Magnetic Force on Moving Charges

[AcidicVision]

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.

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.

 

[tiny-tim]

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.
…
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.

Hi AcidicVision!

I think you're right … assuming they mean the magnetic equator …

not a very good question, is it?