Field Force
Sunday, July 25th, 2010A charged particle moves perpendicular to a magnetic field experiences a force whose magnitude is?
2.7x10 ^-3N. If the same count had to move at the same speed and the angle between the speed and the same magnetic field was 38 degrees, what would be the magnitude of the magnetic force that the charge would experience?
F = Q (v XB), where v XB | BB | sin (theta) is the vector product of the velocity vector q and the magnetic field vector B. Let F = 2.7e-3 N = QVB when theta = 90 degrees (perpendicular). Let f = sin QVB (38) is force when theta = 38 deg. Therefore f = F sin (38) = 0.616 2.7e-3 * = 1.66E-3 Newton is the reduction of force.