Question

The Earth's magnetic field protects us from cosmic rays, which are extremely-high-energy subatomic particles generated by esoteric processes in interesting parts of the universe. To get an idea of how this protection might happen, pretend the earth's magnetic field has a constant value of 5.0 x 10^-5 T (northward) from the ground up to a height of 50 km.

(a) Suppose a high-energy proton (charge +1.6 x 10^-19 C, mass 1.67 x 10^-27 kg) hits the top of the earths magnetic field at some high velocity, call it v, straight. On the diagram, sketch the path it would follow.

(b) For what initial speed v would the proton just make it to the earth's surface?

(c) When the proton from part (b) reaches the surface, what is its velocity (magnitude and direction)?

(d) Would protons moving faster than your answer to (b) reach the ground? Would slower ones hit the ground?

(e) What does happen to the protons that don't hit the ground?

Answer 1

a] The path it would follow will be circular but since the ground is only 50 km from the top, it would only trace out quarter of a circle (that is if it were to hit the ground).

b] If the proton is to hit the ground, the radius of its path is r = 40 km = 40000 m

and the radius is given by:

c] From the above relation of the radius of the path, if the velocity increases, the radius of the path increases and so a faster proton will reach the ground and it will land away from the initial landing position.

If the proton has smallest velocity, the radius is smaller and it might not reach the ground depending on how small the velocity is.

d] The ones that do not hit the ground has nothing to stop its motion and so they will continue to move in a circular path with the constant velocity v.