Calculate the radius of curvature of the path of a beta particle of mass 9.1×10-31 kg and speed 4.0×107 m/s pe?
Calculate the radius of curvature of the path of a beta particle of mass 9.1×10-31 kg and speed 4.0×107 m/s perpendicular to a magnetic field of 0.0140 T.
A charged particle moving perpendicular to a magnetic field experiences a force of F(mag) = q v B. This force is perpendicular to the velocity of the particle so it is a centripetal force: F(centipetal) = mv^2/R. Therefore: q v B = m v^2 /R Divide both side by v : q B = m v / R Now it's easy to solve the equation for R : R = (m v) / (q B) Substituting numerical values we have:
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A charged particle moving perpendicular to a magnetic field experiences a force of F(mag) = q v B. This force is perpendicular to the velocity of the particle so it is a centripetal force: F(centipetal) = mv^2/R. Therefore: q v B = m v^2 /R Divide both side by v : q B = m v / R Now it's easy to solve the equation for R : R = (m v) / (q B) Substituting numerical values we have:
ANSWER: R = 0.016 m or R = 1.6 cm
a individual won't directly experience love, reason love can take time, have faith and to earnestly understand the different inspite of the extra proper issues they have shown or shared...somebody might have helped You yet then that is not a sturdy experience at cases. nevertheless yet, a minimum of you is additionally (experience) grateful by using appreciating the sturdy issues that it had dropped at instruction manual and help, by using not inflicting them any complication or soreness. maximum of all, by using doing the failings you may to enhance what they have contributed into Your existence. Have a sturdy 2d..! sometimes it takes the braveness and power to coach the extra proper way. even though it additionally should be with an truthful purpose, unfastened from malice and egocentric schemes or self serving.