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Chapter 28: Problem 63

The magnetic dipole moment of the Earth is approximately \(8.0 \cdot 10^{22} \mathrm{~A} \mathrm{~m}^{2}\). The source of the Earth's magnetic field is not known; one possibility might be the circulation culating ions move a circular loop of radius \(2500 \mathrm{~km}\). What “current" must they produce to yield the observed field?

Short Answer

Expert verified
Answer: The ions must produce a current of approximately \(4.08\times10^7~\text{A}\) to yield the observed magnetic field of Earth.

Step by step solution

01

Write down the given values.

The magnetic dipole moment of the Earth is given as \(8.0\times10^{22}~\text{A}~\mathrm{m}^{2}\). The radius of the circular loop is given as \(2500~\text{km}\).
02

Convert the radius to meters.

To work with SI units, we need to convert the radius from kilometers to meters: $$r = 2500~\text{km} \times 1000~\mathrm{m}/\text{km} = 2.5\times10^{6}~\text{m}$$
03

Express the magnetic dipole moment formula for a circular loop.

The formula for the magnetic dipole moment (\(\mu\)) of a circular loop with a current \(I\) and a loop radius \(r\) is given by: $$\mu = I \times \pi r^2$$
04

Solve for current I.

We can rearrange the formula for the magnetic dipole moment to solve for the required current: $$I = \frac{\mu}{\pi r^2}$$
05

Substitute the given values and calculate the current.

Now, substitute the given values for \(\mu\) and \(r\) and calculate the current \(I\): $$I = \frac{8.0\times10^{22}~\text{A}~\mathrm{m}^{2}}{\pi (2.5\times10^6~\mathrm{m})^2} \approx 4.08\times10^7~\text{A}$$ So, the ions must produce a current of approximately \(4.08\times10^7~\text{A}\) to yield the observed magnetic field of Earth.

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Most popular questions from this chapter

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