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Old Aug 31st 2013, 10:25 AM   #1
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Join Date: Feb 2013
Location: Greater St. Louis area
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Cosmic ray proton

I'm not really sure if this problem belongs in this forum. It is from Tipler chapter 3 on Quantization of Charge, Light, and Energy.

Problem 4: A cosmic ray proton approaches Earth vertically at the equator, where the horizontal component of Earth's magnetic field is 3.5 X 10^-5 T. If the proton is moving at 3.0X10^6 m/s, what is the ratio of the magnetic force to the gravitational force on the proton?

My answer is Fb/Fg = BQv sin theta / mg and since the velocity is perpendicular to the magnetic force then sin theta = 1.

So Fb/Fg = ((3.5 X 10^-5 T) (1.602176 X 10^-19 C) (3 X 10^6 m/s) ) /
(1.672622 X 10^-27 kg) (9.8 m/s^2)
= 1.0263 X 10^9

Could someone verify if this is the correct answer or not?

Thank you.
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