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A proton high above the equator approaches the Earth moving straight downward with a speed of 355 m/s. Find the acceleration of the proton, given that the magnetic field at its altitude is 4.05 X 10^-5 T.
F = MA
=eVBsin?
a = evB/m
a = ((1.6x10^-9C)(355 m/s)(4.05x10^-5)/(1.673x10^-27)) = 1.38x10^16 m/s^2
My attempt is pretty straight forward, but I think I am missing something. There has to be some relevance to the magnet force near the equator and the angle that the particle is moving, I think its perpendicular to Earth's magnetic lines, but I am not sure.
Thanks.
Homework Equations
F = MA
=eVBsin?
a = evB/m
The Attempt at a Solution
a = ((1.6x10^-9C)(355 m/s)(4.05x10^-5)/(1.673x10^-27)) = 1.38x10^16 m/s^2
My attempt is pretty straight forward, but I think I am missing something. There has to be some relevance to the magnet force near the equator and the angle that the particle is moving, I think its perpendicular to Earth's magnetic lines, but I am not sure.
Thanks.
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