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Earth’s magnetic dipole moment has magnitude 8.00 × 10²² J/T. and is generated by charges flowing in the planet’s molten outer core. How much current, flowing around a circular path of radius 3500 km, would be needed to create a dipole of this magnitude?

User Terik
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Final answer:

To create Earth's magnetic dipole moment of 8.00 × 10²² J/T, a current of approximately 2.06 × 10¶ A flowing around a circular path of radius 3500 km would be required.

Step-by-step explanation:

To determine the amount of current needed to create a magnetic dipole moment of 8.00 × 10²² J/T, we must use the formula for the magnetic dipole moment (μ) of a current loop, which is given by μ = I×A, where I is the current and A is the area of the loop. For a circular loop, the area A is πr², where r is the radius of the loop.

In this case, since the Earth's magnetic dipole moment is 8.00 × 10²² J/T and the radius provided is 3500 km, which we convert to meters (3500 × 10³ m), we can set up the equation 8.00 × 10²² J/T = I × π × (3500 × 10³ m) ². Solving for I, we find:

I = μ / (πr²) = 8.00 × 10²² J/T / [π × (3500 × 10³ m)²] ≈ 2.06 × 10¶ A

Therefore, a current of approximately 2.06 × 10¶ A flowing around a circular path of radius 3500 km would be needed to create a magnetic dipole of the given magnitude.

User Mark Vital
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