HW_10 Solutions

HW_10 Solutions - Physics 2213 Problem 1. HW #10 Solutions...

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Physics 2213 HW #10 Solutions Fall 2009 Problem 1. The force on the electron due to the magnetic field is given by () F ev B = −×  where v and B are the velocity of the electron and the strength of the magnetic field. From this, we see that only the component of the electron’s velocity that is perpendicular to the magnetic field results in a force. We can calculate the perpendicular component of this velocity as 14 19 5 / | | 4 10 / (1.6 10 )(0.60) 4.16 10 Perp v F eB −− = = ×× = × m/s. To obtain a force pointing upward, we see from the right hand rule that this perpendicular component of the velocity should be pointing due North. Also, since this perpendicular component is less than the total speed of the electron, we know that there is a component along the magnetic field of magnitude 2 25 5 5.0 4.16 10 2.76 10 Par v = ×= ± × m/s. (pointing east) From this, we see there are two possible directions in which the electron could have been moving. (b) If instead, the electron was moving due North at a speed of 5 5 10 × m/s, the force on it due to the magnetic field would be 19 5 14 (1.6 10 )(5 10 )(0.60) 4.8 10 F = = × N pointing upward. Problem 2 (a) The force on the cable due to Earth’s magnetic field is given by 4 (100 )(50 )(0.50 10 ) 0.25 F IlB A m T N = = From the right hand rule, this force will point 30 degrees below the horizontal pointing toward the south. (b) The total weight of the cable is given by 2 lg 8900 (0.005) 50 9.8 342 AN ρπ = × ×× = . So, the magnetic force is negligible compared to the weight. 27.19. IDENTIFY: In part (a), apply conservation of energy to the motion of the two nuclei. In part (b) apply 2 /. q vB mv R = SET UP: In part (a), let point 1 be when the two nuclei are far apart and let point 2 be when they are at their closest separation.
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This note was uploaded on 12/25/2009 for the course PHYS 213 at Cornell.

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HW_10 Solutions - Physics 2213 Problem 1. HW #10 Solutions...

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