A proton moving at a speed of 3.8 x 10⁶ m/s cuts across the lines of a magnetic field at an angle of 70 $\degree$.The strength of the field is 0.25 x 10^-4 T.What is the magnitude of the force acting on the proton? (q_p = 1.6 x 10^-19 C)

Question

Quizplus · Accepted Answer

The magnetic force on a moving charge is given by F = qvBsinθ. Substituting the given values: F = (1.6 x 10^-19 C)(3.8 x 10^6 m/s)(0.25 x 10^-4 T)sin(70°) = 1.4 x 10^-17 N.

A Proton Moving at a Speed of 3 °\degree° The Strength of the Field Is 0

A Proton Moving at a Speed of 3 $\degree$ The Strength of the Field Is 0