Question

The magnetic moment of the Earth is approximately 8.00 ×10²²A.m² . Imagine that the planetary magnetic field were caused by the complete magnetization of a huge iron deposit. with density 7900 kg / m³, and approximately 8.50×10²⁸ atoms / m³, (a) How many unpaired electrons, each with a magnetic moment of 9.27 ×10⁻²⁴A.m² , would participate?

Answer 1

$4.315 \times10^{45}$ unpaired electrons,are required each with a magnetic moment of 9.27 ×10⁻²⁴A.m² , to participate .

• Magnetic moment, also called magnetic dipole moment, is a measure of the  tendency of an object to align with a magnetic field.
• Magnetic moment is a vector quantity. Objects tend to be placed so that their magnetic moment vectors are parallel to the lines of magnetic force.
• The direction of the magnetic moment points from the south pole of the magnet to the north pole. The magnetic field produced by a magnet is directly proportional to its magnetic moment.

"Magnetic moment is defined as the magnetic strength and orientation of a magnet or other object that produces a magnetic field."

It is given that magnetic moment of the Earth is approximately 8.00 ×10²²A.m² .

Number of unpaired is found by dividing the magnetic moment of the Earth to the given magnetic moment of 9.27 ×10⁻²⁴A.m²

   $n=\frac{8.00\times10^{22}}{9.27\times10^{-24}} \\\\n=8.63\times10^{45}$

Each iron atom has two unpaired electrons, so the number of iron atoms required is $\frac{1}{2}\times 8.63\times10^{45}=4.315\times10^{45}$ electrons.

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