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Unformatted text preview: song (shs546) Practice HW 12 Solutions weathers (22202) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 0.0 points An electron is accelerated by a 6 . 7 kV poten tial difference. The charge on an electron is 1 . 60218 10 19 C and its mass is 9 . 10939 10 31 kg. How strong a magnetic field must be expe rienced by the electron if its path is a circle of radius 5 . 7 cm? Correct answer: 0 . 00484247 T. Explanation: Let : r = 5 . 7 cm = 0 . 057 m , V = 6 . 7 kV = 6700 V , m = 9 . 10939 10 31 kg , and q = 1 . 60218 10 19 C . From Newtons second law, F = q v B = m v 2 r v = q B r m . The kinetic energy is K = 1 2 m v 2 = q 2 B 2 r 2 2 m = q V , so the magnitude of the magnetic field is B = 1 r radicalBigg 2 V m q = 1 . 057 m radicalBigg 2 (6700 V) (9 . 10939 10 31 kg) (1 . 60218 10 19 C) = . 00484247 T . Dimensional analysis : The voltage must be converted from kV to V and the radius from cm to m. 002 0.0 points One electron collides with a second electron initially at rest. After the collision, the radii of their trajectories are 0 . 008 m and 0 . 017 m. The trajectories are perpendicular to a uni form magnetic field of magnitude 0 . 013 T. Use a charge of 1 . 60218 10 19 C and a mass of 9 . 10939 10 31 kg. Determine the energy of the incident elec tron. Correct answer: 5 . 24671 keV. Explanation: Let : q = 1 . 60218 10 19 C , B = b , m = 9 . 10939 10 31 kg , r 1 = 0 . 008 m , and r 1 = 0 . 017 m . For each electron, q v B = m v 2 r v = q m B r . v 1 = 1 . 60218 10 19 C 9 . 10939 10 31 kg (0 . 013 T) (0 . 008 m) = 1 . 82917 10 7 m / s , and v 2 = 1 . 60218 10 19 C 9 . 10939 10 31 kg (0 . 013 T) (0 . 017 m) = 3 . 88699 10 7 m / s . The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic: T = 1 2 m v 2 1 + 1 2 m v 2 2 = q 2 B 2 2 m bracketleftBig r 2 1 + r 2 2 bracketrightBig = (1 . 60218...
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 Spring '00
 Littler

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