Practice HW 12 solution

# Practice HW 12 solution - song(shs546 – Practice HW 12...

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Unformatted text preview: song (shs546) – Practice HW 12 Solutions – weathers – (22202) 1 This print-out should have 10 questions. Multiple-choice 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 Newton’s 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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## This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Practice HW 12 solution - song(shs546 – Practice HW 12...

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