81. (a) The textbook uses “geomagnetic north” to refer to Earth’s magnetic pole lying in the northern hemisphere. Thus, the electrons are traveling northward. The vertical component of the magnetic field is downward. The right-hand rule indicates that GGvB×is to the west, but since the electron is negatively charged (and GGGFqvB=×), the magnetic force on it is to the east. We combine F = meawith F = evBsin φ. Here, Bsin represents the downward component of Earth’s field (given in the problem). Thus, a = evB /me. Now, the electron speed can be found from its kinetic energy. Since Kmv=122, vKme==×××−−22120 10160 10911 10649 10319317...eVJ eVkgmschchTherefore, ()( )1976221414311.60 10C6.49 10 m s55.0 10 T6.27 10 m s6.3 10 m s .
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.