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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

Two straight long conductors \(\mathrm{AOB}\) and \(\mathrm{COD}\) are perpendicular to each other and carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). The magnitude of the mag. field at a point " \(\mathrm{P}^{n}\) at a distance " \(\mathrm{a}^{\prime \prime}\) from the point "O" in a direction perpendicular to the plane \(\mathrm{ABCD}\) is (a) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}+I_{2}\right)\) (b) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}-I_{2}\right)\) (c) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}\right)^{1 / 2}$ (d) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left[\left(I_{1} I_{2}\right) /\left(I_{1}-I_{2}\right)\right]$

An electron moving with a speed \(y_{0}\) along the positive \(x\) -axis at \(\mathrm{y}=0\) enters a region of uniform magnetic field \(\mathrm{B}^{-}=-\mathrm{B}_{0} \mathrm{k} \wedge\) which exists to the right of y-axis. The electron exits from the region after some time with the speed at co-ordinate y then.

An electron having mass \(9 \times 10^{-31} \mathrm{~kg}\), charge $1.6 \times 10^{-19} \mathrm{C}\( and moving with a velocity of \)10^{6} \mathrm{~m} / \mathrm{s}$ enters a region where mag. field exists. If it describes a circle of radius \(0.10 \mathrm{~m}\), the intensity of magnetic field must be Tesla (a) \(1.8 \times 10^{-4}\) (b) \(5.6 \times \overline{10^{-5}}\) (c) \(14.4 \times 10^{-5}\) (d) \(1.3 \times 10^{-6}\)

The mag. field (B) at the centre of a circular coil of radius "a", through which a current I flows is (a) \(\mathrm{B} \propto \mathrm{c} \mathrm{a}\) (b) \(\mathrm{B} \propto(1 / \mathrm{I})\) (c) \(\mathrm{B} \propto \mathrm{I}\) (d) \(B \propto I^{2}\)

A conducting rod of 1 meter length and \(1 \mathrm{~kg}\) mass is suspended by two vertical wires through its ends. An external magnetic field of 2 Tesla is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is [take $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$ (a) \(0.5 \mathrm{Amp}\) (b) \(15 \mathrm{Amp}\) (c) \(5 \mathrm{Amp}\) (d) \(1.5 \mathrm{Amp}\)
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