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Unformatted text preview: Version PREVIEW HW7 feng (58230) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Charged Particle in a Field 02 001 10.0 points A charged particle is projected with its initial velocity parallel to a uniform magnetic field. What is the resulting path? 1. circular arc. 2. straight line parallel to the field. correct 3. parabolic arc. 4. straight line perpendicular to the field. 5. spiral. Explanation: The force on a moving charge due to a magnetic field is given by vector F = qvectorv vector B . If vectorv and vector B are parallel, then vectorv vector B = 0 . Hence the force on the particle is zero, and the particle continues to move in a straight line parallel to the field. Charge in a Magnetic Field 04 002 10.0 points A particle with charge q and mass m is un- dergoing circular motion with speed v . At t = 0, the particle is moving along the nega- tive x axis in the plane perpendicular to the magnetic field vector B , which points in the positive z direction in the figure below. x y z vectorv vector B Find the direction of the instantaneous ac- celeration hatwide a at t = 0 if q is negative. 1. hatwide a = i 2. hatwide a = k 3. hatwide a = j + k 4. hatwide a = k + i 5. hatwide a = i 6. hatwide a = k + i 7. hatwide a = j 8. hatwide a = k 9. hatwide a = i + j 10. hatwide a = j correct Explanation: The particle is moving along the negative x-axis in this instant vectorv = v i ; since it is moving in a circle, we need to talk about instantaneous direction. The force F B is equal to q vectorv vector B at all times. We know that vector B is pointing in the z direction, so vector B = B k , and therefore vector F B = q v ( i ) B k = q v B ( i k ) = q v B j . The charge q is negative ( q = | q | ) , so vector F B = | q | v B j = | q | v B ( j ) . All quantities are positive, so the actual di- rection in which vector F B points is the negative y direction, or hatwide a = j . Electron in a Magnetic Field 003 10.0 points Version PREVIEW HW7 feng (58230) 2 An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnetic field of 0 . 496 T at right angles to the direction of the electrons motion. The mass of the electron is 9 . 11 10 31 kg and its charge is 1 . 60218 10 19 C. What is the magnitude of the force on the electron due to the magnetic field? Correct answer: 1 . 06853 10 11 N. Explanation: Let : V = 51400 V , B = 0 . 496 T , m = 9 . 11 10 31 kg , q e = 1 . 60218 10 19 C ....
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