Chap_22_soln[1]

Chap_22_soln[1] - 22-10. Picture the Problem: A proton...

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22-10. Picture the Problem: A proton moves straight downwards from a location above the equator, moving at right angles to the magnetic field that is horizontal and points due north. Strategy: Combine Newton's Second Law with the magnetic force (equation 22-1) to find the acceleration of the proton. Solution: Set the magnetic force equal to the mass multiplied by the acceleration and solve for a :    19 5 27 62 1.60 10 C 355 m/s 4.05 10 T sin90 1.673 10 kg 1.38 10 m/s evB a m     Insight: The proton would experience the very same acceleration if it were traveling due east or due west, or any other direction that is perpendicular to the horizontal magnetic field that points due north. 22-21. Picture the Problem: A velocity selector is constructed by forming perpendicular and EB fields as indicated in the diagram at the right. Strategy: The particle will pass through the fields undeflected if the magnetic force exactly balances the electric force. This occurs when the velocity is , vEB as discussed in the text. Solution: Calculate the ratio of the fields: 450 N/C 2.5 km/s 0.18 T E v B Insight: The velocity selector will work the same if both and are reversed. If only one field is reversed, the selected velocity must point in the ˆ x direction. 22-23. Picture the Problem: An electromagnetic flowmeter measures the speed of the blood in an artery. A diagram of the apparatus is shown at the right.
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Strategy: As described in the text, the speed of the blood in the artery is given by , vE B where EVd (equation 20-4), d is the inside diameter of the artery, and B is the magnitude of the magnetic field produced by the instrument.
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This note was uploaded on 10/13/2011 for the course PHY 102 taught by Professor Alexandrakis during the Fall '06 term at FIU.

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Chap_22_soln[1] - 22-10. Picture the Problem: A proton...

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