91Ó°ÊÓ

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

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

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

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

The required field along the dipole axis:

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

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

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

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

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.