Answer to Question #207427 in Astronomy | Astrophysics for eunice

Question #207427

Ø A proton travels with a speed of 5.02´106 m/s in a direction that makes an angle of 60.0° with the direction of a magnetic field of magnitude 0.180 T in the positive x direction. What are the magnitudes of: a) Magnetic force on the proton. b) The proton's acceleration?


1
Expert's answer
2021-06-16T09:52:42-0400

The magnitude:


F=qvBsinθ=1.251013 N.F=qvB\sin\theta=1.25·10^{-13}\text{ N}.

The acceleration:


a=Fm=7.501013 m/s2.a=\frac Fm=7.50·10^{13}\text{ m/s}^2.


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