Question
---Select--- +z axis y axis +y axis z axis direction not pertinent x axis +x axis
(a)
FE = ---Select--- N/C T C N ---Select--- y axis z axis +x axis x axis +z axis direction not pertinent +y axis
FBx = ---Select--- N N/C C T ---Select--- direction not pertinent y axis +y axis +x axis +z axis z axis x axis
FBy = ---Select--- N C N/C T ---Select--- y axis z axis x axis +z axis +x axis direction not pertinent +y axis
(b)
FE = ---Select--- T C N/C N ---Select--- x axis z axis +x axis direction not pertinent y axis +z axis +y axis
FBx = ---Select--- C T N/C N ---Select--- z axis x axis direction not pertinent +y axis +x axis +z axis y axis
FBy = ---Select--- C N N/C T ---Select--- x axis +y axis +z axis z axis direction not pertinent +x axis y axis
(c)
FE = ---Select--- N N/C C T ---Select--- +z axis +x axis +y axis x axis z axis direction not pertinent y axis
FBx = ---Select--- C T N/C N ---Select--- +x axis x axis y axis direction not pertinent +z axis z axis +y axis
FBy = ---Select--- N/C C N T
---Select--- +z axis y axis +y axis z axis direction not pertinent x axis +x axis
Explanation / Answer
we know that a moving charged particle will experience a magnetic force is given by F= qvBsin here = angle between velocity vector and direction magntic field (a)since the particle is at rest it will experience only electric force not magnetic force F_E = eE = (5.78*10^-6 C)*(300 N/C) = 1.734*10^-3 N along +ve x direction F_Bx = 0 F_By = 0 (b)v = 390 m/s F_E = F_E = eE = (5.78*10^-6 C)*(300 N/C) = 1.734*10^-3 N along +ve x direction F_Bx = qvBsin0^0 = 0 N F_By = qvBsin90^0 = (5.78C)(390m/s)(1.5T) = 3.38*10^-3 N (c) F_E = (5.78*10^-6 C)*(300 N/C) = 1.734*10^-3 N along +ve x direction F_Bx = qvBx sin90^0 = 3.764 *10^-3 N positive y axis F_By = qvBsin90^0 =(5.78C)(390m/s)(1.5T) = 3.38*10^-3 N negative x axis