Question

A domain in a ferromagnetic substance is in the form of a cube of side 1 µm. If it contains 8 × 10¹⁰ atoms and each atomic dipole has a dipole moment of 9 × 10^-24 Am² then the magnetisation of the domain is ______.

विकल्प

• 7.2 × 10⁵ Am^-1 

• 7.2 × 10³ Am^-1 

• 7.2 × 10⁹ Am^-1 

• 7.2 × 10¹² Am^-1 

Answer 1

A domain in a ferromagnetic substance is in the form of a cube of side 1 µm. If it contains 8 × 10¹⁰ atoms and each atomic dipole has a dipole moment of 9 × 10^-24 Am² then the magnetisation of the domain is 7.2 × 10⁵ Am^-1.

Explanation:

Given, number of atoms, N = 8 × 10¹⁰ atoms

and dipole moment, M = 9 × 10^-24 Am² 

The maximum dipole moment is given by

M_max = NM

= 8 × 10¹⁰ × 9 × 10^-24 

= 7.2 × 10^-13 Am²

The volume of cubic domain,

V = `"I"^3 = (10^-6)^3   ...[because l = 1 mu"m" = 10^-6 "m"]`

= 10^-18 m³

The magnetisation of domain

`= "M"_"max"/"V" = (7.2 xx 10^-13)/10^-18 = 7.2 xx 10^5  "Am"^-1`