# Electrons In Space

1. Jul 17, 2007

### MJC8719

High above the surface of the Earth, charged particles (such as electrons and protons) can become trapped in the Earth's magnetic field in regions known as Van Allen belts. A typical electron in a Van Allen belt has an energy of 45 keV and travels in a roughly circular orbit with an average radius of 200 m. What is the magnitude of the Earth's magnetic field where such an electron orbits

So, the only equation I can think of that might be useful is r = mv/(qB) where solving for B would give me the magnetic field of the earth....I cannot figure out though, what the velocity of the particle is....

I thought it might be 45keV = 1/2mv^2 with m as the mass of an electron, but solving for v and plugging it into the above equation, did not yield a correct answer....

This problem is listed as an 'easy' one in our book, so I think its just something simple that cannot be clicking....

Any help would be greatly appreciated.

2. Jul 17, 2007

### ice109

did you do the computation right? how much energy is 45keV in joules?

3. Jul 17, 2007

### MJC8719

1 keV = 1.783 x 10-33J of energy....i am pretty sure i did that part right

Our professor also gave us the fact that the mass of e = 511kev/C2

So that would mean
(1/2)mv^2 = 45
v^2 = (45/0.5x511)
v = 0.41967

Then B = (511KeV)(0.41967)/(1.60X10-19)(200)
Which means B = 6.7016e18 T

Which is way to high....plus our answer is supposed to be in uT.

Am I missing a unit conversion somewhere

4. Jul 17, 2007

### ice109

1eV= 1.6e-19J
1.6e-19 (times) 45000= 7.2e-15J

edit

what is (1/2)mv^2=45?????

edit 2

b field of earth is about 10^-4T so like 100uT

Last edited: Jul 17, 2007
5. Jul 17, 2007

### MJC8719

I am trying to solve for the velocity correct....so I can use it in my equation for objects rotating around a circle....

So (1/2)mv^2 = KE = 45keV

I would then convert this all to joules....then I would have the velocity which could then be plugged into the equation: r = mv/(qB) so I could then solve for B...

Maybe I am approaching this wrong then??

6. Jul 17, 2007

### ice109

show all the work step by step because im getting an answer in uT and a much different velocity from you.

7. Jul 17, 2007

### Dick

You are being really sloppy with units. For simplicity convert everything to MKS, eg 511kev/c^2=9.1x10^(-31) kg.

8. Jul 17, 2007

### MJC8719

Thanks guys...i was being sloppy with the units....long day of work, class, then volunteering, plus an exam to study for tonight....much appreciated

9. Jul 17, 2007

### ice109

10. Jul 17, 2007

### Dick

I got a few microtesla. Probably about the same as you.

11. Jul 17, 2007

### ice109

ya but thats not the b field of the earth

12. Jul 17, 2007

### Dick

The earth's magnetic field (at the surface) is tens of microtesla. So we're in the right ballpark and the van Allen belt is not at the surface.

13. Jul 17, 2007

### ice109

oh yea wasn't thinking that it is above the surface

14. Jul 17, 2007

### Dick

Good thing for us, yea? 45KeV electrons aren't healthy.

15. Jul 17, 2007

### ice109

is the van allen belt where the aurora happens?

16. Jul 17, 2007

### Dick

Where aurora happen is where the van Allen belts dump excess electrons when they are disturbed. The electrons follow the field lines down to the poles. The aurora areas are nowhere near the belts proper.