P29_005

# P29_005 - of Earths Feld. Technically, then, the electron...

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5. (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 Feld is downward. The right-hand rule indicates that ~v × ~ B is to the west, but since the electron is negatively charged (and ~ F = q~v × ~ B ), the magnetic force on it is to the east. (b) We combine F = m e a with F = evB sin φ . Here, B sin φ represents the downward component of Ear th ’sFe ld(g ivenintheprob lem ) . Thus , a = evB/m e . Now, the electron speed can be found from its kinetic energy. Since K = 1 2 mv 2 , v = r 2 K m e = s 2(12 . 0 × 10 3 eV)(1 . 60 × 10 19 J / eV) 9 . 11 × 10 31 kg =6 . 49 × 10 7 m / s . Therefore, a = evB m e = (1 . 60 × 10 19 C)(6 . 49 × 10 7 m / s)(55 . 0 × 10 6 T) 9 . 11 × 10 31 kg =6 . 27 ×
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Unformatted text preview: of Earths Feld. Technically, then, the electron should follow a circular arc. However, the deection is so small that many of the technicalities of circular geometry may be ignored, and a calculation along the lines of projectile motion analysis (see Chapter 4) provides an adequate approximation: x = vt = t = x v = . 200 m 6 . 49 10 7 m / s which yields a time of t = 3 . 08 10 9 s. Then, with our y axis oriented eastward, y = 1 2 at 2 = 1 2 ( 6 . 27 10 14 )( 3 . 08 10 9 ) 2 = 0 . 00298 m ....
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