2013 Semifinal Exam Part A 6 The motion is changes at an angle where the static friction is greatest,or when both conditions are equalities: f umg cos0 and a=aR In that case sin0c 1+B =sin0e-4cos0。 or tan Oc Question A3 A beam of muons is maintained in a circular orbit by a uniform magnetic field.Neglect energy loss due to electromagnetic radiation. The mass of the muon is 1.88x 10-28 kg,its charge is-1.602 x 10-19 C,and its half-life is 1.5234s. a.The speed of the muons is much less than the speed of light.It is found that half of the muons decay during each full orbit.What is the magnitude of the magnetic field? b.The experiment is repeated with the same magnetic field,but the speed of the muons is increased;it is no longer much less than the speed of light.Does the fraction of muons which decay during each full orbit increase,decrease,or stay the same? The following facts about special relativity may be useful: The Lorentz factor for a particle moving at speed v is Y=- 1-u2/c2 The Lorentz factor gives the magnitude of time dilation;that is,a clock moving at speed v in a given reference frame runs slow by a factor y in that frame. The momentum of a particle is given by p=Ymi where m does not depend on v. The Lorentz force law in the form 正=g(E+元× d continues to hold. Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 6 The motion is changes at an angle where the static friction is greatest, or when both conditions are equalities: f = µmg cos θ and a = αR In that case sin θc 1 + β = sin θc − µ cos θc or tan θc = 1 β + 1 µ Question A3 A beam of muons is maintained in a circular orbit by a uniform magnetic field. Neglect energy loss due to electromagnetic radiation. The mass of the muon is 1.88 × 10−28 kg, its charge is −1.602 × 10−19 C, and its half-life is 1.523 µs. a. The speed of the muons is much less than the speed of light. It is found that half of the muons decay during each full orbit. What is the magnitude of the magnetic field? b. The experiment is repeated with the same magnetic field, but the speed of the muons is increased; it is no longer much less than the speed of light. Does the fraction of muons which decay during each full orbit increase, decrease, or stay the same? The following facts about special relativity may be useful: • The Lorentz factor for a particle moving at speed v is γ = 1 p 1 − v 2/c2 • The Lorentz factor gives the magnitude of time dilation; that is, a clock moving at speed v in a given reference frame runs slow by a factor γ in that frame. • The momentum of a particle is given by ~p = γm~v where m does not depend on v. • The Lorentz force law in the form d~p dt = q(E~ + ~v × B~ ) continues to hold. Copyright c 2013 American Association of Physics Teachers
2013 Semifinal Exam Part A Solution For brevity we simply present the full relativistic solution. The relationships for circular motion - ai 2nr =T are purely a matter of mathematics,and thus continue to hold under special relativity.Meanwhile, since l is constant for circular motion,is constant as well.Thus we can take magnitudes in the Lorentz force law (and set E=0)to find di m =qB Combining these relationships, 2n m=9B If half of the muons decay during each orbit,in the muons'frame of reference each orbit takes one half-life Ti/2.In the lab frame,then, T=YT and so 2T -m =gB T1/2 B= 2rm qT12 Numerically, B=4.85mT The speed of the muons is irrelevant to the fraction which decay per orbit,even in the relativistic case.(The orbits take longer,but the muons live longer,both by the same factor y.) Question A4 A graduated cylinder is partially filled with water;a rubber duck floats at the surface.Oil is poured into the graduated cylinder at a slow,constant rate,and the volume marks corresponding to the surface of the water and the surface of the oil are recorded as a function of time. Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 7 Solution For brevity we simply present the full relativistic solution. The relationships for circular motion d~v dt = |~v| 2 r 2πr = |~v| T are purely a matter of mathematics, and thus continue to hold under special relativity. Meanwhile, since |~v| is constant for circular motion, γ is constant as well. Thus we can take magnitudes in the Lorentz force law (and set E~ = 0) to find γm d~v dt = q |~v| B Combining these relationships, 2π T γm = qB If half of the muons decay during each orbit, in the muons’ frame of reference each orbit takes one half-life T1/2 . In the lab frame, then, T = γT1/2 and so 2π γT1/2 γm = qB B = 2πm qT1/2 Numerically, B = 4.85 mT The speed of the muons is irrelevant to the fraction which decay per orbit, even in the relativistic case. (The orbits take longer, but the muons live longer, both by the same factor γ.) Question A4 A graduated cylinder is partially filled with water; a rubber duck floats at the surface. Oil is poured into the graduated cylinder at a slow, constant rate, and the volume marks corresponding to the surface of the water and the surface of the oil are recorded as a function of time. Copyright c 2013 American Association of Physics Teachers