Question
El momento magnético de la Tierra es de aproximadamenteT. $8.00 \times 10^{27} \mathrm{~A} \cdot \mathrm{m}^2$. Imagine que el campo magnético planetario fuese causado por la magnetización completa de un enorme depósito de hierro con densidad de $7900 \mathrm{~kg} / \mathrm{m}^3$ y aproximadamente $8.50 \times 10^{35}$ átomos de hierro $/ \mathrm{m}^3$. (a) ¿Cuántos electrones no apareados participarian cada uno con un momento magnético de $9.27 \times 10^{-24} \mathrm{~A} \cdot \mathrm{m}^2$ ? (b) Con dos electrones no apareados por átomo de hierro, ¿̇cuántos kilogramos de hierro tendría el depósito?
Step 1
The Earth's magnetic moment is given as \(8.00 \times 10^{27} \, \mathrm{A} \cdot \mathrm{m}^2\). Each unpaired electron contributes a magnetic moment of \(9.27 \times 10^{-24} \, \mathrm{A} \cdot \mathrm{m}^2\). Show more…
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The magnetic moment of the Earth is approximately $8.00 \times 10^{22} \mathrm{A} \cdot \mathrm{m}^{2} .$ Imagine that the planetary magnetic field were caused by the complete magnetization of a huge iron deposit with density 7900 $\mathrm{kg} / \mathrm{m}^{3}$ and approximately 8.50 $\times 10^{28}$ iron atoms/m^{3} . (a) How many unpaired electrons, each with a magnetic moment of 9.27 $\times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2}$ , would participate? (b) At two unpaired electrons per iron atom, how many kilograms of iron would be present in the deposit?
The magnetic moment of the Earth is approximately $8.00 \times 10^{22} \mathrm{~A} \cdot \mathrm{m}^{2}$ (a) Imagine that the planetary magnetic field were caused by the complete magnetization of a huge iron deposit. How many unpaired electrons would participate? (b) At two unpaired electrons per iron atom, how many kilograms of iron would compose the deposit? Iron has a density of $7900 \mathrm{~kg} / \mathrm{m}^{3}$ and approximately $8.50 \times 10^{28}$ iron atoms $/ \mathrm{m}^{3}$
The magnetic moment of the Earth is approximately $8.00 \times 10^{22} \mathrm{A} \cdot \mathrm{m}^{2}$ (a) Imagine that the planetary magnetic field were caused by the complete magnetization of a huge iron deposit. How many unpaired electrons would participate? (b) At two unpaired electrons per iron atom, how many kilograms of iron would compose the deposit? Iron has a density of 7900 $\mathrm{kg} / \mathrm{m}^{3}$ and approximately $8.50 \times 10^{28}$ iron atoms $/ \mathrm{m}^{3}$ .
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