Question

Density of iron is 7.8 g/cc and induced magnetic field in iron is 1 tesla then find magnetic dipole moment of each iron atom.

• A. $57\times {10}^{-24}A{m}^2$
• B. $9.5\times {10}^{-24}A{m}^2$
• C. $21\times {10}^{-24}A{m}^2$
• D. $37\times {10}^{-24}A{m}^2$

Answer 1

The correct option is B $9.5\times {10}^{-24}A{m}^2$

Given: $densityofiron,\rho =7.8g/cc=7800kg/m^3$

$magneticfield,B=1T$

To find: magnetic moment,$m=?$

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

So,

$volumeofiron,V=\frac{massofiron}{\rho }$

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

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

As we know that,

$magneticfield={\mu }_0\ast magnetisation$

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

So, $magneticfield={\mu }_0\frac{magneticmoment}{volume}$

$==>B={\mu }_0\frac{m}{V}$

$==>m=\frac{BV}{{\mu }_0}$

$==>m=\frac{1\ast 0.01191794872\ast {10}^{-27}}{4\pi \ast {10}^{-7}}$

$==>m=0.00094888127\ast {10}^{-20}$

$==>m=9.4888127\ast {10}^{-24}$

$==>m=9.5\ast {10}^{-24}A{m}^2$

hence,

The correct opt : B