Question

The exchange coupling mentioned in Module as being responsible for ferromagnetism is not the mutual magnetic interaction between two elementary magnetic dipoles. To show this, calculate (a) the magnitude of the magnetic field a distance of $$10$$ nm away, along the dipole axis, from an atom with magnetic dipole moment $$1.5 \times 10^{-23} J/T$$ (cobalt), and (b) 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 the kinetic energy of $$0.040\, eV$$, what can you conclude?

Answer 1

(a) The field of a dipole along its axis is given by Eq. 30-29: $$ B =\dfrac{μ_0}{2\pi}\,\dfrac{\mu}{z^3}$$, where μ is the dipole moment and z is the distance from the dipole. Thus,
$$B=\dfrac{(4\pi \times 10^{-7}\,T.m/A)(1.5 \times 10^{-23} \,J/T)}{2\pi(10 \times 10^{-9}\,m)}=3.0 \times 10^{-6}\,T$$
(b) The energy of a magnetic dipole $$\vec μ$$ in a magnetic field $$\vec B$$ is given by
$$U=-\vec \mu.\vec B=-\mu\cos \phi$$
where $$φ$$ is the angle between the dipole moment and the field. The energy required to turn it end-for-end (from $$φ = 0°$$ to $$φ = 180°$$) is
$$ΔU=2\mu B=2(1.5 \times 10^{-23}\,J/T)(3.0 \times 10^{-6}T)=9.0 \times 10^{-29}\,J=5.6 \times 10^{-10}\,eV$$
The mean kinetic energy of translation at room temperature is about $$0.04\, eV$$. Thus, if dipole-dipole interactions were responsible for aligning dipoles, collisions would easily randomize the directions of the moments and they would not remain aligned.