The domain in a ferromagnetic material is in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$10 	imes 10^4 A m^{-1}$

Vedclass · Answer

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

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):

Similar Questions

$A$ magnetizing field of $2 \times 10^3 \, A/m$ produces a magnetic flux density of $8 \pi \, T$ in an iron rod. The relative permeability of the rod will be -

$A$ domain in a ferromagnetic substance is in the form of a cube of side $1 \mu m$. If it contains $8 \times 10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24} \ A \cdot m^{2}$,then the magnetisation of the domain is

Give the unit and dimensional formula of magnetisation.

The dimensions of magnetic intensity are

$A$ 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 $A/m$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module