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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 = 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 m q B correct 2. T = q B m 3. T = 2 q B m gilvin (jg47854) 17 Magnetic fields 2 meyers (21235) 2 4. T = 2 mB q 5. T = q B m 6. T = 2 m q B 7. T = mB q 8. T = 2 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 r v , which is intuitive since the particle traverses a distance 2 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 r v = 2 m q B ....
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This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.
 Spring '11
 doc
 Magnetism, Work

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