Question

The exchange coupling mentioned in Section 32-11 as being responsible for ferromagnetism is not the mutual magnetic interaction between two elementary magnetic dipoles. To show this, (a) Calculate the magnitude of the magnetic field a distance of $\mathbf{10}\mathbf{}\mathbf{nm}$away, along the dipole axis, from an atom with magnetic dipole moment $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{T}$(cobalt), and (b) Calculate the minimum energy required to turn a second identical dipole end for end in this field. (c) By comparing the latter with the mean translational kinetic energy of $\mathbf{0}\mathbf{.}\mathbf{040}\mathbf{}\mathbf{eV}$, what can you conclude?

Short answer

1. The magnitude of the magnetic fieldis,$\mathrm{B}=3.0\times {10}^{-6}\mathrm{T}$
2. The minimum energy required is,${\mathrm{U}}_{\mathrm{minimum}}=5.6\times {10}^{-10}\mathrm{eV}$
3. The collision would easily randomize the direction of moments, and they would not remain aligned.

Step-by-step solution

Listing the given quantities

${\mathrm{\mu }}_{\mathrm{B}}=1.5\times {10}^{-23}\mathrm{J}/\mathrm{T}$

$\begin{array}{c}\mathrm{r}=10\mathrm{nm}\\ =10\times {10}^{-9}\mathrm{m}\end{array}$

Understanding the concepts of magnetic field

Here, we need to use the equation of magnetic field due to magnetic dipole moment. The minimum energy can be calculated using the equation of energy related to magnetic dipole moment and magnetic field.

Formulae:

Magnetic field due to the dipole moment ${\text{μ}}_{\text{B}}$at a distance r is

$\mathrm{B}=\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}\left(\frac{{\mathrm{\mu }}_{\mathrm{B}}}{{\mathrm{r}}^3}\right)$

(a) Calculations of the magnitude of the magnetic field 

The required field along the dipole axis:

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}\left(\frac{{\mathrm{\mu }}_{\mathrm{B}}}{{\mathrm{r}}^3}\right)\\ =\frac{4\mathrm{\pi }\times {10}^{-7}}{2\mathrm{\pi }}\left(\frac{1.5\times {10}^{-23}}{{\left({10}^{-8}\right)}^3}\right)\\ =3.0\times {10}^{-6}\text{T}\end{array}$

The magnitude of the magnetic field is,$\mathrm{B}=3.0\times {10}^{-6}\mathrm{T}$

(b) Calculations of the minimum energy required to turn the dipole end in the above calculated magnetic field

$\begin{array}{c}{\mathrm{U}}_{\mathrm{minimum}}={\mathrm{\mu }}_{\mathrm{B}}\mathrm{B}(\cos \left({\mathrm{\varphi }}_2\right)-\cos \left({\mathrm{\varphi }}_1\right))\\ =1.5\times {10}^{-23}\times 3.0\times {10}^{-6}\times (\cos \left(0\right)-\cos \left(180\right))\\ =9.0\times {10}^{-29}\mathrm{J}\\ =\frac{9.0\times {10}^{-29}}{1.6\times {10}^{-19}}\\ =5.6\times {10}^{-10}\mathrm{eV}\end{array}$

The minimum energy required is,${\mathrm{U}}_{\mathrm{minimum}}=5.6\times {10}^{-10}\mathrm{eV}$

(c) Explanation

As the mean translational kinetic energy (0.04 eV) is much larger than the required energy of the aligning dipoles, if dipole – dipole interactions were responsible for aligning dipoles, the collision would easily disturb the direction of moments, and they would not remain aligned.