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Chapter 13: Problem 1983

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) \(7.2 \times 10^{3}\)

Short Answer

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

Step by step solution

01

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)
02

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
03

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.

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Most popular questions from this chapter

Read the assertion and reason carefully to mark the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not the correct explanation of the assertion, (C) If assertion is true but reason is false. (D) If the assertion and reason both the false. (E) If assertion is false but reason is true. Assertion: Diamagnetic materials can exhibit magnetism. Reason: Diamagnetic materials have permanent magnetic dipole moment.

A magnetic field \(B^{-}=\) Bo \(j \wedge\) exists in the region \(a

The forces existing between two parallel current carrying conductors is \(\mathrm{F}\). If the current in each conductor is doubled, then the value of force will be (a) \(2 \mathrm{~F}\) (b) \(4 \mathrm{~F}\) (c) \(5 \mathrm{~F}\) (d) \((\mathrm{F} / 2)\)

For the mag. field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be (a) \(0^{\circ}\) (b) \(90^{\circ}\) (c) \(180^{\circ}\) (d) \(45^{\circ}\)

Two iclentical short bar magnets, each having magnetic moment \(\mathrm{M}\) are placed a distance of \(2 \mathrm{~d}\) apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is. (a) $\sqrt{2}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (b) $\sqrt{3}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (c) $\sqrt{4}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (d) $\sqrt{5}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$
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