The magnitude of the dipole moment associated with an atom of iron in an iron bar is $\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{T}$. Assume that all the atoms in the bar, which is$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$long and has a cross-sectional area of$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{cm}}^{\mathbf{2}}$, have their dipole moments aligned. (a) What is the dipole moment of the bar? (b) What torque must be exerted to hold this magnet perpendicular to an external field of magnitude$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{T}$? (The density of iron is$\mathbf{7}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$.)

$\begin{array}{c}{\text{μ}}_{\text{Fe}}=2.1\times {10}^{-23}\text{J}/\text{T}\\ \text{A}=1.0{\text{cm}}^2\\ \text{B}=1.5\text{T}\\ \text{ρ}=7.9\text{g}/{\text{cm}}^3\end{array}$

We can find the number of atoms in iron atom by using the mass of the iron bar, the molar mass and Avogadro’s number. Then, find the dipole moment of the bar using the dipole of moment of the atom and the number of atoms in the iron bar. Torque can be calculated by using the dipole moment of the bar and the magnetic field.

Formula:

Number of atoms in the iron bar is$\mathbf{N}\mathbf{=}\mathbf{\left(}\frac{\mathbf{m}}{\mathbf{M}}\mathbf{\right)}\mathbf{\times }{\mathbf{N}}_{\mathbf{A}}$

Here,

$\mathbf{m}\mathbf{}\mathbf{-}$Mass of iron bar.

$\mathbf{M}\mathbf{-}\mathbf{molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{iron}\mathbf{.}$

${\mathbf{N}}_{\mathbf{A}}\mathbf{-}\mathbf{}\mathbf{Avogadro}\mathbf{\text{'}}\mathbf{s}\mathbf{}\mathbf{number}\mathbf{.}$

Dipole moment of the iron bar:$\mathbf{\mu }\mathbf{=}{\mathbf{\mu }}_{\mathbf{Fe}}\mathbf{\times }\mathbf{N}$

${\mathbf{\mu }}_{\mathbf{Fe}}\mathbf{-}\mathbf{}\mathbf{dipole}\mathbf{}\mathbf{moment}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{atoms}\mathbf{}\mathbf{in}\mathbf{}\mathbf{the}\mathbf{}\mathbf{iron}\mathbf{}\mathbf{bar}\mathbf{.}$

(a)

$\begin{array}{c}\mathrm{Mass}\left(\mathrm{m}\right)=\mathrm{Density}\left(\mathrm{\rho }\right)\times \mathrm{volume}\left(\mathrm{V}\right)\\ \mathrm{m}=7.9\times 1.0\times 5.0\\ =39.5\mathrm{g}\end{array}$

The number of atoms in the iron bar is

$\begin{array}{c}\mathrm{N}=\left(\frac{\mathrm{m}}{\mathrm{M}}\right)\times {\mathrm{N}}_{\mathrm{A}}\\ =\frac{39.5}{55.847}\times 6.023\times {10}^{23}\\ =4.3\times {10}^{23}\mathrm{atoms}\end{array}$

The dipole moment of the iron bar:

$\begin{array}{c}\mathrm{\mu }={\mathrm{\mu }}_{\mathrm{Fe}}\times \mathrm{N}\\ =2.1\times {10}^{-23}\times (4.26\times {10}^{23})\\ =8.9\mathrm{A}\cdot {\mathrm{m}}^2\end{array}$

The dipole moment of the bar is$\mathrm{\mu }=8.9\mathrm{A}\cdot {\mathrm{m}}^2$

(b)

$\begin{array}{c}\mathrm{\tau }=\text{μBsin}\left(\text{θ}\right)\\ =8.9\times 1.5\times \sin \left(90\right)\\ =13\mathrm{N}.\mathrm{m}\end{array}$

The torque required is $\mathrm{T}=13\text{N.m}$