Homwork Combined

# Homwork Combined - Kapoor(mk9499 homework19 Turner(60230...

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Kapoor (mk9499) – homework19 – Turner – (60230) 1 This print-out should have 12 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 In an experiment designed to measure the strength oF a uniForm magnetic feld produced by a set oF coils, electrons are accelerated From rest through a potential di±erence 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 feld is perpendic- ular to the beam, fnd the magnitude oF the magnetic feld. 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 2 = | e | V v = ± 2 | e | V m e = ± 2 (1 . 60218 × 10 - 19 C) (331 V) 9 . 10939 × 10 - 31 kg . 07905 × 10 7 m / s . ²rom conservation oF energy, the increase in the electrons’ kinetic energy must equal the change in their potential energy | e | V : F = ev B = 2 r B = | e | r = (9 . 10939 × 10 - 31 kg) (1 . 60218 × 10 - 19 C) × (1 . 07905 × 10 7 m / s) (0 . 0465 m) = 0 . 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: ²or the angular velocity oF the electron we obtain ω = v r = 1 . 07905 × 10 7 m / s 0 . 0465 m = 2 . 32053 × 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 8 . 3 MeV using a magnetic feld 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 . 6T , q . 60218 × 10 - 19 C , E =8 . 3 MeV = (8 . 3 × 10 6 eV) × (1 . 602 × 10 - 19 J / eV) , . 32966 × 10 - 12 J , and m . 67262 × 10 - 27 kg . The speed oF the proton is v = ² 2 E m = ± 2 (1 . 32966 × 10 - 12 J) (1 . 67262 × 10 - 27 kg) =3 . 98737 × 10 7 m / s ,

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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 = qB = m ± 2 E m = 2 mE = ² 2 (1 . 32966 × 10 - 12 J) (1 . 60218 × 10 - 19 C) (0 . 6 T) × ³ (1 . 67262 × 10 - 27 kg) = 0 . 693781 m . 004 10.0 points A mass spectrometer consists of an acceler- ating potential (to give the ion momentum) and a uniform magnetic Feld (to momentum analyze the ion). The mass spectrometer using a beam of doubly ionized sodium atoms has the follow- ing settings: the magnetic Feld is 0 . 0815 T; the charge of an atom is 3 . 20435 × 10 - 19 C; the radius of the orbit is 0 . 0537 m; and the potential di±erence is 159 V. Calculate the mass of a sodium atom. Correct answer: 1 . 93009 × 10 - 26 kg. Explanation: Let : q =3 . 20435 × 10 - 19 C , r =0 . 0537 m , V = 159 V , and B . 0815 T . The kinetic energy is 1 2 2 = qV v = ± 2 m . The atoms move in a circular path, so q v B = m v 2 r q B r = = m ± m = ² 2 q V m m = 2 r 2 2 V = (3 . 20435 × 10 - 19 C) 2 (0 . 0815 T) × (0 . 0815 T) 2 (0 . 0537 m) 2 = 1 . 93009 × 10 - 26 kg . 005 10.0 points A wire carrying a current 20 A has a length 0 . 1 m between the pole faces of a magnet at an angle 60 (see the Fgure). The magnetic Feld is approximately uniform at 0 . 5 T. We ignore the Feld beyond the pole pieces.
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## This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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Homwork Combined - Kapoor(mk9499 homework19 Turner(60230...

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