solution_pdf - garza (olg97) Homework No. 6 jimenez...

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garza (olg97) – Homework No. 6 – jimenez – (1302-01) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points The magnetic feld oF the Earth at a certain location is directed vertically downward and has a magnitude oF 7 . 1 × 10 - 5 T. A proton is moving horizontally toward the west in this feld with a speed oF 2 . 4 × 10 6 m / s. The charge on the proton is 1 . 60218 × 10 - 19 C and its mass is 1 . 67262 × 10 - 27 kg. What is the magnitude oF the magnetic Force the feld exerts on this charge? Correct answer: 2 . 73011 × 10 - 17 N. Explanation: Let : q =1 . 60218 × 10 - 19 C , v =2 . 4 × 10 6 m / s , and B =7 . 1 × 10 - 5 T . The Lorentz Force is F = q v B = (1 . 60218 × 10 - 19 C) (2 . 4 × 10 6 m / s) × (7 . 1 × 10 - 5 T) = 2 . 73011 × 10 - 17 N . 002 (part 2 oF 2) 10.0 points What is the radius oF the circular arc Followed by this proton? Correct answer: 352 . 891 m. Explanation: Let : m . 67262 × 10 - 27 kg . The Lorentz Force is the centripetal Force oF the circular motion: q v B = m v 2 r . r = mv qB = (1 . 67262 × 10 - 27 kg) (2 . 4 × 10 6 m / s) (1 . 60218 × 10 - 19 C) (7 . 1 × 10 - 5 T) = 352 . 891 m . 003 10.0 points A3 . 09013 μ C charged particle with a kinetic energy oF 0 . 0407885 J is placed in a uniForm magnetic feld oF magnitude 0 . 0663511 T. IF the particle moves in a circular path oF radius 1 . 96214 m, fnd its mass. Correct answer: 1 . 984 × 10 - 12 kg. Explanation: Let : q =3 . 09013 μ C = 3 . 09013 × 10 - 6 C , K =0 . 0407885 J , B . 0663511 T , and R . 96214 m . Kinetic energy is K = 1 2 2 . The magnetic feld supplies the centripetal Force, so F c = F mag 2 R = q v B = q RB m 2 v 2 =( ) 2 m ± 1 2 2 ² = 1 2 ( ) 2 mK = 1 2 ( ) 2 m = ( q B R ) 2 2K = ( 3 . 09013 × 10 - 6 C ) 2 (0 . 0663511 T) 2 (1 . 96214 m) 2 2 (0 . 0407885 J) = 1 . 984 × 10 - 12 kg . 004 (part 1 oF 2) 10.0 points An electron moves in a circular path perpen- dicular to a constant magnetic feld oF magni- tude 1 . 13 mT. The charge on an electron is q e .
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garza (olg97) – Homework No. 6 – jimenez – (1302-01) 2 If the angular momentum of the elec- tron about the center of the circle is 4 . 88 × 10 - 25 J · s, Fnd the radius of the circu- lar path. Correct answer: 5 . 19206 cm. Explanation: Let : q =1 . 602 × 10 - 19 C , L =4 . 88 × 10 - 25 J · s , and B . 13 mT = 0 . 00113 T . The magnetic force on a moving charge is q v B = mv 2 R , q RB = mv , and the angular momentum is L = mv R = qR 2 B R = ± L qB = ± 4 . 88 × 10 - 25 J · s (1 . 602 × 10 - 19 C) (0 . 00113 T) × ² 100 cm 1m ³ = 5 . 19206 cm . 005 (part 2 of 2) 10.0 points ±ind the speed of the electron. Correct answer: 1 . 03183 × 10 7 m / s. Explanation: Let : m =9 . 109 × 10 - 31 kg . L = v = L mR = 4 . 88 × 10 - 25 J · s (9 . 109 × 10 - 31 kg) (5 . 19206 cm) × ² 100 cm ³ = 1 . 03183 × 10 7 m / s . 006 10.0 points An electron is accelerated by a 2 . 8 kV poten- tial di²erence. The charge on an electron is 1 . 60218 × 10 - 19 C and its mass is 9 . 10939 × 10 - 31 kg. How strong a magnetic Feld must be expe- rienced by the electron if its path is a circle of radius 5 . 3 cm? Correct answer: 0 . 00336672 T. Explanation: Let : r =5 . 3 cm = 0 . 053 m , V =2 . 8 kV = 2800 V , m . 10939 × 10 - 31 kg , and q . 60218 × 10 - 19 C . ±rom Newton’s second law, F = q v B = 2 r v = q B r m .
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solution_pdf - garza (olg97) Homework No. 6 jimenez...

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