Ask Question
14 June, 07:59

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.50 107 m/s and experiences an acceleration of 2.20 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

+2
Answers (1)
  1. 14 June, 08:24
    0
    9.175 x 10∧-3

    Explanation:

    since acceleration is in positve X direction the magnetic field must be in negative Y direction

    acceleration to right hand thumb rule.

    B = fm/qvsinO = ma/qvsin0

    B = (1.67 x 10∧-27) (2.20 x 10∧13) / (1.60 x 10∧-19) (2.50 x 10∧7) sin90

    B = 3.67 x 10∧-14 / 4 x 10∧-12

    = 9.175 x 10∧-3

    B = 9.175 x 10∧-3 in negative Y direction
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.50 107 m/s and experiences an acceleration of 2.20 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers