SOLUTIONS FOR THEORETICAL COMPETITION Theoretical Question 1 (10 points) 1A (3.5 points)

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1 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page 1/10 SOLUTIONS FOR THEORETICAL COMPETITION Theoretical Question 1 (10 points) 1A (.5 points) m v Mu m = + w + ( u w ) + mgr It is elementar to show that the jump of the cube appears at the puck position depicted in the picture on the left hand side. Let u be the cube velocit of mass M at this time moment and let w be the horizontal relative velocit of the puck of mass m with respect to the cube. Since the friction in the sstem is totall absent, the horizontal projection of the total momentum of the sstem is conserved, mv = Mu+ m( u w), (1) as well as with the total mechanical energ,. () In the instant frame of reference associated with the cube, the puck moves with the velocit w along the circle of radius R and its equation of motion projected on the radial direction is given b mw N + mg =. () R It is rather obvious that the condition of the cube s jump from the plane of the table is found, according to Newtons third law, as N = Mg. (4) Solving the set of equations (1)-(4), the puck velocit is obtained as v = M m gr m + M. (5) The minimal velocit of the puck is derived from relation (5) b differentiating over M / m, vmin u mg N w = gr (6) and it is achieved at the mass ratio M / m=. (7) Marking scheme Content points 1 Formula (1) 0.5 Formula () 0.5 Formula () Formula (4) Formula (5) Formula (6) Formula (7) 0.5 1В (4 points) We can replace the infinite circuit of current sources b an effective current source with an emf ε and an internal resistance r. Thus, we obtain the circuit shown in the figure on the left hand side. Then, we disconnect the resistance R, add another two current sources and connect back the resistance R. Hence, we obtain the circuit shown in the figure on the right hand side. Since the

2 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page /10 number of cells with the sources is infinite, then both circuits should be equivalent at an value of R. ε 1, r 1 ε, r R ε, r ε, r R It can be shown from the direct current laws that the following two statements are valid: 1. Let us take two current sources with ε 1, r1 and ε,r, connected in series. Then, the can be replaced b a single source with ε = ε1+ ε and r = r1+ r.. Let us take two current sources with ε 1, r1 and ε, r, connected in parallel. Then, the can be replaced b a single source with ε = ( ε1 r + ε1 r) /( r1+ r) and r = rr 1 /( r1+ r). Now, appling 1 and to the circuit shown on the right hand side, we should obtain the circuit shown on the left hand side, thus the following relations must be satisfied: ( ε + ε1) r + ε( r+ r1) ε =, (1) r+ r1+ r r( r+ r1) r =. () r + r1 + r Solution is given b ε 1 4r ε = ε =.0, () r 1 r 1 4r r = = r Ω. (4) 1 Therefore, the current flowing through the resistance R is found as ε I = = 1.0 A. (5) R+ r Marking scheme Content Points 1 Equivalent circuit 1,0 Rule Rule Formula (1) Formula () Formula () Formula (4) Formula (5) 0.5

3 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page /10 1C (.5 points) All the ras eradiated from the point A have to pass through the point A after refraction in the lens; all the ras eradiated from the point have to pass through the point after refraction in the lens. Ras passing through the optical center of the lens do not change direction. Therefore, the point of intersection of lines AA and is the optical center O of the lens. If a ra passes through both the points A and, then it should necessaril pass through the points A and. Consequentl, the point of the intersection of lines A and A lies in the plane of the lens. Thus, the plane of the lens passes through the points O and C. The main optical ais of the lens passes through its optical center and is perpendicular to the plane of the lens, Further constructions are traditional: we draw ra D through the point which is parallel to the main optical ais, and after refraction in the lens the ra (or its etension) should pass through. From its continuation to the intersection with the main optical ais, we find one of the main focuses F1. Similarl, we find the second main focus F. The drawing above shows that the lens is concave (diverging).

4 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page 4/10 Theoretical Question (10 points) Electrical conductivit of metals Ohms law 1. [1 point] In accordance with the Joule-Lentz law, the heat power released in the conductor is found as U P =, (1) R which means that the specific heat power P is written as U U P R RSl () With the aid of l 1 l U R = ρ = and E =, S σ S l () one gets P = σ E. (4) The Drude model. [1 point] The second law of Newton for the electron motion in a constant electric field is read as ma= F= ee. (5) It follows from Eq.(5) that for the time interval τ the electron passes the distance aτ s =, (6) which means that the module of the average velocit of the electron is s aτ eeτ u = = =, (7) τ m or, in the vector form, eτ u= E. (8) m. [1 point] The current densit depends on the electron number densit, its electric charge, and its average velocit as follows: enτ j= neu= E, (9) m which is Ohm s law with the specific conductivit found as enτ σ =. (10) m 4. [1 point] Each electron transfers its kinetic energ at the end of the acceleration, i.e. at the moment of collision with an ion,

5 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page 5/10 muma m eeτ Ek = = m. (11) definition there are n electrons in the cubic meter of the conductor, and each of them transfers its kinetic energ (11) for the time interval τ. Thus, the total specific energ Q transferred b electrons to the crstal lattice in the unit of volume and in the unit of time, nek nmu e n Q = τ E σ E τ = τ = m =. (1) This epression coincides with Eq.(4), thus proving the validit of the Joule-Lenz law in the Drude model. Magnetoresistance 5. [1 point] In the presence of magnetic field the equation of motion for the electron is written as du m e e = E u. (1) The projections on the coordinate aes are found as du m = ee + eu, (14) du m = eu, (15) du m z = 0. (16) Eq.(16) shows that the electron trajector lies in XY plane. Substituting u = u + E/ into Eqs. (14)-(15), we obtain du m = eu, (17) du m = eu. (18) Solutions to Eqs. (17) and (18) are derived as harmonic oscillations of the form u = Acos( ωt+ α), (19) u = Asin( ωt+ α), (0) or, in terms of the previous variables, u = Acos( ωt+ α), (1) u = u, E u = Asin( ωt+ α), () where ω = e / m. From initial conditions u = 0 and u = 0, we determine the constants A= E/ and α = π /. Substitution into Eqs. (1) and () ields ( ) E sin e u t = t m, () E e u ( t) = 1 cos t m (4)

6 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page 6/10 6. [ points] At small magnitude of the magnetic field induction, Eq. () takes the form ee e E u = t t m 6m. (5) The displacement of the electron along the OX ais over the time interval τ equals ee e E 4 s = τ τ, (6) m 4m and the average speed is found as s ee e E uav = = τ τ. (7) τ m 4m Thus, we are able to determine the relative deviation of the specific conductivit as σ neuav ( ) neuav ( = 0) 1 e τ = = σ neuav ( = 0) 1 m, (8) and, therefore, 1 eτ µ = 1 m, ν =. (9) The Hall effect 7. [0.5 points] The Lorentz force acting on the electrons is directed downward, therefore the negative charge is accumulated near the bottom face. 8. [1.5 points] Since the electrons are accumulated near the bottom face of the bar, the Hall electric field is oppositel directed with respect to the OY ais. Hence, the electron equation of motion (1) is rewritten as du m = ee + eu, (0) du m = eeh eu, (1) du m z = 0. () Again, the electron trajector lies in the XY plane. Making substitution u = u + E/ in Eqs. (0) and (1), one gets u = u E /, H du m = eu, () du m = eu. (4) Solutions to Eqs. () and (4) are again derived as harmonic oscillations of the form u = Acos( ωt+ α), (5) u = Asin( ωt+ α), (6) or, in terms of the previous variables, EH u = Acos( ωt+ α) +, (7)

7 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page 7/10 E u = Asin( ωt+ α). (8) From initial conditions u = 0 and u = 0, we obtain the following final solution E e EH e u ( t) = sin t+ 1 cos t m m (9) EH e E e u ( t) = sin t 1 cos t m m (40) 9. [1 point] At small magnitudes of the magnetic field induction, the condition for zero final displacement ( τ ) = 0 along the OY ais at the time moment τ or τ eeτ u() t = 0 EH =, (41) m 0 E H j =. (4) ne Marking scheme Content points 1 The Joule-Lenz law (1) 0.5 The specific heat power () 0.5 Formulae () Final result (4) Equation of motion (5) Path (6) Average speed (7) ector of the average velocit (8) Current densit (9) Specific conductivit (10) Kinetic energ of electrons (11) Total heat transferred (1) Equation of motion (1) 0,5 14 Equations of motion (14)-(16) 0,5 15 elocit () 0,5 16 elocit (4) 0,5 17 Epansion of the velocit (5) 0,5 18 Displacement (6) 0,5 19 Average speed (7) Final result (9) *0.5 1 The correct face stated 0.5 Equations of motion (0)-() 0.5 elocit (9) elocit (40) The Hall electric field strength (41) The Hall electric field strength (4) 0.5

8 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page 8/10 Theoretical Question 1 [1 point] The constant C is found from the condition that the total number of particles is equal to N : Nn = N. (1) n= 1 Substituting the epression for the oltzmann distribution function and obtaining summation, we get ep kt N = Nn Cep n C = = n= 1 n= 1 1 ep () 1 ep kt Nn = N ep n kt ep [ points] The internal energ of the gas is a sum of the kinetic energies of all atoms: ep kt U = EnNn Cnε ep n C = = = n= 1 n= 1 1 ep () ε = N 1 ep In the classical limit kt >> ε, the argument of the eponent is small, it is thus justifiable to use the ε approimate formula ep 1. In this case, we obtain kt U= Nk T. (4) At low temperatures, the eponent itself is small, ep << 1, hence ε U = N Nε 1 ep ε + kt. (5) 1 ep [ points] The molar heat capacit at fied volume is found as U C =. (6) T In the most general case we derive

9 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page 9/10 C ep U N kt Aε ε = = ep = R T kt kt 1 ep 1 ep Б. (7) In order to approimate epressions in two limiting cases it is easier to use the epansions deduced in Subproblem. In the high temperature limit, we get kt >> ε U= N kt C = R. (8) A i.e. the molar heat capacit is a constant. Here N A is the Avogadro constant, NAk = R stands for the universal gas constant. At law temperatures, U = NAε 1+ ep kt. ε C = NAε ep = R ep Fig.1 kt kt (9) It is seen that the molar heat capacit goes to zero as the temperature vanishes. The schematic plot is drawn in figure 1. 4 [ points] Calculation of the gas pressure can be conducted in different was. For eample, the average force eerted on the wall b a single atom is equal to the ratio of the moment transferred to the time interval between two consecutive collisions, p mvn mvn En fn = = = =. (10) τ L L L vn To determine the pressure it is necessar to summarize those forces Nn fn n U P= = NE n n =. (11) S SL n= 1 Substituting the formula for the internal gas energ (), we obtain N ε P =. (1) 1 ep In the two limiting cases the above obtained epressions for the internal energ should be used. At kt >> ε N P= kt, (1) i.e. the pressure is proportional to the absolute temperature. At low temperatures, we have

10 II International Zhautkov Olmpiad/Theoretical Competition/Solutions Page 10/10 Nε P = 1+ ep. (14) kt At temperatures going to zero, the pressure tends to a constant value P 0 N ε =. (15) The schematic plot of the pressure against the temperature is shown in figure.. Marking scheme Fig. Contents points 1 Normalizing condition (1) 0,5 Calculation of the number of particles () 0,5 General epression for the internal energ U 0,5 4 Calculation of the internal energ U () 1,0 5 Calculation of the classical limit of U (4) 0,5 6 Calculation of the low temperature limit of U (5) 1,0 7 General epression for the molar heat capacit С_ (6) 0,5 8 Calculation of the molar heat capacit С_ (7) 1,0 9 Calculation of the classical limit of С_ (8) 0,5 10 Calculation of the low temperature limit of С_ (9) 0,5 11 Schematic plot for С_ 0,5 1 General epression for average force (10) 0,5 1 General epression for P (11) 0,5 1 Calculation of the pressure P (1) 0,5 14 Calculation of the classical limit of P (1) 0,5 15 Calculation of the low temperature limit of P (14) 0,5 16 Schematic plot P 0,5 1