CH 19 - Magnetism-solutions

# CH 19 - Magnetism-solutions - This print-out should have 8...

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Unformatted text preview: This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron is accelerated by a 6 . 8 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 . 8 cm? Correct answer: 0 . 00479436 T. Explanation: Let : r = 5 . 8 cm = 0 . 058 m , V = 6 . 8 kV = 6800 V , m = 9 . 10939 × 10- 31 kg , and q = 1 . 60218 × 10- 19 C . From Newton’s second law, F = q v B = mv 2 r v = q B r m . The kinetic energy is K = 1 2 mv 2 = q 2 B 2 r 2 2 m = q V , so the magnitude of the magnetic field is B = 1 r 2 V m q = 1 . 058 m × 2 (6800 V) (9 . 10939 × 10- 31 kg) (1 . 60218 × 10- 19 C) = . 00479436 T . Dimensional analysis : The voltage must be converted from kV to V and the radius from cm to m. 002 10.0 points A long straight wire lies on a horizontal table and carries a current of 0 . 92 μ A. A proton moves parallel to the wire (opposite the cur- rent) with a constant velocity of 15200 m / s at a distance d above the wire....
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## This note was uploaded on 10/03/2010 for the course PH 102 taught by Professor Staff during the Fall '08 term at Alabama.

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CH 19 - Magnetism-solutions - This print-out should have 8...

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