Altitude where Earth's magnetic field will no longer affect a compassby magnetics Tags: altitude, compass, earth, magnetic field 

#1
Apr1413, 08:42 PM

P: 32

Hi,
The Earth's magnetic field at the surface is roughly 0.5 Gauss or 0.05 mT. If the Earth's field strength diminishes in proportion to the inverse square of the distance, surely the field is going to be negligible at 10,000m altitude. Yet as far as I am aware it will still have an effect on the plane's compass traveling at such an altitude. Can someone please help me out with the maths to explain this scenario? Thank you. 



#2
Apr1413, 09:54 PM

Sci Advisor
PF Gold
P: 2,242

Hi magnetics
maybe you dont realise that the Earth's magnetic field surrounds the earth and extends out into space 1000's of kilometres ? At 10,000m or even 100,000m you are still deep within the field The magneto tail stretches 100's of 1000's of km "down stream" of the Earth ( ie the opposite direction to the sun as seen in the above pic do some googling on the Earth's magnetic field for tons of info and images :) Dave 



#4
Apr1413, 11:33 PM

P: 32

Altitude where Earth's magnetic field will no longer affect a compass
Thank you Dave,
I was aware that the magnetosphere extends many km into space. But I would like to see the maths used to calculate the approximate strength of the earth's field at 10km altitude. With regard to the sensitivity of a compass, I suspect it's about the quality of the compass and the force required to overcome the friction to turn the needle? On the face of it, I would have thought it very difficult to make a compass sensitive enough to turn in a very weak magnetic field 10km up? That's where I was looking for some help in the maths department. Thanks, 



#5
Apr1513, 04:46 AM

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P: 2,242

I dont know for sure without reading up about it
But I would expect the field would be stronger at that height for for some distance after all you are still deep within the field .... We both really need an expert answer on this one :) Dave 



#6
Apr1513, 05:10 AM

P: 551

If the field at R is 0.5 Gauss, then it's going to be 1/4th of that at 2R, i.e., ~6370km above the surface. Compare with your 10km high scenario. 



#7
Apr1513, 06:40 AM

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P: 2,242

lots of searching google tonite and quite difficult to find specific answers
but in this PDF file .... http://www.uio.no/studier/emner/matn...r%202_2708.pdf and deep into the info I found this ..... As long as we are located on the earth’s surface, r=R and the quantity (R/r)^{3} equals 1. But if we travel away from the earth’s surface, r increases, and the dipole field decreases. The reduction behaves like the third power of the distance; i.e., for r=2R (R ≈ 6500km) the field is just about 0.125 (12.5%) of the field at the earth’s surface. Dave 



#8
Apr1513, 07:10 AM

P: 32

Wow, thanks for the effort Dave. That's starting to make sense now.
If at 2R (6,500km) the field is still at 12.5%, then at R+10km it must be very close to 99.5%. {Although, according to Bandersnatch, shouldn't it be 1/4 (25%) and not 1/8 (12.5%)} I wonder how this would relate to a very powerful spherical neodymium magnet of say r = 15mm with a surface magnetic field strength of 3,500 Gauss or 0.35T? 



#9
Apr1513, 05:01 PM

P: 114

The field is a dipole one so presumably would fall off as 1/r^3 once you were far enough away from the generator structure?




#10
Apr1513, 05:55 PM

P: 551

Lately I'm getting better and better at putting my foot in my mouth. 


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