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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) z , where 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 = 4 10 14 N. As shown in the figure there are two values of giving sin the same value. sin = 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 = 124 and = 180- 124 = 56 . (b) v (alternate) q F = v B B B maximum force direction giving v v (c) For maximum force sin = 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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