PART 4. Theoretical Competition

Transcription

1 PART 4 Theoretical Copetition Exa coission page 98 Probles in English page 99 Solutions in English page 106 Probles in three other languages and back-translations of these page 117 Exaples of student papers page 130 Photos fro the grading process page Exaple of «Old Masters» original theoretical work (Fro: The collected papers of Albert Einstein, Vol 4, 1995)

2 98 Per Chr Heer Chief exainer Coission for the Theoretical Copetiton: Per Chr Heer Alex Hansen Eivind Hiis Hauge Kjell Mork Kåre Olaussen Norwegian University of Science and Technology, Trondhei & Torgeir Engeland Yuri Galperin Anne Holt Asbjørn Kildal Leif Veseth University of Oslo

3 7 th INTERNATIONAL PHYSICS OLYMPIAD OSLO, NORWAY THEORETICAL COMPETITION JULY Tie available: 5 hours READ THIS FIRST : 1 Use only the pen provided Use only the arked side of the paper 3 Each proble should be answered on separate sheets 4 In your answers please use priarily equations and nubers, and as little text as possible 5 Write at the top of every sheet in your report: Your candidate nuber (IPhO identification nuber) The proble nuber and section identification, eg /a Nuber each sheet consecutively 6 Write on the front page the total nuber of sheets in your report This set of probles consists of 7 pages

4 PROBLEM 1 (The five parts of this proble are unrelated) a) Five 1Ω resistances are connected as shown in the figure The resistance in the conducting wires (fully drawn lines) is negligible 100 Deterine the resulting resistance R between A and B (1 point) b) A skier starts fro rest at point A and slides down the hill, without turning or braking The friction coefficient is µ When he stops at point B, his horizontal displaceent is s What is the height difference h between points A and B? (The velocity of the skier is sall so that the additional pressure on the snow due to the curvature can be neglected Neglect also the friction of air and the dependence of µ on the velocity of the skier) (15 points) c) A therally insulated piece of etal is heated under atospheric pressure by an electric current so that it receives electric energy at a constant power P This leads to an increase of the absolute teperature T of the etal with tie t as follows: [ 1 ] Tt () = T + at ( t) 0 0 Here a, t 0 and T 0 are constants Deterine the heat capacity Cp ( T)of the etal (teperature dependent in the teperature range of the experient) ( points) 14

5 d) A black plane surface at a constant high teperature T h is parallel to another black plane surface at a constant lower teperature T l Between the plates is vacuu In order to reduce the heat flow due to radiation, a heat shield consisting of two thin black plates, therally isolated fro each other, is placed between the war and the cold surfaces and parallel to these After soe tie stationary conditions are obtained 101 By what factor ξ is the stationary heat flow reduced due to the presence of the heat shield? Neglect end effects due to the finite size of the surfaces (15 points) e) Two straight and very long nonagnetic conductors C + and C, insulated fro each other, carry a current I in the positive and the negative z direction, respectively The cross sections of the conductors (hatched in the figure) are liited by circles of diaeter D in the x-y plane, with a distance D/ between the centres Thereby the resulting cross sections each have an area 1 1 ( 1 π + 8 3D ) The current in each conductor is uniforly distributed over the cross section Deterine the agnetic field B(x,y) in the space between the conductors (4 points)

6 PROBLEM The space between a pair of coaxial cylindrical conductors is evacuated The radius of the inner cylinder is a, and the inner radius of the outer cylinder is b, as shown in the figure below The outer cylinder, called the anode, ay be given a positive potential V relative to the inner cylinder A static hoogeneous agnetic field r B parallel to the cylinder axis, directed out of the plane of the figure, is also present Induced charges in the conductors are neglected We study the dynaics of electrons with rest ass and charge _ e The electrons are released at the surface of the inner cylinder 10 a) First the potential V is turned on, but r B = 0 An electron is set free with negligible velocity at the surface of the inner cylinder Deterine its speed v when it hits the anode Give the answer both when a non-relativistic treatent is sufficient, and when it is not (1 point) For the reaining parts of this proble a non-relativistic treatent suffices b) Now V = 0, but the hoogeneous agnetic field r B is present An electron starts out with an initial velocity r v 0 in the radial direction For agnetic fields larger than a critical value B c, the electron will not reach the anode Make a sketch of the trajectory of the electron when B is slightly ore than B c Deterine B c ( points) Fro now on both the potential V and the hoogeneous agnetic field r B are present

7 c) The agnetic field will give the electron a non-zero angular oentu L with respect to the cylinder axis Write down an equation for the rate of change dl/dt of the angular oentu Show that this equation iplies that L kebr is constant during the otion, where k is a definite pure nuber Here r is the distance fro the cylinder axis Deterine the value of k (3 points) d) Consider an electron, released fro the inner cylinder with negligible velocity, that does not reach the anode, but has a axial distance fro the cylinder axis equal to r Deterine the speed v at the point where the radial distance is axial, in ters of r (1 point) e) We are interested in using the agnetic field to regulate the electron current to the anode For B larger than a critical agnetic field B c, an electron, released with negligible velocity, will not reach the anode Deterine B c (1 point) 103 f) If the electrons are set free by heating the inner cylinder an electron will in general have an initial nonzero velocity at the surface of the inner cylinder The coponent of the initial velocity parallel to r B is v B, the coponents orthogonal to r B are vr (in the radial direction) and v ϕ (in the aziuthal direction, ie orthogonal to the radial direction) Deterine for this situation the critical agnetic field B c for reaching the anode ( points)

8 PROBLEM 3 In this proble we consider soe gross features of the agnitude of id-ocean tides on earth We siplify the proble by aking the following assuptions: 104 (i) The earth and the oon are considered to be an isolated syste, (ii) the distance between the oon and the earth is assued to be constant, (iii) the earth is assued to be copletely covered by an ocean, (iv) the dynaic effects of the rotation of the earth around its axis are neglected, and (v) the gravitational attraction of the earth can be deterined as if all ass were concentrated at the centre of the earth The following data are given: Mass of the earth: M = kg Mass of the oon: M = kg Radius of the earth: R = Distance between centre of the earth and centre of the oon: L = The gravitational constant: G = kg -1 s - a) The oon and the earth rotate with angular velocity ω about their coon centre of ass, C How far is C fro the centre of the earth? (Denote this distance by l) Deterine the nuerical value of ω ( points) We now use a frae of reference that is co-rotating with the oon and the center of the earth around C In this frae of reference the shape of the liquid surface of the earth is static

9 In the plane P through C and orthogonal to the axis of rotation the position of a point ass on the liquid surface of the earth can be described by polar coordinates r, ϕ as shown in the figure Here r is the distance fro the centre of the earth We will study the shape r (ϕ) = R + h (ϕ) of the liquid surface of the earth in the plane P b) Consider a ass point (ass ) on the liquid surface of the earth (in the plane P) In our frae of reference it is acted upon by a centrifugal force and by gravitational forces fro the oon and the earth Write down an expression for the potential energy corresponding to these three forces 105 Note: Any force F(r), radially directed with respect to soe origin, is the negative derivative of a spherically syetric potential energy V(r): Fr () = V () r (3 points) c) Find, in ters of the given quantities M, M, etc, the approxiate for h(ϕ) of the tidal bulge What is the difference in eters between high tide and low tide in this odel? You ay use the approxiate expression 1 1+ a acos 1 1+ acos + a ( 3cos 1), valid for a uch less than unity In this analysis ake siplifying approxiations whenever they are reasonable (5 points)

10 7 th INTERNATIONAL PHYSICS OLYMPIAD OSLO, NORWAY 106 THEORETICAL COMPETITION JULY 1996 Solution Proble 1 a) The syste of resistances can be redrawn as shown in the figure: The equivalent drawing of the circuit shows that the resistance between point c and point A is 05Ω, and the sae between point d and point B The resistance between points A and B thus consists of two connections in parallel: the direct 1Ω connection and a connection consisting of two 05Ω resistances in series, in other words two parallel 1Ω connections This yields R = 05 Ω

11 b) For a sufficiently short horizontal displaceent s the path can be considered straight If the corresponding length of the path eleent is L, the friction force is given by µ g s L and the work done by the friction force equals force ties displaceent: µ g s L = µ g s L 107 Adding up, we find that along the whole path the total work done by friction forces i µ g s By energy conservation this ust equal the decrease g h in potential energy of the skier Hence h = µs c) Let the teperature increase in a sall tie interval dt be dt During this tie interval the etal receives an energy P dt The heat capacity is the ratio between the energy supplied and the teperature increase: C = Pdt dt = P p dt dt The experiental results correspond to Hence dt dt T0 a T a a t t 34 / = + = T 0 [ 1 ( 0 )] T C P 4P at T 3 = = dt dt 0 p 4 3 (Coent: At low, but not extreely low, teperatures heat capacities of etals follow such a T 3 law)

12 d) Under stationary conditions the net heat flow is the sae everywhere: J = σ( T T ) 4 4 J = σ( T T ) h J = σ( T T l ) 1 Adding these three equations we get 4 4 3J = σ( T T ) = J 0, h where J 0 is the heat flow in the absence of the heat shield Thus ξ = J/J 0 takes the value ξ = 1/3 l e) The agnetic field can be deterined as the superposition of the fields of two cylindrical conductors, since the effects of the currents in the area of intersection cancel Each of the cylindrical conductors ust carry a larger current I, deterined so that the fraction I of it is carried by the actual cross section (the oon-shaped area) The ratio between the currents I and I equals the ratio between the cross section areas: I D I = π 3 ( ) π = π 4 D 6π Inside one cylindrical conductor carrying a current I Apère s law yields at a distance r fro the axis an aziuthal field B φ = µ 0 πr I π r π D 4 = µ 0Ir πd

13 The cartesian coponents of this are B B y µ Iy x = φ = 0 ; B B x µ Ix y = = 0 φ r πd r π D For the superposed fields, the currents are axes are located at x = D/4 I and the corresponding cylinder The two x-coponents add up to zero, while the y-coponents yield µ 0 B D I x D I x D µ 0I 6µ I y = + = = 0 [ ( / 4) ( / 4)], π π D ( π + 3 3) D ie, a constant field The direction is along the positive y-axis 109 Solution Proble a) The potential energy gain ev is converted into kinetic energy Thus 1 v = ev c 1 v c c = ev (non-relativistically) (relativistically) Hence ev (non - relativistically) v = c c 1 ( ) (relativistically) c + ev (1) b) When V = 0 the electron oves in a hoogeneous static agnetic field The agnetic Lorentz force acts orthogonal to the velocity and the electron will ove in a circle The initial velocity is tangential to the circle The radius R of the orbit (the cyclotron radius ) is deterined by equating the centripetal force and the Lorentz force:

14 ie v ebv = R 0 0, v B = 0 er () 110 Fro the figure we see that in the critical case the radius R of the circle satisfies By squaring we obtain a + R = b br + R, ie R = ( b a )/b Insertion of this value for the radius into the expression () gives the critical field B c a + R = b - R v0 bv0 = = er ( b a ) e c) The change in angular oentu with tie is produced by a torque Here r the aziuthal coponent F φ of the Lorentz force F = ( e) B v provides a torque F φ r It is only the radial coponent v r = dr/dt of the velocity that provides an aziuthal Lorentz force Hence which can be rewritten as dl dt = ebr dr, dt d dt ( L ebr ) = 0

15 Hence C = L 1 ebr (3) is constant during the otion The diensionless nuber k in the proble text is thus k = 1/ d) We evaluate the constant C, equation (3), at the surface of the inner cylinder and at the axial distance r : 1 0 eba = vr ebr 1 which gives v eb( r a ) = r (4) 111 Alternative solution: One ay first deterine the electric potential V(r) as function of the radial distance In cylindrical geoetry the field falls off inversely proportional to r, which requires a logarithic potential, V(s) = c 1 ln r + c When the two constants are deterined to yield V(a) = 0 and V(b) = V we have V( r) = V ln( r / a) ln( b/ a) The gain in potential energy, sv(r ), is converted into kinetic energy: Thus 1 v = ev ln( r / a) ln( b/ a) ev r a v = ln( / ) ln( b/ a) (5) (4) and (5) see to be different answers This is only apparent since r is not an independent paraeter, but deterined by B and V so that the two answers are identical e) For the critical agnetic field the axial distance r equals b, the radius of the outer cylinder, and the speed at the turning point is then eb( b a ) v = b

16 Since the Lorentz force does no work, the corresponding kinetic energy equals ev (question a): 1 v v = ev The last two equations are consistent when eb( b a ) b = ev The critical agnetic field for current cut-off is therefore 11 B c = b V b a e f) The Lorentz force has no coponent parallel to the agnetic field, and consequently the velocity coponent v B is constant under the otion The corresponding displaceent parallel to the cylinder axis has no relevance for the question of reaching the anode Let v denote the final aziuthal speed of an electron that barely reaches the anode Conservation of energy iplies that giving 1 v v v v v 1 ( B + φ + r ) + ev = ( B + ), v = v + v + ev / r φ (6) Evaluating the constant C in (3) at both cylinder surfaces for the critical situation we have 1 1 v a eb a = vb eb b φ c c Insertion of the value (6) for the velocity v yields the critical field B c = vb ( va φ ) b = eb ( a ) eb ( a ) [ vr + vφ + ev vφ a b] / /

17 Solution Proble 3 a) With the centre of the earth as origin, let the centre of ass C be located at l The distance l is deterined by M l = M (L - l), which gives M l = M + M L = , less than R, and thus inside the earth (1) The centrifugal force ust balance the gravitational attraction between the oon and the earth: 113 M l G MM ω =, L which gives GM ω = = Ll GM ( + M) L = s () (This corresponds to a period π/ω = 7 days) We have used (1) to eliinate l b) The potential energy of the ass point consists of three contributions: (1) Potential energy because of rotation (in the rotating frae of reference, see the proble text), 1 ω r1, where r 1 is the distance fro C This corresponds to the centrifugal force ω r 1, directed outwards fro C () Gravitational attraction to the earth, G M r (3) Gravitational attraction to the oon,

18 G M r, r where is the distance fro the oon r Describing the position of by polar coordinates r, φ in the plane orthogonal to the axis of rotation (see figure), we have r r r = (r l) = r rlcosφ + l Adding the three potential energy contributions, we obtain r 1 V( r) = ω (r rlcosφ + l ) G M G M r r (3) Here l is given by (1) and r r rr r = (L r) = L Lr + r = L 1 + (r L) (r L) cosφ c) Since the ratio r/l = a is very sall, we ay use the expansion 1 1+ a acosφ 1 = 1+ acos φ + a ( 3cos φ 1) Insertion into the expression (3) for the potential energy gives GM GM r 1 Vr (, φ) = ω r ( 3cos φ 1), 3 r L apart fro a constant We have used that (4) rl GM r ω cosφ cos φ = 0, L when the value of ω, equation (), is inserted

19 The for of the liquid surface is such that a ass point has the sae energy V everywhere on the surface (This is equivalent to requiring no net force tangential to the surface) Putting r = R + h, where the tide h is uch saller than R, we have approxiately h 1 h = 1 r R+ h = R ( ) =, 1+ ( h R) R R R R as well as r = R + Rh + h R + Rh Inserting this, and the value () of ω into (4), we have GM ( + M) R GM Vr h L R h GMr (, φ) = + ( 3cos φ 1), 3 3 L again apart fro a constant The agnitude of the first ter on the right-hand side of (5) is a factor (5) 115 ( M + M ) R 10 M L 3 5 saller than the second ter, thus negligible If the reaining two ters in equation (5) copensate each other, ie, Mr R h = ( 3cos φ 1), 3 ML then the ass point has the sae energy everywhere on the surface Here r can safely be approxiated by R, giving the tidal bulge 4 MR h = ( 3cos φ 1) 3 ML 4 3 The largest value hax = MR ML occurs for φ = 0 or π, in the direction of the oon or in the opposite direction, while the sallest value

20 h M R ML in = corresponds to φ = π/ or 3π/ The difference between high tide and low tide is therefore MR hax hin = = ML (The values for high and low tide are deterined up to an additive constant, but the difference is of course independent of this) 116 Photo: Arnt Inge Vistnes Here we see the Exa Officer, Michael Peachey (in the iddle), with his helper Rod Jory (at the left), both fro Australia, as well as the Chief exainer, Per Chr Heer The picture was taken in a silent oent during the theory exaination Michael and Rod had a lot of experience fro the 1995 IPhO in Canberra, so their help was very effective and highly appreciated!