Calculating Earth's Core Current: 6.2E8 A

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The discussion revolves around calculating the electric current produced by charges flowing in Earth's molten outer core, using the equation M = IA, where M is the magnetic dipole moment and A is the area. The magnetic dipole moment of Earth is given as 8E22 J/T, and the area is calculated using the radius of 3200 km, resulting in an area of approximately 1.29E14 m². The current is then calculated as I = M/A, yielding a value of 6.2E8 A. Participants clarify that the area should be considered as the circular path of the charges rather than the entire sphere of the Earth. The initial confusion about the shape of the area is addressed, emphasizing the importance of correctly applying the formula.
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Hello everyone, I'm missing this problem and i aksed the professor if i was doing it wrong and he said, nope it looks like you just use that equation and solve for I. Here is the equation:
M = IA, where M stands for the moment; A stands for Area.
So you know the Moment is 1022 J/T. You also know the area is 4PIr^2;
A = 4PI(3200E3)^2;
A = 1.29E14;
I = M/A
I = (8E22)/1.29E14 = 620155038.8 = 6.2E8
Am i doing this right? It says its wrong!

The magnetic dipole moment of Earth is 8E22 J/T. Assume that this is produced by charges flowing in Earth's molten outer core. If the radius of their circular path is 3200 km, calculate the current they produce.
 
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The area enclosed by a circle is \pi r^2
 
Ohh my bad, i thought the Earth was more of a sphere than a circle, thank you!
 
mr_coffee said:
Ohh my bad, i thought the Earth was more of a sphere than a circle, thank you!

Sure, the Earth is a sphere. But we're not talking about the Earth here! I think you need to take another look at the basic concept.

mr_coffee said:
If the radius of their circular path is 3200 km...

We're talking about the area bounded by the circular path of the charges.
 
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