A cosmic-ray proton traveling at half the speed of light is heading directly tow

Question

A cosmic-ray proton traveling at half the speed of light is heading directly toward the center of the Earth in

the plane of the Earth’s equator. Will it hit the Earth? Assume that the Earth’s magnetic field is uniform

over the planet’s equatorial plane with a magnitude of 50.0 μT, extending out 1.30 × 107 m from the surface

of the Earth. Assume that the field is zero at greater distances. Calculate the radius of curvature of the

proton’s path in the magnetic field. Ignore relativistic effects.

Explanation / Answer

r = mv/qB = 1.67*10^-27*(1.5*10^8)/(1.6*10^-19*50*10^-6) = 31312.5 m = 3.13*10^4 m <-------radius of curvature of proton.

So, it wont hit the surface of earth....

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