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Unformatted text preview: Kapoor (mk9499) – homework19 – Turner – (60230) 1 This printout should have 12 questions. Multiplechoice 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 331 V. The resulting electron beam travels in a circle with a radius of 4 . 65 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 . 00131937 T. Explanation: Let : e = 1 . 60218 × 10 − 19 C , r = 4 . 65 cm = 0 . 0465 m , V = 331 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) (331 V) 9 . 10939 × 10 − 31 kg = 1 . 07905 × 10 7 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) × (1 . 07905 × 10 7 m / s) (0 . 0465 m) = . 00131937 T . 002 (part 2 of 2) 10.0 points What is the angular velocity of the electrons? Correct answer: 2 . 32053 × 10 8 rad / s. Explanation: For the angular velocity of the electron we obtain ω = v r = 1 . 07905 × 10 7 m / s . 0465 m = 2 . 32053 × 10 8 rad / s . 003 10.0 points Hint: Use nonrelativistic mechanics to work this problem. A cyclotron is designed to accelerate pro tons to energies of 8 . 3 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 . 693781 m. Explanation: Let : B = 0 . 6 T , q = 1 . 60218 × 10 − 19 C , E = 8 . 3 MeV = (8 . 3 × 10 6 eV) × (1 . 602 × 10 − 19 J / eV) , = 1 . 32966 × 10 − 12 J , and m = 1 . 67262 × 10 − 27 kg . The speed of the proton is v = radicalbigg 2 E m = radicalBigg 2 (1 . 32966 × 10 − 12 J) (1 . 67262 × 10 − 27 kg) = 3 . 98737 × 10 7 m / s , Kapoor (mk9499) – 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 . 32966 × 10 − 12 J) (1 . 60218 × 10 − 19 C) (0 . 6 T) × radicalBig (1 . 67262 × 10 − 27 kg) = . 693781 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 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

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