2017 USA Physics Olympiad Exam Part A 6 Since the total entropy remains constant,TiT2T3 is constant;this is a direct generalization of the result for Tf found in part (a).Energy is also conserved,as it makes no sense to leave stored energy unused,so Ti+T2+T3 is constant. When one object is at temperature TH,the other two must be at the same lower temperature To,or else further work could be extracted from their temperature difference,so T+T2+T3 TH +2To,TT2T3 THTo. Plugging in temperatures with values divided by 100 for convenience,and eliminating To gives TH(7-TH)2=36 We know that TH 1 is one (spurious)solution,since this is the minimum possible final temperature as found in part (b).The other roots are Ty=4 and TH =9 by the quadratic formula.The solution TH=9 is impossible by energy conservation,so TH=400K. It is also possible to solve the problem more explicitly.For example,one can run a Carnot cycle between the first two objects until they are at the same temperature,then run a Carnot cycle in reverse between the last two objects using the stored work.At this point,the first two objects will no longer be at the same temperature,so we can repeat the procedure;this yields an infinite series for TH.Some students did this,and took only the first term of the series.This yields a fairly good approximation of TH395K. Another explicit method is to continuously switch between running one Carnot engine forward and another Carnot engine in reverse;this yields three differential equations for T1,T2,and T3.Solving the equations and setting T1=T2 yields T3=TH. Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 6 Since the total entropy remains constant, T1T2T3 is constant; this is a direct generalization of the result for Tf found in part (a). Energy is also conserved, as it makes no sense to leave stored energy unused, so T1 + T2 + T3 is constant. When one object is at temperature TH, the other two must be at the same lower temperature T0, or else further work could be extracted from their temperature difference, so T1 + T2 + T3 = TH + 2T0, T1T2T3 = THT 2 0 . Plugging in temperatures with values divided by 100 for convenience, and eliminating T0 gives TH(7 − TH) 2 = 36. We know that TH = 1 is one (spurious) solution, since this is the minimum possible final temperature as found in part (b). The other roots are TH = 4 and TH = 9 by the quadratic formula. The solution TH = 9 is impossible by energy conservation, so TH = 400K. It is also possible to solve the problem more explicitly. For example, one can run a Carnot cycle between the first two objects until they are at the same temperature, then run a Carnot cycle in reverse between the last two objects using the stored work. At this point, the first two objects will no longer be at the same temperature, so we can repeat the procedure; this yields an infinite series for TH. Some students did this, and took only the first term of the series. This yields a fairly good approximation of TH ≈ 395K. Another explicit method is to continuously switch between running one Carnot engine forward and another Carnot engine in reverse; this yields three differential equations for T1, T2, and T3. Solving the equations and setting T1 = T2 yields T3 = TH. Copyright c 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A Question A3 A ship can be thought of as a symmetric arrangement of soft iron.In the presence of an external magnetic field,the soft iron will become magnetized,creating a second,weaker magnetic field.We want to examine the effect of the ship's field on the ship's compass,which will be located in the middle of the ship. Let the strength of the Earth's magnetic field near the ship be Be,and the orientation of the field be horizontal,pointing directly toward true north. The Earth's magnetic field Be will magnetize the ship,which will then create a second magnetic field Bs in the vicinity of the ship's compass given by B。=Be(-Kcos06+Ksin0s）》 where Ko and Ks are positive constants,6 is the angle between the heading of the ship and magnetic north,measured clockwise,b and s are unit vectors pointing in the forward direction of the ship (bow)and directly right of the forward direction(starboard),respectively. Because of the ship's magnetic field,the ship's compass will no longer necessarily point North. a.Derive an expression for the deviation of the compass,60,from north as a function of Ko, Ks;and 0. Solution We add the fields to get the local field.The northward component is Bnorth Be-BeKb cos0 cos0-BeKs sin 0 sin 0 while the eastward component is Beast =-BeKb sin 0 cos0+BeKs cos0 sin 0 The deviation is given by sin0 cos0 tan60=(K,-K)1-Kocos20-K sin20 This form is particularly nice,because as we'll see below,Ko and Ks are small enough to ignore in the denominator. b.Assuming that Ko and K's are both much smaller than one,at what heading(s)0 will the deviation 0 be largest? Solution By inspection,0=45°will yield the largest deviation.It's also acceptable to list45°，l35°， 225°，and315°. Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 7 Question A3 A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We want to examine the effect of the ship’s field on the ship’s compass, which will be located in the middle of the ship. Let the strength of the Earth’s magnetic field near the ship be Be, and the orientation of the field be horizontal, pointing directly toward true north. The Earth’s magnetic field Be will magnetize the ship, which will then create a second magnetic field Bs in the vicinity of the ship’s compass given by B~ s = Be −Kb cos θ bˆ + Ks sin θ ˆs where Kb and Ks are positive constants, θ is the angle between the heading of the ship and magnetic north, measured clockwise, bˆ and ˆs are unit vectors pointing in the forward direction of the ship (bow) and directly right of the forward direction (starboard), respectively. Because of the ship’s magnetic field, the ship’s compass will no longer necessarily point North. a. Derive an expression for the deviation of the compass, δθ, from north as a function of Kb, Ks, and θ. Solution We add the fields to get the local field. The northward component is Bnorth = Be − BeKb cos θ cos θ − BeKs sin θ sin θ while the eastward component is Beast = −BeKb sin θ cos θ + BeKs cos θ sin θ The deviation is given by tan δθ = (Ks − Kb) sin θ cos θ 1 − Kb cos2 θ − Ks sin2 θ . This form is particularly nice, because as we’ll see below, Kb and Ks are small enough to ignore in the denominator. b. Assuming that Kb and Ks are both much smaller than one, at what heading(s) θ will the deviation δθ be largest? Solution By inspection, θ = 45◦ will yield the largest deviation. It’s also acceptable to list 45◦ , 135◦ , 225◦ , and 315◦ . Copyright c 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 8 A pair of iron balls placed in the same horizontal plane as the compass but a distance d away can be used to help correct for the error caused by the induced magnetism of the ship. A binnacle,protecting the ship's compass in the center,with two soft iron spheres to help correct for errors in the compass heading.The use of the spheres was suggested by Lord Kelvin. Just like the ship,the iron balls will become magnetic because of the Earth's field Be.As spheres,the balls will individually act like dipoles.A dipole can be thought of as the field produced by two magnetic monopoles of strength +m at two different points. The magnetic field of a single pole is i=士m2 where the positive sign is for a north pole and the negative for a south pole.The dipole magnetic field is the sum of the two fields:a north pole at y=+a/2 and a south pole at y =-a/2,where the y axis is horizontal and pointing north.a is a small distance much smaller than the radius of the iron balls;in general a =KiBe where Ki is a constant that depends on the size of the iron sphere. North B a iron ball c.Derive an expression for the magnetic field B;from the iron a distance d>a from the center of the ball.Note that there will be a component directed radially away from the ball and a Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 8 A pair of iron balls placed in the same horizontal plane as the compass but a distance d away can be used to help correct for the error caused by the induced magnetism of the ship. A binnacle, protecting the ship’s compass in the center, with two soft iron spheres to help correct for errors in the compass heading. The use of the spheres was suggested by Lord Kelvin. Just like the ship, the iron balls will become magnetic because of the Earth’s field Be. As spheres, the balls will individually act like dipoles. A dipole can be thought of as the field produced by two magnetic monopoles of strength ±m at two different points. The magnetic field of a single pole is B~ = ±m ˆr r 2 where the positive sign is for a north pole and the negative for a south pole. The dipole magnetic field is the sum of the two fields: a north pole at y = +a/2 and a south pole at y = −a/2, where the y axis is horizontal and pointing north. a is a small distance much smaller than the radius of the iron balls; in general a = KiBe where Ki is a constant that depends on the size of the iron sphere. φ North B~ i iron ball a c. Derive an expression for the magnetic field B~ i from the iron a distance d a from the center of the ball. Note that there will be a component directed radially away from the ball and a Copyright c 2017 American Association of Physics Teachers