If a charge of magnitude 4.96x10-17C, with speed 3.74x106 m/s, and mass 5.72x10-25kg moves within the magnetic field of magnitude 2.79x10-2T, what is the resulting radius of the path (in meters) the particle traces out?

Physics for Scientists and Engineers
10th Edition
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter1: Physics And Measurement
Section: Chapter Questions
Problem 26P: Review. Prove that one solution of the equation 2.00x43.00x3+5.00x=70.0 is x = 2.22.
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When a charged particle moves perpendicularly to the direction of a uniform magnetic field, the direction of the magnetic force is perpendicular to both the direction of the magnetic field and the direction of the velocity of the charged particle.  Accordingly, the charge begins to move in a circular path.  Since the direction of the magnetic force is toward the center of the circular path the charged particle moves along, the magnetic force is a centripetal force (recall centripetal forces from chapter 5 of the text, covered in PHY 2010).  A schematic of such motion is shown below.  The "X"s represent the magnetic field that is directed into the plane of this screen.  The direction of velocity and magnetic force lie within the plane of this screen and are at right angles to one another, as shown below.

 

If a charge of magnitude 4.96x10-17C, with speed 3.74x106 m/s, and mass 5.72x10-25kg moves within the magnetic field of magnitude 2.79x10-2T, what is the resulting radius of the path (in meters) the particle traces out?

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