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 to a height of 50 km.

(a) For what initial speed v would a proton of charge 1.6 x 10^-19 C and mass 1.67 x 10^-27 kg just make it to the earth's surface?

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

(c) What happens to the protons that are too slow that don't hit the ground?

Answer 1

part A

the path that proton follows wll be circular

if it has to hit the gorund, it raiud r = 50 km

so using mv^2/r = q v B

v = q B r/m

v = (1.6*10^-19 * 5*10^-5 * 50000)/(9.11*10^-27)

v = 4.4 *10^7 m/s

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speed is 4.4*10^7 m ' and direction will be raidally circular path

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part C:

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.