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PH214_2010SPRING_EXAM1_PROFSOLN_[0]

# PH214_2010SPRING_EXAM1_PROFSOLN_[0] - Ph 214 Spring 2010...

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Ph 214, Spring 2010 Exam # 1 Question #1 SOLUTIONS 1. [30 pts] Mass spectroscopy: how heavy is that ion? (a) [5 pts] Show that the deflection angle is given by θ = q B 0 d m v . The chamber is essentially a segment of a cyclotron and the particle follows a circular trajectory while it is inside the chamber. The radius of the trajectory is obtained by equating the centripetal force with the Lorentz force: mv 2 R = qvB 0 R = mv qB 0 . [2 pts for correctly determining radius of orbit] We identify θ as the angle defined by the intersection of the line representing the original (undeviated) trajectory and the line tan- gent to the trajectory at the moment the particle leaves the cham- ber (see figure). From the diagram it is clear to see that this angle is the same as the angle subtended by the circular segment of the particle’s trajectory. We can therefore write sin θ = d R and, applying the small angle approximation, we can conclude that θ sin θ = d R = qB 0 d mv . [3 pts: 1pt for defining θ , 1 pt for small angle approximation, 1 pt for final expression] (b) [10 pts] Calculate the fractional energy loss Δ E E 0 where Δ E is the total energy radiated by each nucleus while inside the chamber and E 0 = 1 2 m v 2 is the energy of one nucleus. The particle is accelerated while in the chamber and therefore radiates electromagnetic waves. The total power radiated is given by the Larmor formula: P L = q 2 a 2 6 π ǫ 0 c 3 . The magnitude of the acceleration of our particle is constant and is given by the Lorentz force: m | vectora | = qvB 0 → | vectora | = qvB 0 m = v 2 d θ . Substituting this back into the Larmor formula we obtain P L = q 2 v 4 θ 2 6 π ǫ 0 c 3 d 2 . [3 pts for the correct expression for P L ] The energy lost due to radiation per nucleus is simply given by the product of the radiated power (energy per unit time) and the length of time a single nucleus spends in the chamber. For small values of θ , the arc length of the trajectory is approximately equal to the length of the chamber d . Therefore, the length of time a single nucleus spends in the chamber is given by Δ t d v . This implies that Δ E = P L Δ t = q 2 6 π ǫ 0 d v 3 c 3 θ 2 . [5 pts for correct expressions for Δ t and Δ E ] This allows us to conclude that the fractional energy loss is Δ E E = Δ E 1 2 mv 2 = q 2 3 π ǫ 0 d m v 2 v 3 c 3 θ 2 . [2 pts] (c) [15 pts] For a single nucleus entering the chamber at t = 0, give complete expressions (amplitude, direction, time dependence) for the electric field, magnetic field and Poynting vector of the radiation pulse detected by observers P and Q , located a distance r >> d away from the chamber in the ˆ x and ˆ z directions, respectively. 1

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Ph 214, Spring 2010 Exam # 1 Question #1 SOLUTIONS The radiated electric field from an accelerated point charge is given by vector E rad ( vector r, t ) = - q vectora ( t - r/c ) 4 π ǫ 0 c 2 r . For small angle deflection in the chamber we obtain vectora = - q v B 0 m ˆ z = - v 2 R ˆ z .
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