 points) Now we assume that in addition to the permanent magnets we have also plates made from a ferromag- netic material,similar to what is used in transformer cores.In the situations we're concerned with,it can be considered to have a constant but very large relative permeability10 Hint 5.Large permeability means that magnetic field lines near the outside surface of an object made of the material are nearly perpendicular to the surface.This is similar to the behavior of electric field lines near the outside surface of a conductor. Theory D22 Q1-3 NLD-S-2 T-1 Q-3 English(Official) 1.0pt B.1 A spherical magnet from part A.4 is at a distance s=6 from a thick infinite ferromagnetic plate(see the answer sheet).The magnetization of the sphere is oriented perpendicular to the plate.Sketch the field lines in the cross-section shown in the answer sheet.In that figure,three points(denoted as 1,2,and 3) are marked;you need to show field lines passing through each of these points in their full length,i.e.as much as fits into the figure. 1.0pt B.2 Now the spherical magnet is brought into direct contact with the plate.Which direction is taken by the magnetization vector of the spherical magnet at a sta- ble equilibrium and what is the normal force between the plate and the magnet? Mark the correct direction(s)with a tick in the corresponding box in the answer sheet.Incorrect ticks will reduce your score. Now a magnet from part A.1 is placed between two thick circular ferromagnetic 1.5pt B.3 plates of diameter D=2d so that the flat faces of the magnet are pressed against the plates and all three discs are coaxial.Find the magnetic force F acting on each plate.Hint:You may neglect the magnetic field both outside the ferromagnetic plates and outside the gap between them. Part C.(Anti)ferromagnetic order(2 points) The magnetic properties of materials are due to the magnetic dipole moments of electrons and atomic nuclei.If the dipole moments orient themselves parallel to each other,the field created by them is magni- fied-these are ferromagnetic materials.On the other hand,if for each dipole moment there is another antiparallel dipole moment nearby,the fields cancel out-these are anti-ferromagnetic materials. In what follows,we consider a very large number of spherical magnets of part A.4,arranged at the nodes of a two-dimensional lattice;see real photos of stable equilibrium configurations below.Assume that all the magnetization vectors lie in the plane of the figure.Consider in your calculations only nearest- neighbour interactions(on the figure of C.1,each magnet has four nearest neighbours,and on the figure of C.2-six). Theory IPh22 NLD-S-2 T-1 Q-4 Q1-4 English(Official) c.1 Show the magnetization directions of the magnets in the figure below.You 0.8pt are not required to prove that the configuration you suggested is the only pos- sibility.You still need to justify that the configuration you suggested is indeed stable.Find the energy needed to pull one magnet out of this lattice from some- where in the middle of the lattice,assuming the other magnets are kept sta- tionary.Does this configuration correspond to the order of ferromagnetic or antiferromagnetic materials? C.2 Answer the same questions as in task C.1 for the configuration shown in the 1.2pt figure below. Theory 2h③22 NLD-S-2 T-2 Q-1 Q2-1 English(Official) James Webb Space Telescope(12 points) This is a question on the physics of the James Webb Space Telescope.Light from a star strikes the primary mirror,with an area of Amir25m2,and reflects off of a secondary mirror.The focal length of the system is f=130m.The light is focused into the ISIM(Integrated Science Instrument Module),which contains the CCD(charged-coupled device)cameras. 间 ronan Spxcecrah Bus Stnracken Image Credit:NASA Part A.Imaging a Star(1.8 points) The nearest Red Giant is 89 light-years distant,has a temperature of Ttar=3600K,and a diameter of d。=1.7×10m. A.1 Calculate the diameter of a focused image of the star on the CCD camera imag- 0.4pt ing surface. A.2 Estimate the diameter of a diffraction central maximum on the CCD camera 0.4pt imaging surface.Assume a wavelength of =800 nm,which is the strongest intensity wavelength from the red giant star. A.3 If the CCD is not cooled and can lose heat only by radiating from the top of 1.0pt the imaging surface,what would be the equilibrium temperature of the CCD at the location of the image of the red giant star?Assume the CCD surface is a blackbody.Provide a formula and a numerical estimate. Part B.Counting Photons(1.8 points) The absorption of a photon by the CCD camera leads to the emission of an electron within the apparatus. This occurs only if the photon has sufficient energy to excite an electron across an energy gap AE Assume that every photon with sufficient energy succeeds.There is also leakage of electrons across the gap caused by the temperature of the CCD camera;this is the dark current i and is measured in the Theory 022 NLD-S-2 T-2 Q-2 Q2-2 English(Official) number of electrons per second.It is a function of temperature according to id ine-AEal/6knT (10 where io is a constant. 102 101 e 100 10-1 10-2 106.1012 0.130.140.150.160.17 1/T(1/K0 The graph shows how dark current varies with temperature.The units for dark current,e/s should be thought of as counting a number of electrons per second. B.1 From the dark current graph,provide an order of magnitude estimate for the 0.4pt temperature of a distant source of thermal photons that would just be capable of exciting an electron on the pixel. The electrons are collected in a capacitor,and after an exposure time r,the electrons are counted.There are three main sources of uncertainty in the process:a fixed uncertainty in the counting process called read out noise;a Poisson distribution error associated with the dark current,and a Poisson distribution error associated with the detected incoming photons.Poisson distribution errors are equal to the square root of the number of counts associated with a process.The measured photon count is equal to the number of electrons in the capacitor,minus the number of electrons associated with the dark current. B.2 Write an expression for the total count uncertainty if there is a readout noise 0.4pt o,a dark current id,an incoming photon rate p,and an exposure time7. For remaining questions in this part assume the exposure time isr=10s and the read out noise is a fixed a.14. B.3 Assume an operating temperature ofTo=7.5K.Calculate the minimum photon 0.Spt rate p so that the photon count is ten times the count uncertainty. Theory NLD-S-2 T-2 Q-3 Q2-3 English(Official) B.4 Assuming all photons are just capable of exciting an electron across the band 0.5pt gap,what is the intensity of the source of photons found in B.3 on the primary mirror?Express your answer in W/m2 Part C.Passive Cooling(4.4 points) An infrared CCD camera must be kept at a low temperature.The first tool is a shield to protect from the sun's radiation. The sun-shield consists of five separated reflective layers in thin sheets(black);radiant energy(gray)from the sun is incident on the first sheet on left,and some energy escapes between every pair of sheets. ΨΨ44 Schematic of energy flow:the vertical lines(black)are the sheets,the flow of energy(gray)is from the left to the right,however,between sheets,some energy flows up and out into space. On the left is a simple model of two adjacent sheets 1 and 2 separated by a distance h.The sheets are not connected,and the perimeter is open to space.Assume the sheets are parallel. Thermal radiation can be exchanged between the sheets,and thermal radiation can escape through the perimeter gap.On the right,the perimeter gap has been shaded to help visualize. Theory Ph 22 NLD-S-2 T-2 Q-4 Q2-4 English(Official) Assume the following simplifications: Sheets are square,each with area Asheet=200m2. Sheets are parallel and separated by h=25 cm along the perimeter. Sheets have constant emissivitye1.Assume that all reflections off of sheet surfaces are diffuse. Sheets are thin with temperature on the front and back surfaces equal and uniform. The fraction of radiant flux emitted by a sheet that is absorbed by the adjacent sheet is o<1.This means that if sheet 1 in the figure above emits an amount of heat Q,toward sheet 2 then sheet 2 will absorb an amount aQ from sheet 1. The amount of radiant flux ejected out of the perimeter gap between two sheets is approximated as BQ12 where aQ2 is the net flux between the two sheets.The fraction B<1.This is equivalent to saying that the heat loss to space between two sheets is proportional to the net heat exchange between the sheets.This is a rough approximation for this problem. Background temperature of space is negligible. C.1 Derive expressions for the equilibrium temperatures of the first sheet and fifth 2.4pt sheet in terms of the incident solar radiation intensity In.the constants a and B,and any necessary physical constants.To simplify your expression,you may define additional constants in terms of a and B,etc. C.2 Derive numerical estimates for o and B from the information about the sheet 1.6pt geometry assuming an emissivity e=0.05.You are encouraged to consider the rectangular box model of the sheets above,where the perimeter area effec- tively acts as a perfect absorber of radiant energy. C.3 Numerically determine the temperatures of sheet 1 and sheet 5.The solar in- 0.4pt tensity is Io 1360W/m2. Part D.Cryo-cooler(4 points) The last stage of the cooling system directly cools the CCD camera.A closed cycle refrigeration system has a supply pipe line feeding helium gas at constant pressure Pmoving through a sponge like porous plug into a pipe with constant pressure P2.The pipe carries the gas to cool the CCD.The helium gas then passes through a pump before returning to the supply line. A,T P2.Ta gas flow gas flow Porous Plug Helium gas supplied on the left at well defined pressure P and temperature T is forced through the plug to well defined pressure P and temperature Ta,where it is carried away on the right. As the gas moves through the porous plug,viscous friction with the narrow walls of the channels in the sponge becomes an important effect;however,no heat is transferred to or from the gas during the Theory ③22 NLD-S-2 T-2 Q-5 Q2-5 English(Official) process.The bulk speed of the gas in region 2 is only marginally greater than the bulk speed in region 1 Helium is not an ideal gas,but does remain in a gaseous state throughout this process D.1 Consider a mole of gas that passes from left to right through the plug. Complete the table in your answer sheet by writing'>'or'<'to identify the quan- 1.0pt tity that must be greater,'='to identify quantities that must be equal,or?if it is not possible to know which is greater or equal without more information. D.2 Identify a conserved quantity constructed from U(internal energy).P(pres- sure),and V(volume)as a mole of gas moves through the plug:show work on 0.6pt how you derived this conserved quantity. Your answer sheets have graphs of internal energy per mass against volume per mass for helium with isotherms and lines of constant entropy. D.3 Assuming that V=0.100m3/kg and T 7.5K,use the graph to find a nu-1.4pt merical value for the conserved quantity that you found in Part D.2.Show the construction on the graph! D.4 Find the maximum possible temperature for T.Show the construction on the graph! 0.8pt D.5 Assuming your value for the maximum T found in D.4,find a numerical value 0.2pt for P.