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.

Question

Physics

Taffy

5 November, 20:42

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.

+5

Answers (1)

Know the Answer?

Meredith Gardner · Answer 1

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