(Figure 27.44) - Sorry for the poor eraser quality, the imagedoesn\'t look exact
ID: 1755103 • Letter: #
Question
(Figure 27.44) - Sorry for the poor eraser quality, the imagedoesn't look exactly like what I customized.
Figure 27.44 shows the paths of several particles in a bubblechamber. (See DQ 27.14) The two spirals near the top ofthe photo come from two particles that were created at the sameinstant due to a high-energy gamma ray. (a) What can youconclude about the signs of the charges of these twoparticles, assuming that the magnetic field is perpendicular to theplan of the photograph and pointing into the paper? (b) Which ofthe two particles (the right one or the left one) had more initialmomentum? How do you know? (c) Why do the paths spiralinward? What causes this to happen?
Explanation / Answer
PART A>> We know that the two particles haveopposite signs for several reasons. Firstly, Conservation ofCharge says that since the gamma ray before the creation has nocharge (gammas are neutral), then the total charge of the twoparticles after creation must also be zero. Thus, the twoparticles must have opposite charge in order to cancel out and givea total charge of zero. Secondly, the two particles curve around in oppositedirections. Thus, they must have opposite signs for theircharges. Using the right-hand three-finger triad rule, we find that theparticle on the right is a positive charge, and the particle on theleft is a negative charge. Point index finger indirection to B field (perpendicular to paper), point thumb indirection of initial velocity vector (basically up toward the topof the page), and your middle finger points in the direction of theforce (and the curvature of the spiral) for a positivecharge. For a negative charge, use the left hand rule, orreverse the direction of your middle finger when using the righthand rule. PART B>> The one on the right has more momentumsince it has a larger radius of curvature in your sketch. F = q v B = mv2 / r = ( m v) v / r The quantity in the parentheses is momentum, and the single vin the far right and in q v B cancels out to give: q B = ( momentum) / r Since both q and B are constant, this result saysthat (mommentum) is directly proportional to radius ofcurvatuve, r. A larger r means a larger momentum. PART C >> The two charged particlescreated begin to interact strongly (because they are charged) withthe matter they are passing through, losing energy byelectromagnetic interactions and collisions. Thus, they beginto slow down and v decreases. Using F = q v B, wesee that since q and B are both constant, F is directlyproportional to v. Since v decreases due to loss of KE via collisions withmatter, then F must also decrease. A smaller F meansless centripetal force to pull the particle toward the center oftheir circular spiral, and thus the radius of the spiraldecreases.Related Questions
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