Question

Density of iron is 7800 kg/m${}^3$ and induced magnetic field in iron is 8 x10${}^5$ A/m then find magnetic dipole moment of each iron atom is (nearly)

• A. $6\times 10$ ^{-24} $Am$ ^{2} $
• B. $9\times 10$ ^{-22} $Am$ ^{2} $
• C. 10${}^{-23}$Am${}^2$
• D. $4\times 10$ ^{-24} $Am$ ^{2} $

Answer 1

The correct option is C 10${}^{-23}$Am${}^2$

Given: density of iron,$\rho =7800kg/m^3$

induced magnetic field,$M=8\ast {10}^{5}A/m$

To find: dipole moment,$m=?$

Solution: As we know that,
$massofiron=56a.m.u.=56\ast 1.66\ast {10}^{-27}kg$

So, volume of iron,$V=\frac{massofiron}{\rho }$

$==>V=\frac{56\ast 1.66\ast {10}^{-27}}{7800}$

$==>V=0.01191794872\ast {10}^{-27}{m}^3$

As we know that,

$magnetisation=\frac{magneticmoment}{volume}$

$==>M=\frac{m}{V}$

$==>m=MV$

$==>m=8\ast {10}^5\ast 0.01191794872\ast {10}^{-27}$

$==>m=0.0953\ast {10}^{-22}$

$==>m\simeq 0.1\ast {10}^{-22}A{m}^2$

$==>m\simeq {10}^{-23}A{m}^2$

hence,

The correct opt: C