Calculating Earth's Total Electric Current from Cosmic Ray Protons

[Saladsamurai]

[SOLVED] Cosmic Ray Protons

Homework Statement

Earth's atmosphere is constantly being bombarded with cosmic ray protons that originate somewhere in space. If the protons all pass through the atmosphere, each square meter of Earth's surface would intercept protons at at the average rate of 1500 protons per second. What would be the electric current intercepted by the total surface area of the entire planet?

Homework Equations

Surface Area of sphere=4*pi*r^2

The Attempt at a Solution

Okay, I don't even understand what this question is asking.
It wants to know the total # of protons the surface will intercept. I get that. But then it gives a rate protons per second, but does not give a tine interval.

I am lost on this one :/

 

[Doc Al]

All you need is the rate at which charge is intercepted, which is given. (What's the definition of current?)

 

[BishopUser]

From how I am understanding it. Current is defined as flow of charge (charge/time). So if they are asking for current it would be on a per unit time basis. I think the main part of the problem is how to convert protons/sec into charge/sec.

 

[Shooting Star]

I think the main part of the problem is how to convert protons/sec into charge/sec.

Why would that be a problem? Does mankind not know the charge of a proton yet?

 

[Saladsamurai]

All you need is the rate at which charge is intercepted, which is given. (What's the definition of current?)

All I can find for a definition of current is i=\frac{dq}{dt} where i is in ampheres and dq is in Coulomb's. . . hmmm. So I would need to convert 1500*4\pi R_{earth}^2 into Coulombs?

 

[BishopUser]

Why would that be a problem? Does mankind not know the charge of a proton yet?

uhh mankind does, but maybe some students don't (as witnessed by this thread)?

 

[Doc Al]

All I can find for a definition of current is i=\frac{dq}{dt} where i is in ampheres and dq is in Coulomb's. . . hmmm. So I would need to convert 1500*4\pi R_{earth}^2 into Coulombs?

You're on the right track. 1500*4\pi R_{earth}^2 gives you protons per second. What's the charge of a proton? The current would be in Coulombs/second or Amps.

 

[Saladsamurai]

You're on the right track. The current would be in Coulombs/second or Amps. What's the charge of a proton?

Isn't it just 1e? I don't follow. . . or are you going to follow that question with "now how many e are in one Coulomb" . . . cause if you are, I would have to say 6.25(10^18) and then I would have to do a simple chain conversion and get the right answer!

Thanks,
Casey

Now someone please help me with my trig before I burn down this town! J/K, but seriously please someone check out my finding the angle thread.

 

[Doc Al]

I would have said: What's the proton charge in Coulombs? e = 1.60217646 × 10-19 C. (But same difference.)

 

[Shooting Star]

uhh mankind does, but maybe some students don't (as witnessed by this thread)?

Point taken (somewhat) . I am sure that the student knew the charge of a proton. The problem lied in correlating the number/sec and the corresponding charge/sec.

 

[Astronuc]

As Doc mentioned, you're on the right track.

If N particles/sec, each of charge e⁺, pass some point, the current I = Ne.

 

[Saladsamurai]

I would have said: What's the proton charge in Coulombs? e = 1.60217646 × 10-19 C. (But same difference.)

Yeah. I did too many extracurricular activities in my late-teens/ early twenties. . . I cannot afford to memorize all these conversions, so I just memorize one and derive the others from it!