Question

3 From a point on the equator where the earth magnetic field is 50 microT , pointing due north parallel to the surface of the earth, a proton is launched straight up, with a speed of 3 x10^6 / m/ s a. Calculate the magnitude of the magnetic force on the proton. b. What is the magnitude of the acceleration of the proton? c. The proton will follow a curved path until it will hit the surface of the earth again. What is the maximum height that the proton will reach? What is its speed at that point? d. Relative to the launch point, where will the proton hit the surface, and what is its velocity (speed and direction) at that point?

Answer 1

Magnetic force = 2.4* 10^ -17 Newtons

acceleration = 1.437*10^10 m/ s²

maximum height = 313 m

velocity at highest point is zero.

point of landing is at 12525m from point of launch.

velocity at landing = 9* 10^6 m/s.