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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

Consider an electron to be a uniformly dense sphere of charge, with a total charge of \(-e=-1.60 \cdot 10^{-19} \mathrm{C}\) spinning at an angular frequency, \(\omega\). a) Write an expression for its classical angular momentum of rotation, \(L\) b) Write an expression for its magnetic dipole moment, \(\mu\). c) Find the ratio, \(\gamma_{e}=\mu / L\), known as the gyromagnetic ratio.

A hairpin configuration is formed of two semiinfinite straight wires that are \(2.00 \mathrm{~cm}\) apart and joined by a semicircular piece of wire (whose radius must be \(1.00 \mathrm{~cm}\) and whose center is at the origin of \(x y z\) -coordinates). The top straight wire is along the line \(y=1.00 \mathrm{~cm},\) and the bottom straight wire is along the line \(y=-1.00 \mathrm{~cm} ;\) these two wires are in the left side \((x<0)\) of the \(x y\) -plane. The current in the hairpin is \(3.00 \mathrm{~A},\) and it is directed toward the right in the top wire, clockwise around the semicircle, and to the left in the bottom wire. Find the magnetic field at the origin of the coordinate system.

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