(A) The volume of the domain $V$ is given by the cube of its side length: $V = (2 \times 10^{-6} m)^3 = 8 \times 10^{-18} m^3$. The net magnetic dipol

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(A) The volume of the domain $V$ is given by the cube of its side length: $V = (2 \times 10^{-6} m)^3 = 8 \times 10^{-18} m^3$. The net magnetic dipole moment $M_{\text{net}}$ is the product of the number of atoms and the dipole moment per atom: $M_{\text{net}} = (9 \times 10^{10}) \times (9 \times 10^{-24} A m^2) = 81 \times 10^{-14} A m^2$. Magnetization $I$ is defined as the net magnetic moment per unit volume: $I = \frac{M_{\text{net}}}{V} = \frac{81 \times 10^{-14} A m^2}{8 \times 10^{-18} m^3}$. Calculating the value: $I = 10.125 \times 10^4 A m^{-1} \approx 10 \times 10^4 A m^{-1}$.

Class 12 Physics Magnetism and Matter Magnetization, Magnetic Induction Susceptibility

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The domain in a ferromagnetic material is in the form of a cube of side $2 \mu m$. The number of atoms in that domain is $9 \times 10^{10}$ and each atom has a dipole moment of $9 \times 10^{-24} A m^2$. The magnetization of the domain is (approximately):

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A
$10 \times 10^4 A m^{-1}$
B
$8 \times 10^4 A m^{-1}$
C
$12 \times 10^4 A m^{-1}$
D
$9 \times 10^4 A m^{-1}$

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Magnetization, Magnetic Induction Susceptibility

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$A$ substance has a mass of $0.075 \, kg$ and a density of $7500 \, kg/m^3$. If its magnetic dipole moment is $8 \times 10^{-7} \, A \cdot m^2$,what is its magnetization in $A/m$?

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The magnetic field $B$ and the magnetic intensity $H$ in a material are found to be $1.6 \ T$ and $1000 \ A/m$ respectively. Calculate the relative permeability $\mu_{r}$ of the material.

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An iron bar having a cross-sectional area of $2 \times 10^{-5} \ m^2$ and a magnetising field of $2400 \ A/m$ produces a magnetic flux of $2.4 \pi \times 10^{-5} \ Wb$. What will be the value of permeability $(\mu)$ and susceptibility $(\chi)$ of the bar? (Given $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$)

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$A$ rod of ferromagnetic material with dimensions $10 \, cm \times 0.5 \, cm \times 0.2 \, cm$ is placed in a magnetic field of strength $0.5 \times 10^4 \, A/m$. As a result,a magnetic moment of $5 \, A \cdot m^2$ is produced in the rod. The value of the total magnetic induction will be (in $Tesla$):

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$A$ magnetizing field of $5000 \ A/m$ produces a magnetic flux of $4 \times 10^{-5} \ Wb$ in an iron rod of cross-sectional area $0.4 \ cm^2$. The permeability of the rod in $Wb/(A \cdot m)$ is:

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The domain in a ferromagnetic material is in the form of a cube of side $2 \mu m$. The number of atoms in that domain is $9 \times 10^{10}$ and each atom has a dipole moment of $9 \times 10^{-24} A m^2$. The magnetization of the domain is (approximately):

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$A$ substance has a mass of $0.075 \, kg$ and a density of $7500 \, kg/m^3$. If its magnetic dipole moment is $8 \times 10^{-7} \, A \cdot m^2$,what is its magnetization in $A/m$?

The magnetic field $B$ and the magnetic intensity $H$ in a material are found to be $1.6 \ T$ and $1000 \ A/m$ respectively. Calculate the relative permeability $\mu_{r}$ of the material.

$A$ magnetizing field of $5000 \ A/m$ produces a magnetic flux of $4 \times 10^{-5} \ Wb$ in an iron rod of cross-sectional area $0.4 \ cm^2$. The permeability of the rod in $Wb/(A \cdot m)$ is:

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