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qz5sp09sol - PHY2054 Discussion-Spring 09 Quiz 5(Chapter...

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February 24, 2009 PHY2054 Discussion-Spring ‘09 Quiz 5 (Chapter 19.1-19.6) Name: UFID: *1. (2.5pts) An electron is accelerated through a potential difference of 1500 V from rest and then enters a region where there is a uniform 2.00 T magnetic field. The magnetic field makes 45.0º to the direction of the electron’s motion. What is the direction and magnitude of the magnetic force on the electron? While the electron is accelerated by an electric field, the mechanical energy of the electron is conserved. Thus we have ΔE = 0 (1/2)mv 2 -eΔV = 0 v = √(2eΔV/m) = √[2×1.6×10 -19 ×1500/(9.11×10 -31 )] = 2.30×10 7 m/s The magnetic field makes 45º to the direction of electron’s motion, thus the magnitude of the magnetic force is F = qvBsin45º = 1.6×10 -19 ×2.3×10 7 ×2×sin45º = 5.20×10 -12 N To find the direction of magnetic force on a negative charge, point your fingers opposite the motion of the charge and curl your fingers in the direction of the field, then your thumb points in the direction of the force. It is out of the page.
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