20_InstSolManual_PDF_Part6

# 20_InstSolManual_PDF_Part6 - 20-6 Chapter 20 20.12 Set Up v...

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20.12. Set Up: and E is constant, so v B is constant. Solve: and 20.13. Set Up: Eq. (20.4) says A proton has mass and charge Solve: 20.14. Set Up: Since the particle moves perpendicular to the uniform magnetic field, the radius of its path is The magnetic force is perpendicular to both and The alpha particle has charge Solve: (a) The alpha particle moves in a circular arc of diameter (b) For a very short time interval the displacement of the particle is in the direction of the velocity. The magnetic force is always perpendicular to this direction so it does no work. The work-energy theorem therefore says that the kinetic energy of the particle, and hence its speed, is constant. (c) The acceleration is We can also use and the result of part (a) to calculate the same result. The acceleration is perpendicular to and and so is horizontal, toward the center of curvature of the particle’s path. (d) The unbalanced force is perpendicular to so it changes the direction of but not its magnitude, which is the speed.
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