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Unformatted text preview: Physics 208, Assignment 4, Solutions, 2/14/09, RT Problem 1. (a) F = q v × B = (4 × 10 − 14 N) ˆ z , where ˆ z points upwards (out from the page). For the force to point up, v must lie in the plane of the paper, with F = qvB sin θ = 4 × 10 − 14 N. As shown in the figure there are two values of θ giving sin θ the same value. sin θ = F qvB = 4 × 10 − 14 (1 . 6 × 10 − 19 ) × (5 × 10 5 ) × . 6 = 0 . 833 , where all quantities are in S.I. units. The possible values are θ = 124 ◦ and θ = 180 124 = 56 ◦ . (b) v (alternate) q F = v B B B maximum force direction giving v θ v (c) For maximum force sin θ = 1 and F = qvB = (1 . 6 × 10 − 19 ) × (5 × 10 5 ) × . 6 = 4 . 8 × 10 − 14 N . Problem 2. (a) To cancel the electric force the magnetic force has to be oppositely directed. So B has to be perpendicular to the plane of the paper. Since both F e and F m are proportional to the charge q , the condition for cancellation is v × B = E , as shown in the figure. B points into the plane of the paper, irrespective of the sign of q . v B d E v B E V + 1 (b) Voltage V , electric field E , and plate separation d are related by V = Ed , where V > 0 gives the electric field direction shown. V = Ed = vBd = (80 × 10 3 m / s) × (0 . 15 T) × (0 . 025 m) = 300 V . (c) As explained above the cancellation condition is independent of both the charge and mass of the particle. But for the separator to function usefully the voltage V has to large enough, and the velocity difference...
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This note was uploaded on 09/01/2009 for the course PHYS 208 taught by Professor Amadeuri during the Spring '08 term at Cornell.
 Spring '08
 AMADEURI
 Force

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