337sin2sin2 (0.018 m)(4.6 10 A)(3.4 10 T)sin 206.0 10 N.vFiBds aiBθθπ−−−===××°=×∫πWe note that i, B, and θhave the same value for every segment and so can be factored from the integral. 49. Consider an infinitesimal segment of the loop, of length ds. The magnetic field is perpendicular to the segment, so the magnetic force on it has magnitude dF = iB ds. The horizontal component of the force has magnitude (cos)hdFiBds=and points inward toward the center of the loop. The vertical component has magnitude
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.