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Unformatted text preview: gilvin (jg47854) – 17 Magnetic fields 2 – meyers – (21235) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This homework is due Friday, April 15, at midnight Tucson time. 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 227 V. The resulting electron beam travels in a circle with a radius of 12 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 . 000423386 T. Explanation: Let : e = 1 . 60218 × 10 19 C , r = 12 cm = 0 . 12 m , V = 227 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) (227 V) 9 . 10939 × 10 31 kg = 8 . 93591 × 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 = e v B = mv 2 r B = mv  e  r = (9 . 10939 × 10 31 kg) (1 . 60218 × 10 19 C) × (8 . 93591 × 10 6 m / s) (0 . 12 m) = . 000423386 T . 002 (part 2 of 2) 10.0 points What is the angular velocity of the electrons? Correct answer: 7 . 44659 × 10 7 rad / s. Explanation: For the angular velocity of the electron we obtain ω = v r = 8 . 93591 × 10 6 m / s . 12 m = 7 . 44659 × 10 7 rad / s . 003 10.0 points A particle with a positive charge q and mass m is undergoing circular motion with speed v . At t = 0, the particle is moving along the negative x axis in the plane perpendicular to the magnetic field vector B , which points in the positive z direction as shown in the figure below. x y z vectorv vector B Find the period of the circular motion; i.e. , the time takes for the particle to complete one revolution. 1. T = 2 π m q B correct 2. T = q B m 3. T = 2 π q B m gilvin (jg47854) – 17 Magnetic fields 2 – meyers – (21235) 2 4. T = 2 mB q 5. T = π q B m 6. T = 2 m q B 7. T = mB q 8. T = 2 π mB q 9. T = m q B 10. T = 2 q B m Explanation: The force on a charge q in magnetic field is vector F B = qvectorv × vector B . The centripetal force F c is provided by F B , so mv 2 r = q v B v r = q B m . The period of oscillation is T = 2 π r v , which is intuitive since the particle traverses a distance 2 π r in a revolution and, moving at speed v , takes the time T to do so. We know the ratio v r , so T = 2 π r v = 2 π m q B ....
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 Spring '11
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 Electron, Magnetism, Force, Work, Magnetic Field, µ

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