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February 7th, 2011, 01:31 AM  #1 
Senior Member Joined: Oct 2009 Posts: 895 Thanks: 1  A proton moves perpendicular to a uniform magnetic fieldB at
A proton moves perpendicular to a uniform magnetic field B at 2.20 X 10^7 m/s and experiences an acceleration of 2.00 X 10^13 m/s2 in the +x direction when its velocity is in the +z direction. 1 ) Determine the magnitude and direction of the field. 2) drwa the case  I try The force here is the maximum due the charged particle moves perpendicular to the field the magnitude of the field Therefore, F ? qvB sin ? B = F/qvsinQ = (Is correct ?) Now direction B out 2) 

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fieldb, magnetic, moves, perpendicular, proton, uniform 
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