A proton moving at 6.10 x 106 m/s through a magnetic field of magnitude 1.71 T e

Question

A proton moving at 6.10 x 106 m/s through a magnetic field of magnitude 1.71 T experiences a magnetic force of magnitude 7.70 x 10-13 N. What is the angle between the proton's velocity and the field? The magnitude of the force on a moving charge in a magnetic field is FB = qvB sin theta, and, solving for the angle, we have the following. Theta = sin-1 [FB/qvB] = sin-1 when we evaluate this expression for the angle, we obtain two possible values as answers. Giving the smaller angle first, we have

Explanation / Answer

F = BqvSin

7.70*10^-13 = (1.71*6.10*10^6*1.6*10^-19)*Sin

= 27.47 , 180+27.47

= 27.47 , 207.47

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