Problem 1983 - JEE Physics

Question

A domain in a ferromagnetic substance is in the form of a cube of side length \(1 \mu \mathrm{m}\). If it contains \(8 \times 10^{10}\) atoms and each atomic dipole has a dipole moment of \(9 \times 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\) then magnetization of the domain is \(\mathrm{A} \mathrm{m}^{-1}\) a) \(7.2 \times 10^{5}\) (b) \(7.2 \times 10^{3}\) (c) \(7.2 \times 10^{5}\) (d) \(7.2 \times 10^{3}\)

Step-by-Step Solution

Answer

The magnetization of the domain is \(7.2 \times 10^5 A/m\).

Step 1: Calculate the total magnetic moment

To find the total magnetic moment, we need to multiply the number of atoms by the dipole moment of each atom: Total magnetic moment = (number of atoms) × (dipole moment of each atom) Total magnetic moment = (8 × 10^{10}) × (9 × 10^{-24} A·m^2)

Step 2: Calculate the volume of the domain

As the domain is cube-shaped, we can calculate its volume by taking the cube of its side length. The side length of the domain is given as 1 µm, which is equal to 10^{-6} m. Thus, the volume of the domain is given by: Volume = (side length)^3 Volume = (10^{-6} m)^3 Volume = 10^{-18} m^3

Step 3: Calculate the magnetization of the domain

Now that we have the total magnetic moment and the volume of the domain, we can calculate the magnetization (M) with the formula: M = total magnetic moment / volume M = [(8 × 10^{10}) × (9 × 10^{-24} A·m^2)] / (10^{-18} m^3) M = 7.2 × 10^5 A/m Therefore, the magnetization of the domain is 7.2 × 10^5 A/m, which corresponds to option (c) in the given choices.

Key Concepts

Understanding Ferromagnetic Substances

Ferromagnetic substances are materials that can become magnetized all on their own. Examples of such materials include iron, cobalt, and nickel. These materials have domains - tiny regions where the magnetic moments of atoms are aligned in the same direction. This alignment causes the whole domain to act like a big magnet.
Even without an external magnetic field, these domains can retain magnetization. When a ferromagnetic substance is exposed to a magnetic field, all the individual domains attempt to align according to the field, thus strengthening the material's overall magnetization. Once removed from the field, some of the alignment stays, which is why these materials can stay magnetized.
In our exercise, the domain in question is a cube consisting of many aligned atomic dipoles, which contribute to the overall magnetization of the substance.

Decoding Magnetic Dipole Moment

A magnetic dipole moment is a property of atoms that makes them act like tiny magnets. It's essentially a measure of the strength and orientation of a magnetic source.
This concept is crucial when considering how each atom within a material contributes to its magnetism. The dipole moment is determined by factors such as the current loop formed by the path of electrons around the atom's nucleus. In simple terms, you can imagine each atom as a small bar magnet with a north and a south pole.
In our exercise, each atom in the domain has a magnetic dipole moment of \(9 \times 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\), which means every one of the \(8 \times 10^{10}\) atoms contributes to the total magnetic potential of the domain, enhancing its magnetization.

Cube Volume Calculation Explained

The volume of a cube can be found using a simple mathematical equation. A cube's volume is calculated by raising the length of one of its sides to the third power.
This makes perfect sense because a cube has equal length, width, and height. In our exercise, the side of the cube is given as \(1 \mu \text{m}\), or \(10^{-6} \text{m}\).
To find the volume, you use the formula:

• Volume = (side length)^3
• Substitute the given side length: Volume = \((10^{-6} \text{m})^3 = 10^{-18} \text{m}^3\)

This volume measurement is key as it helps us figure out the magnetization of the domain by considering how much space the magnetic moment is distributed over.

Magnetic Moment Calculation Uncovered

To find the total magnetic moment of a domain, you multiply the number of atoms by the magnetic dipole moment of each atom. This is crucial because the total magnetic moment provides insight into the overall magnetic properties of the substance.
In our example, the calculation is straightforward: Multiply \(8 \times 10^{10}\) atoms by each atom's dipole moment, \(9 \times 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\).

• Total Magnetic Moment = \((8 \times 10^{10}) \times (9 \times 10^{-24} \mathrm{~A} \mathrm{~m}^{2})\)
• This results in a total magnetic moment of \(7.2 \times 10^{-13} \mathrm{~A} \mathrm{~m}^{2}\).

This value is then divided by the volume of the domain to find the magnetization. These calculations allow for understanding how densely packed the magnetic property is within the substance, helping to determine how effectively it can be magnetized.