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Date Posted: 7/14/2008 4:07:19 PM  Status: Live
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Course Textbook Chapter Problem
Calculus Based Physics Physics for Scientists and Engineers A Strategic Approach 1st by Knight 32 30
Question Details:
A proton moves in the magnetic field B_vec \,=\:0.50 \hat{ i }\;{\rm T} with a speed of 1.0 \times 10^{7}\;{\rm m}/{\rm s} in the directions shown in the figure. For each, what is magnetic force F_vec on the proton?  (Intro 1 figure) 
 
Express vector \vec{F} in the form F_x, F_y, F_z, where the x, y, and z components are separated by commas.
 
Express vector \vec{F} in the form F_x, F_y, F_z, where the x, y, and z components are separated by commas.
 
knight_Figure_32_30.jpg
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Date Posted: 7/24/2008 8:36:22 AM  Status: Live
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Response:
 
 
          Given that the  magnetic field induction is  B = 0.50 T
                                 speed of  the  proton is  V = 1.0*107 m/s 
  ---------------------------------------------------------------------------
 
      figure A :
 
                   the force on the prticle moving with velocity is  F = B*q*Vsinθ
 
                  where  q = 1.602*10-19C is the charge of the proton 
                              θ = 450 is  the angle between the magnetic field and velocity vector
 
                  then   the   force is  F = (0.50 T)(1.602*10-19C )(1.0*107 m/s)sin45
                                                   = (--------  N ) 
         direction is along +ve Y-axis 
                                                     
   prt  B
                        θ = 1800 is  the angle between the magnetic field and velocity vector
 
                  then   the   force is  F = (0.50 T)(1.602*10-19C )(1.0*107 m/s)sin180
                                                   =0 N   
 



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