Tutorial Exercise A proton moving at 6.10 times 106 m/s through a magnetic field

Question

Tutorial Exercise A proton moving at 6.10 times 106 m/s through a magnetic field of magnitude 1.71 T experiences a magnetic force of magnitude 7.70 times 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. When we evaluate this expression for the angle, we obtain two possible values as answers. Giving the smaller angle first, we have theta = Your response differs from the correct answer by more than 10%. Double check your calculation, degree.

Explanation / Answer

The second answer will be 360-27.47 = 332.53 degrees.

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