Hw19_pdf - mcconnell (kam2342) homework19 Turner (60230) 1...

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Unformatted text preview: mcconnell (kam2342) homework19 Turner (60230) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points In an experiment designed to measure the strength of a uniform magnetic field produced by a set of coils, electrons are accelerated from rest through a potential difference of 202 V. The resulting electron beam travels in a circle with a radius of 3 . 64 cm. The charge on an electron is 1 . 60218 10 19 C and its mass is 9 . 10939 10 31 kg. Assuming the magnetic field is perpendic- ular to the beam, find the magnitude of the magnetic field. Correct answer: 0 . 00131668 T. Explanation: Let : e = 1 . 60218 10 19 C , r = 3 . 64 cm = 0 . 0364 m , V = 202 V , and m = m e = 9 . 10939 10 31 kg . Since K i = 0 and K f = 1 2 mv 2 , we have 1 2 mv 2 = | e | V v = radicalBigg 2 | e | V m e = radicalBigg 2 (1 . 60218 10 19 C) (202 V) 9 . 10939 10 31 kg = 8 . 42949 10 6 m / s . From conservation of energy, the increase in the electrons kinetic energy must equal the change in their potential energy | e | V : F = ev B = mv 2 r B = mv | e | r = (9 . 10939 10 31 kg) (1 . 60218 10 19 C) (8 . 42949 10 6 m / s) (0 . 0364 m) = . 00131668 T . 002 (part 2 of 2) 10.0 points What is the angular velocity of the electrons? Correct answer: 2 . 31579 10 8 rad / s. Explanation: For the angular velocity of the electron we obtain = v r = 8 . 42949 10 6 m / s . 0364 m = 2 . 31579 10 8 rad / s . 003 10.0 points Hint: Use non-relativistic mechanics to work this problem. A cyclotron is designed to accelerate pro- tons to energies of 6 . 7 MeV using a magnetic field of 0 . 6 T. The charge on the proton is 1 . 60218 10 19 C and its mass is 1 . 67262 10 27 kg. What is the required radius of the cy- clotron? Correct answer: 0 . 623334 m. Explanation: Let : B = 0 . 6 T , q = 1 . 60218 10 19 C , E = 6 . 7 MeV = (6 . 7 10 6 eV) (1 . 602 10 19 J / eV) , = 1 . 07334 10 12 J , and m = 1 . 67262 10 27 kg . The speed of the proton is v = radicalbigg 2 E m = radicalBigg 2 (1 . 07334 10 12 J) (1 . 67262 10 27 kg) = 3 . 58249 10 7 m / s , mcconnell (kam2342) homework19 Turner (60230) 2 where E is the kinetic energy of the proton. The magnetic force supplies the centripetal acceleration, so q v B = mv 2 r r = mv q B = m q B radicalbigg 2 E m = 2 mE q B = radicalbig 2 (1 . 07334 10 12 J) (1 . 60218 10 19 C) (0 . 6 T) radicalBig (1 . 67262 10 27 kg) = . 623334 m . 004 10.0 points A mass spectrometer consists of an acceler- ating potential (to give the ion momentum) and a uniform magnetic field (to momentum analyze the ion)....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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Hw19_pdf - mcconnell (kam2342) homework19 Turner (60230) 1...

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