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. 2. Relevant equations F = MA =eVBsin? a = evB/m 3. 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 im 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 earths magnetic lines, but im not sure. Thanks.