2017 USA Physics Olympiad Exam AAPT UNITED STATES PHYSICS TEAM AIP 2017 USA Physics Olympiad Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor This examination consists of two parts:Part A has four questions and is allowed 90 minutes; Part B has two questions and is allowed 90 minutes. The first page that follows is a cover sheet.Examinees may keep the cover sheet for both parts of the exam. The parts are then identified by the center header on each page.Examinees are only allowed to do one part at a time,and may not work on other parts,even if they have time remaining. .Allow 90 minutes to complete Part A.Do not let students look at Part B.Collect the answers to Part A before allowing the examinee to begin Part B.Examinees are allowed a 10 to 15 minutes break between parts A and B. .Allow 90 minutes to complete Part B.Do not let students go back to Part A. Ideally the test supervisor will divide the question paper into 3 parts:the cover sheet (page 2), Part A (pages 3-13),Part B(pages 15-23).Examinees should be provided parts A and B individually,although they may keep the cover sheet.The answer sheets should be printed single sided! The supervisor must collect all examination questions,including the cover sheet,at the end of the exam,as well as any scratch paper used by the examinees.Examinees may not take the exam questions.The examination questions may be returned to the students after April 15.2017. Examinees are allowed calculators,but they may not use symbolic math,programming,or graphic features of these calculators.Calculators may not be shared and their memory must be cleared of data and programs.Cell phones,PDA's or cameras may not be used during the exam or while the exam papers are present.Examinees may not use any tables,books, or collections of formulas. Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam 1 AAPT AIP 2017 UNITED STATES PHYSICS TEAM USA Physics Olympiad Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of two parts: Part A has four questions and is allowed 90 minutes; Part B has two questions and is allowed 90 minutes. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-13), Part B (pages 15-23). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided! • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 15, 2017. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. Copyright c 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2017 USA Physics Olympiad Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Work Part A first.You have 90 minutes to complete all four problems.Each question is worth 25 points.Do not look at Part B during this time. After you have completed Part A you may take a break. Then work Part B.You have 90 minutes to complete both problems.Each question is worth 50 points.Do not look at Part A during this time. Show all your work.Partial credit will be given.Do not write on the back of any page.Do not write anything that you wish graded on the question sheets. Start each question on a new sheet of paper.Put your proctor's AAPT ID,your AAPT ID, your name,the question number and the page number/total pages for this problem,in the upper right hand corner of each page.For example, Doe,Jamie student AAPT ID# proctor AAPT ID# A1-1/3 A hand-held calculator may be used.Its memory must be cleared of data and programs.You may use only the basic functions found on a simple scientific calculator.Calculators may not be shared.Cell phones,PDA's or cameras may not be used during the exam or while the exam papers are present.You may not use any tables,books,or collections of formulas. Questions with the same point value are not necessarily of the same difficulty. In order to maintain exam security,do not communicate any information about the questions (or their answers/solutions)on this contest until after April 8,2017. Possibly Useful Information.You may use this sheet for both parts of the exam. g=9.8 N/kg G=6.67×10-11N.m2/kg2 k=1/4re0=8.99×109N.m2/C2 km=40/4π=10-7T·m/A c=3.00×10ǒm/s kB=1.38×10-23J/K NA=6.02×1023(mol)-1 R=NAB=8.31J/(mol·K) o=5.67×10-8J/(s·m2.K4) e=1.602×10-19C 1eV=1.602×10-19J h=6.63×10-34J.s=4.14×10-15eV.s me=9.109×10-31kg=0.511MeV/c2(1+x)n≈1+n.r for<1 mp 1.673 x 10-27 kg 938 MeV/c2 In(1+)for 1 sin0≈0-a03forl9l<1 cos0≈1-号02forl0<1 Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Cover Sheet 2 AAPT AIP 2017 UNITED STATES PHYSICS TEAM USA Physics Olympiad Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. • After you have completed Part A you may take a break. • Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. • Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. • Start each question on a new sheet of paper. Put your proctor’s AAPT ID, your AAPT ID, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, Doe, Jamie student AAPT ID # proctor AAPT ID # A1 - 1/3 • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Questions with the same point value are not necessarily of the same difficulty. • In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 8, 2017. Possibly Useful Information. You may use this sheet for both parts of the exam. g = 9.8 N/kg G = 6.67 × 10−11 N · m2/kg2 k = 1/4π0 = 8.99 × 109 N · m2/C 2 km = µ0/4π = 10−7 T · m/A c = 3.00 × 108 m/s kB = 1.38 × 10−23 J/K NA = 6.02 × 1023 (mol)−1 R = NAkB = 8.31 J/(mol · K) σ = 5.67 × 10−8 J/(s · m2 · K4 ) e = 1.602 × 10−19 C 1 eV = 1.602 × 10−19 J h = 6.63 × 10−34 J · s = 4.14 × 10−15 eV · s me = 9.109 × 10−31 kg = 0.511 MeV/c 2 (1 + x) n ≈ 1 + nx for |x| 1 mp = 1.673 × 10−27 kg = 938 MeV/c 2 ln(1 + x) ≈ x for |x| 1 sin θ ≈ θ − 1 6 θ 3 for |θ| 1 cos θ ≈ 1 − 1 2 θ 2 for |θ| 1 Copyright c 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 3 Part A Question Al A pair of wedges are located on a horizontal surface.The coefficient of friction (both sliding and static)between the wedges is u,the coefficient of friction between the bottom wedge B and the horizontal surface is u,and the angle of the wedge is 0.The mass of the top wedge A is m,and the mass of the bottom wedge B is M=2m.A horizontal force F directed to the left is applied to the bottom wedge as shown in the figure. A B Determine the range of values for F so that the the top wedge does not slip on the bottom wedge.Express your answer(s)in terms of any or all of m,g,0,and u. Solution Solution 1.Assume the block does not slip.Considering the horizontal forces on the entire system gives F-3mg=3ma→a=3 F -ug. When F is small,the top wedge wants to slide downward,so static friction points up the ramp. Considering the horizontal and vertical forces on the block gives N cos0+f sin0 mg,Nsin0-f cos0 ma. When the minimal force is applied,the friction is maximal,f=uN.Eliminating N gives sin9-μcos0 ma mg- cos0+usin0 and plugging in our first equation gives Fmin =3mg sin0-μcos0\ u+ cos0+μsin0 =3mg (1+μ2)tan0 1+μtan9 When F is large,the top wedge wants to slide upward,so static friction points down the ramp,and N cos0-f sin0 mg, N sin0+f cos0 ma. Now setting f=uN gives Fmax =3mg sin 6+u cos 2μ+(1-r2)tan0 cos0-μsinθ =3mg 1-μtan0 Therefore,naively the range of forces so that the block will not slip is F∈[Fmin,Fmax: Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 3 Part A Question A1 A pair of wedges are located on a horizontal surface. The coefficient of friction (both sliding and static) between the wedges is µ, the coefficient of friction between the bottom wedge B and the horizontal surface is µ, and the angle of the wedge is θ. The mass of the top wedge A is m, and the mass of the bottom wedge B is M = 2m. A horizontal force F directed to the left is applied to the bottom wedge as shown in the figure. B A F θ Determine the range of values for F so that the the top wedge does not slip on the bottom wedge. Express your answer(s) in terms of any or all of m, g, θ, and µ. Solution Solution 1. Assume the block does not slip. Considering the horizontal forces on the entire system gives F − 3µmg = 3ma ⇒ a = F 3m − µg. When F is small, the top wedge wants to slide downward, so static friction points up the ramp. Considering the horizontal and vertical forces on the block gives N cos θ + f sin θ = mg, N sin θ − f cos θ = ma. When the minimal force is applied, the friction is maximal, f = µN. Eliminating N gives ma = mg sin θ − µ cos θ cos θ + µ sin θ and plugging in our first equation gives Fmin = 3mg µ + sin θ − µ cos θ cos θ + µ sin θ = 3mg (1 + µ 2 ) tan θ 1 + µ tan θ . When F is large, the top wedge wants to slide upward, so static friction points down the ramp, and N cos θ − f sin θ = mg, N sin θ + f cos θ = ma. Now setting f = µN gives Fmax = 3mg µ + sin θ + µ cos θ cos θ − µ sin θ = 3mg 2µ + (1 − µ 2 ) tan θ 1 − µ tan θ . Therefore, naively the range of forces so that the block will not slip is F ∈ [Fmin, Fmax]. Copyright c 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A However,to get full credit,students must account for two edge cases.First,when u>tan,no force is required at all to keep the block in place,so the minimum force is zero.Second,when u>cot0,the block will not slip up under any circumstances,so there is no maximal force. Solution 2.The problem can also be solved geometrically.In general,the no slip condition is u>tano where o is the angle between the vertical and the normal to the plane.Working in the noninertial reference frame of the plane,the fictitious force due to the acceleration is equivalent to a tilting of the gravity vector by an angle tanB= g where,as in solution 1, F a≠3m -μg Then the top block will not slip as long as 0-Bu,otherwise the answer is simply amin =0.At the maximum acceleration amax,B=0+o,which gives dmax tanf+μ 9 1-utan This is only meaningful for cot 6>u,otherwise the answer is simply amax=oo. Question A2 Consider two objects with equal heat capacities C and initial temperatures T and T2.A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine. a.Find the final temperature Tf of the two objects,and the total work W done by the engine. Solution Since a Carnot engine is reversible,it produces no entropy, dS1 +ds2 d+d2=0. T T2 Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 4 However, to get full credit, students must account for two edge cases. First, when µ > tan θ, no force is required at all to keep the block in place, so the minimum force is zero. Second, when µ > cot θ, the block will not slip up under any circumstances, so there is no maximal force. Solution 2. The problem can also be solved geometrically. In general, the no slip condition is µ > tan φ where φ is the angle between the vertical and the normal to the plane. Working in the noninertial reference frame of the plane, the fictitious force due to the acceleration is equivalent to a tilting of the gravity vector by an angle tan β = a g where, as in solution 1, a = F 3m − µg. Then the top block will not slip as long as |θ−β| ≤ φ. At the minimum acceleration amin, β = θ−φ, and taking the tangent of both sides gives amin g = tan θ − µ 1 + µ tan θ . This is only meaningful for tan θ > µ, otherwise the answer is simply amin = 0. At the maximum acceleration amax, β = θ + φ, which gives amax g = tan θ + µ 1 − µ tan θ . This is only meaningful for cot θ > µ, otherwise the answer is simply amax = ∞. Question A2 Consider two objects with equal heat capacities C and initial temperatures T1 and T2. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine. a. Find the final temperature Tf of the two objects, and the total work W done by the engine. Solution Since a Carnot engine is reversible, it produces no entropy, dS1 + dS2 = dQ1 T1 + dQ2 T2 = 0. Copyright c 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 5 By the definition of heat capacity,dQi=CdTi,so dT T Integrating this equation shows that TT is constant,so the final temperature is Ti=VTT2 The change in thermal energy of the objects is C(G-I)+C(4-)=C[2V西-T- By the First Law of Thermodynamics,the missing energy has been used to do work,so W=C[口+五-2V西 Now consider three objects with equal and constant heat capacity at initial temperatures T1 100 K,T2=300 K,and T3 300 K.Suppose we wish to raise the temperature of the third object. To do this,we could run a Carnot engine between the first and second objects,extracting work W.This work can then be dissipated as heat to raise the temperature of the third object.Even better,it can be stored and used to run a Carnot engine between the first and third object in reverse,which pumps heat into the third object. Assume that all work produced by running engines can be stored and used without dissipation b.Find the minimum temperature TL to which the first object can be lowered. Solution By the Second Law of Thermodynamics,we must have TL=100K.Otherwise,we would have a process whose sole effect was a net transfer of heat from a cold body to a warm one. c.Find the maximum temperature TH to which the third object can be raised. Solution The entropy of an object with constant heat capacity is dl =CInT. Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 5 By the definition of heat capacity, dQi = CdTi , so dT1 T1 = − dT2 T2 . Integrating this equation shows that T1T2 is constant, so the final temperature is Tf = p T1T2. The change in thermal energy of the objects is C(Tf − T1) + C(Tf − T2) = C h 2 p T1T2 − T1 − T2 i . By the First Law of Thermodynamics, the missing energy has been used to do work, so W = C h T1 + T2 − 2 p T1T2 i . Now consider three objects with equal and constant heat capacity at initial temperatures T1 = 100 K, T2 = 300 K, and T3 = 300 K. Suppose we wish to raise the temperature of the third object. To do this, we could run a Carnot engine between the first and second objects, extracting work W. This work can then be dissipated as heat to raise the temperature of the third object. Even better, it can be stored and used to run a Carnot engine between the first and third object in reverse, which pumps heat into the third object. Assume that all work produced by running engines can be stored and used without dissipation. b. Find the minimum temperature TL to which the first object can be lowered. Solution By the Second Law of Thermodynamics, we must have TL = 100K. Otherwise, we would have a process whose sole effect was a net transfer of heat from a cold body to a warm one. c. Find the maximum temperature TH to which the third object can be raised. Solution The entropy of an object with constant heat capacity is S = Z dQ T = C Z dT T = C ln T. Copyright c 2017 American Association of Physics Teachers
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
2017 USA Physics Olympiad Exam Part A 9 component directed tangent to a circle of radius d around the ball,so using polar coordinates is recommended. Solution This problem is not nearly as difficult as it looks. North 中 Consider the colored triangle above.The black side has length a.The angle between the green and black sides is o,so the length of the red side is asino and the length of the green side is a coso. The magnetic field strength from one magnetic pole a distance d away is given by 1 B=上mP The sum of the two fields has two components.The angular component is a measure of the "opening"of the triangle formed by the two vectors,and since the two vectors basically have the same length,we can use similar triangles to conclude asin。Bo d 户B B。=m8sm6=B,m d sin o. As expected,this component vanishes for=0. The radial component is given by the difference in the lengths of the two field vectors,or a+ m2x d2 d' where r=a coso is the length of the green side,so Ki Br =2Be" cosΦ. d That wasn't so bad,was it? d.If placed directly to the right and left of the ship compass,the iron balls can be located at a distance d to cancel out the error in the magnetic heading for any angle(s)where 60 is largest. Assuming that this is done,find the resulting expression for the combined deviation 60 due to the ship and the balls for the magnetic heading for all angles 0. Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 9 component directed tangent to a circle of radius d around the ball, so using polar coordinates is recommended. Solution This problem is not nearly as difficult as it looks. φ North a Consider the colored triangle above. The black side has length a. The angle between the green and black sides is φ, so the length of the red side is a sin φ and the length of the green side is a cos φ. The magnetic field strength from one magnetic pole a distance d away is given by B = ±m 1 d 2 The sum of the two fields has two components. The angular component is a measure of the “opening” of the triangle formed by the two vectors, and since the two vectors basically have the same length, we can use similar triangles to conclude a sin φ d ≈ Bφ B ⇒ Bφ = m a d 3 sin φ = Be mKi d 3 sin φ. As expected, this component vanishes for φ = 0. The radial component is given by the difference in the lengths of the two field vectors, or Br = m 1 d 2 − 1 (d + x) 2 = m d 2 1 − 1 (1 + x/d) 2 ≈ m d 2 2x d , where x = a cos φ is the length of the green side, so Br = 2Be mKi d 3 cos φ. That wasn’t so bad, was it? d. If placed directly to the right and left of the ship compass, the iron balls can be located at a distance d to cancel out the error in the magnetic heading for any angle(s) where δθ is largest. Assuming that this is done, find the resulting expression for the combined deviation δθ due to the ship and the balls for the magnetic heading for all angles θ. Copyright c 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 10 Solution Note that the two iron balls create a magnetic field near the compass that behaves like that of the ship as a whole.There is a component directed toward the bow given by Bo =-2Be sin o cos0 and a component directed toward the starboard given by Bs 2Br cos oo sin where the factors of 2 are because there are two balls.Note that 0 is the ship heading while o is the angle between North and the location of the compass relative to one of the balls.Thus, if the field is corrected for the maximum angles it will necessarily cancel out the induced ship field for all of the angles,so that 60=0 for all 0.Effectively,this means placing the balls to make Ko=K's. Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 10 Solution Note that the two iron balls create a magnetic field near the compass that behaves like that of the ship as a whole. There is a component directed toward the bow given by Bb = −2Bθ ∝ sin φ ∝ cos θ and a component directed toward the starboard given by Bs = 2Br ∝ cos φ ∝ sin θ where the factors of 2 are because there are two balls. Note that θ is the ship heading while φ is the angle between North and the location of the compass relative to one of the balls. Thus, if the field is corrected for the maximum angles it will necessarily cancel out the induced ship field for all of the angles, so that δθ = 0 for all θ. Effectively, this means placing the balls to make Kb = Ks. Copyright c 2017 American Association of Physics Teachers
