Current through Earth's center to produce magnetic field

  • Thread starter aldofbg
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1. The field strength of the earth is 5x10^-5 T at the surface. Space shuttles orbit the earth at a speed of 7860 m/s . They orbit at an altitude of 322 km. If we assume that a single current loop at the earths radius is generating the magnetic field, then what is the current? Determine the magnetic field at the space shuttles altitude.

2. Radius of Earth= 6.38x10^6 m
Amperes law= SBdl=uI (the S is my integral sign)
V=kQ/r



3. SBdl=uI
B is the magnetic field at the earths surface. dl becomes the earths circumference which is 2piR. I then solved for I by plugging in values and ending up with the formula I=(Bx2piR)/u= 1.6x10^9 A
I then used that value and solved in reverse for B using R as the distance from earths center to the shuttle height getting 4.77x10^-5 T.
This was an exam question and my professor marked it as wrong. I don't understand why and I was hoping someone could explain what I did wrong. Thanks
 

Answers and Replies

  • #2
TSny
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You used the formula for the B field of a long straight current. The problem refers to a "single current loop at the earth's radius". It's not clear to me what that means.
 

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