The force on a charged particle moving in a direction perpendicular to a magnetic field is 0.50 N. What is the force on the particle in the same field if the angle between the particle's velocity vector and the field is 50°?
  1. 0.25 N
  2. 0.32 N
  3. 0.38 N
  4. 0.60 N
Explanation
Correct Response: C. The force on a charge moving through a magnetic field is proportional to the component of the velocity vector that is perpendicular to the field. In this case, the velocity was originally perpendicular to the field, so the force on the particle could be represented by the equation F = qvB = 0.50 N. After the direction of the velocity changed to 50° relative to the field, the component that is perpendicular to the field is equal to vsin 50°, and the force is now F = qvBsin 50° = 0.50 N sin 50° = 0.38 N. Answer choice A is half of the original force. Answer choice B would be obtained if cos 50° were incorrectly used in place of sin 50°. Answer choice D would be obtained by multiplying the original force by tan 50°.
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