A proton moving at 5.50 x 106 m/s through a magnetic field of magnitude 1.77 T experiences a magnetic force of magnitude 7.30 x 10-13 N. What is the angle between the proton's velocity and the field? Step 1 The magnitude of the force on a moving charge in a magnetic field is FB = qvB sin 8, and, solving for the angle, we have the following. 8 sin -1 = sin-¹ B qvB 1.60 x 10-19 c)([ x 10-13 N |× 106 m/s) (

A proton moving at 5.50 x 106 m/s through a magnetic field of magnitude 1.77 T experiences a magnetic force of magnitude 7.30 x 10-13 N. What is the angle between the proton's velocity and the field? Step 1 The magnitude of the force on a moving charge in a magnetic field is FB = qvB sin 8, and, solving for the angle, we have the following. 8 sin -1 = sin-¹ B qvB 1.60 x 10-19 c)([ x 10-13 N |× 106 m/s) (

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter22: Magnetism

Section: Chapter Questions

Problem 9PE: (a) A cosmic ray proton moving toward the Earth at 5.00107m/s experiences a magnetic force of...

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