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.
