### Learning Objectives

By the end of this section, you will be able to:

- Classify magnetic materials as paramagnetic, diamagnetic, or ferromagnetic, based on their response to a magnetic field
- Sketch how magnetic dipoles align with the magnetic field in each type of substance
- Define hysteresis and magnetic susceptibility, which determines the type of magnetic material

Why are certain materials magnetic and others not? And why do certain substances become magnetized by a field, whereas others are unaffected? To answer such questions, we need an understanding of magnetism on a microscopic level.

Within an atom, every electron travels in an orbit and spins on an internal axis. Both types of motion produce current loops and therefore magnetic dipoles. For a particular atom, the net magnetic dipole moment is the vector sum of the magnetic dipole moments. Values of $\mu $ for several types of atoms are given in Table 12.1. Notice that some atoms have a zero net dipole moment and that the magnitudes of the nonvanishing moments are typically ${10}^{\mathrm{-23}}\text{A}\xb7{\text{m}}^{2}.$

Atom | Magnetic Moment $\left({10}^{\mathrm{-24}}\phantom{\rule{0.2em}{0ex}}\text{A}\xb7{\text{m}}^{2}\right)$ |
---|---|

H | 9.27 |

He | 0 |

Li | 9.27 |

O | 13.9 |

Na | 9.27 |

S | 13.9 |

A handful of matter has approximately ${10}^{26}$ atoms and ions, each with its magnetic dipole moment. If no external magnetic field is present, the magnetic dipoles are randomly oriented—as many are pointed up as down, as many are pointed east as west, and so on. Consequently, the net magnetic dipole moment of the sample is zero. However, if the sample is placed in a magnetic field, these dipoles tend to align with the field (see Equation 12.14), and this alignment determines how the sample responds to the field. On the basis of this response, a material is said to be either paramagnetic, ferromagnetic, or diamagnetic.

In a paramagnetic material, only a small fraction (roughly one-third) of the magnetic dipoles are aligned with the applied field. Since each dipole produces its own magnetic field, this alignment contributes an extra magnetic field, which enhances the applied field. When a ferromagnetic material is placed in a magnetic field, its magnetic dipoles also become aligned; furthermore, they become locked together so that a permanent magnetization results, even when the field is turned off or reversed. This permanent magnetization happens in ferromagnetic materials but not paramagnetic materials. Diamagnetic materials are composed of atoms that have no net magnetic dipole moment. However, when a diamagnetic material is placed in a magnetic field, a magnetic dipole moment is directed opposite to the applied field and therefore produces a magnetic field that opposes the applied field. We now consider each type of material in greater detail.

### Paramagnetic Materials

For simplicity, we assume our sample is a long, cylindrical piece that completely fills the interior of a long, tightly wound solenoid. When there is no current in the solenoid, the magnetic dipoles in the sample are randomly oriented and produce no net magnetic field. With a solenoid current, the magnetic field due to the solenoid exerts a torque on the dipoles that tends to align them with the field. In competition with the aligning torque are thermal collisions that tend to randomize the orientations of the dipoles. The relative importance of these two competing processes can be estimated by comparing the energies involved. From Equation 12.14, the energy difference between a magnetic dipole aligned with and against a magnetic field is ${U}_{B}=2\mu B.$ If $\mu =9.3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-24}}\text{A}\xb7{\text{m}}^{2}$ (the value of atomic hydrogen) and *B* = 1.0 T, then

At a room temperature of $27\phantom{\rule{0.2em}{0ex}}\text{\xb0C},$ the thermal energy per atom is

which is about 220 times greater than ${U}_{B}.$ Clearly, energy exchanges in thermal collisions can seriously interfere with the alignment of the magnetic dipoles. As a result, only a small fraction of the dipoles is aligned at any instant.

The four sketches of Figure 12.23 furnish a simple model of this alignment process. In part (a), before the field of the solenoid (not shown) containing the paramagnetic sample is applied, the magnetic dipoles are randomly oriented and there is no net magnetic dipole moment associated with the material. With the introduction of the field, a partial alignment of the dipoles takes place, as depicted in part (b). The component of the net magnetic dipole moment that is perpendicular to the field vanishes. We may then represent the sample by part (c), which shows a collection of magnetic dipoles completely aligned with the field. By treating these dipoles as current loops, we can picture the dipole alignment as equivalent to a current around the surface of the material, as in part (d). This fictitious surface current produces its own magnetic field, which enhances the field of the solenoid.

We can express the total magnetic field $\overrightarrow{B}$ in the material as

where ${\overrightarrow{B}}_{0}$ is the field due to the current ${I}_{0}$ in the solenoid and ${\overrightarrow{B}}_{m}$ is the field due to the surface current ${I}_{m}$ around the sample. Now ${\overrightarrow{B}}_{m}$ is usually proportional to ${\overrightarrow{B}}_{0},$ a fact we express by

where $\chi $ is a dimensionless quantity called the magnetic susceptibility. Values of $\chi $ for some paramagnetic materials are given in Table 12.2. Since the alignment of magnetic dipoles is so weak, $\chi $ is very small for paramagnetic materials. By combining Equation 12.34 and Equation 12.35, we obtain:

For a sample within an infinite solenoid, this becomes

This expression tells us that the insertion of a paramagnetic material into a solenoid increases the field by a factor of $(1+\chi ).$ However, since $\chi $ is so small, the field isn’t enhanced very much.

The quantity

is called the magnetic permeability of a material. In terms of $\mu ,$ Equation 12.37 can be written as

for the filled solenoid.

Paramagnetic Materials | $\chi $ | Diamagnetic Materials | $\chi $ |
---|---|---|---|

Aluminum | $2.2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}$ | Bismuth | $\mathrm{-1.7}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}$ |

Calcium | $1.4\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}$ | Carbon (diamond) | $\mathrm{-2.2}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}$ |

Chromium | $3.1\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}$ | Copper | $\mathrm{-9.7}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-6}}$ |

Magnesium | $1.2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}$ | Lead | $\mathrm{-1.8}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}$ |

Oxygen gas (1 atm) | $1.8\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-6}}$ | Mercury | $\mathrm{-2.8}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}$ |

Oxygen liquid (90 K) | $3.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-3}}$ | Hydrogen gas (1 atm) | $\mathrm{-2.2}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-9}}$ |

Tungsten | $6.8\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}$ | Nitrogen gas (1 atm) | $\mathrm{-6.7}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-9}}$ |

Air (1 atm) | $3.6\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-7}}$ | Water | $\mathrm{-9.1}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-6}}$ |

### Diamagnetic Materials

A magnetic field always induces a magnetic dipole in an atom. This induced dipole points opposite to the applied field, so its magnetic field is also directed opposite to the applied field. In paramagnetic and ferromagnetic materials, the induced magnetic dipole is masked by much stronger permanent magnetic dipoles of the atoms. However, in diamagnetic materials, whose atoms have no permanent magnetic dipole moments, the effect of the induced dipole is observable.

We can now describe the magnetic effects of diamagnetic materials with the same model developed for paramagnetic materials. In this case, however, the fictitious surface current flows opposite to the solenoid current, and the magnetic susceptibility $\chi $ is negative. Values of $\chi $ for some diamagnetic materials are also given in Table 12.2.

### Interactive

Water is a common diamagnetic material. Animals are mostly composed of water. Experiments have been performed on frogs and mice in diverging magnetic fields. The water molecules are repelled from the applied magnetic field against gravity until the animal reaches an equilibrium. The result is that the animal is levitated by the magnetic field.

### Ferromagnetic Materials

Common magnets are made of a ferromagnetic material such as iron or one of its alloys. Experiments reveal that a ferromagnetic material consists of tiny regions known as magnetic domains. Their volumes typically range from ${10}^{\mathrm{-12}}$ to ${10}^{\mathrm{-8}}{\text{m}}^{3},$ and they contain about ${10}^{17}$ to ${10}^{21}$ atoms. Within a domain, the magnetic dipoles are rigidly aligned in the same direction by coupling among the atoms. This coupling, which is due to quantum mechanical effects, is so strong that even thermal agitation at room temperature cannot break it. The result is that each domain has a net dipole moment. Some materials have weaker coupling and are ferromagnetic only at lower temperatures.

If the domains in a ferromagnetic sample are randomly oriented, as shown in Figure 12.24, the sample has no net magnetic dipole moment and is said to be unmagnetized. Suppose that we fill the volume of a solenoid with an unmagnetized ferromagnetic sample. When the magnetic field ${\overrightarrow{B}}_{0}$ of the solenoid is turned on, the dipole moments of the domains rotate so that they align somewhat with the field, as depicted in Figure 12.24. In addition, the aligned domains tend to increase in size at the expense of unaligned ones. The net effect of these two processes is the creation of a net magnetic dipole moment for the ferromagnet that is directed along the applied magnetic field. This net magnetic dipole moment is much larger than that of a paramagnetic sample, and the domains, with their large numbers of atoms, do not become misaligned by thermal agitation. Consequently, the field due to the alignment of the domains is quite large.

Besides iron, only four elements contain the magnetic domains needed to exhibit ferromagnetic behavior: cobalt, nickel, gadolinium, and dysprosium. Many alloys of these elements are also ferromagnetic. Ferromagnetic materials can be described using Equation 12.34 through Equation 12.39, the paramagnetic equations. However, the value of $\chi $ for ferromagnetic material is usually on the order of ${10}^{3}$ to ${10}^{4},$ and it also depends on the history of the magnetic field to which the material has been subject. A typical plot of *B* (the total field in the material) versus ${B}_{0}$ (the applied field) for an initially unmagnetized piece of iron is shown in Figure 12.25. Some sample numbers are (1) for ${B}_{0}=1.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\text{T},B=0.60\phantom{\rule{0.2em}{0ex}}\text{T},$ and $\chi =\left(\raisebox{1ex}{$0.60$}\!\left/ \!\raisebox{-1ex}{$1.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}$}\right.\right)-1\approx 6.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}$; (2) for ${B}_{0}=6.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\text{T},B=1.5\phantom{\rule{0.2em}{0ex}}\text{T},$ and $\chi =\left(\raisebox{1ex}{$1.5$}\!\left/ \!\raisebox{-1ex}{$6.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}$}\right.\right)-1\approx 2.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}.$

When ${B}_{0}$ is varied over a range of positive and negative values, *B* is found to behave as shown in Figure 12.26. Note that the same ${B}_{0}$ (corresponding to the same current in the solenoid) can produce different values of *B* in the material. The magnetic field *B* produced in a ferromagnetic material by an applied field ${B}_{0}$ depends on the magnetic history of the material. This effect is called hysteresis, and the curve of Figure 12.26 is called a hysteresis loop. Notice that *B* does not disappear when ${B}_{0}=0$ (i.e., when the current in the solenoid is turned off). The iron stays magnetized, which means that it has become a permanent magnet.

Like the paramagnetic sample of Figure 12.23, the partial alignment of the domains in a ferromagnet is equivalent to a current flowing around the surface. A bar magnet can therefore be pictured as a tightly wound solenoid with a large current circulating through its coils (the surface current). You can see in Figure 12.27 that this model fits quite well. The fields of the bar magnet and the finite solenoid are strikingly similar. The figure also shows how the poles of the bar magnet are identified. To form closed loops, the field lines outside the magnet leave the north (N) pole and enter the south (S) pole, whereas inside the magnet, they leave S and enter N.

Ferromagnetic materials are found in computer hard disk drives and permanent data storage devices (Figure 12.28). A material used in your hard disk drives is called a spin valve, which has alternating layers of ferromagnetic (aligning with the external magnetic field) and antiferromagnetic (each atom is aligned opposite to the next) metals. It was observed that a significant change in resistance was discovered based on whether an applied magnetic field was on the spin valve or not. This large change in resistance creates a quick and consistent way for recording or reading information by an applied current.

### Example 12.10

#### Iron Core in a Coil

A long coil is tightly wound around an iron cylinder whose magnetization curve is shown in Figure 12.25. (a) If $n=20$ turns per centimeter, what is the applied field ${B}_{0}$ when ${I}_{0}=0.20\phantom{\rule{0.2em}{0ex}}\text{A}?$ (b) What is the net magnetic field for this same current? (c) What is the magnetic susceptibility in this case?#### Strategy

(a) The magnetic field of a solenoid is calculated using Equation 12.28. (b) The graph is read to determine the net magnetic field for this same current. (c) The magnetic susceptibility is calculated using Equation 12.37.#### Solution

- The applied field ${B}_{0}$ of the coil is
$$\begin{array}{ccc}\hfill {B}_{0}& =\hfill & {\mu}_{0}n{I}_{0}=(4\pi \phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-7}}\text{T}\xb7\text{m/A})(2000\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\text{m})(0.20\phantom{\rule{0.2em}{0ex}}\text{A})\hfill \\ \hfill {B}_{0}& =\hfill & 5.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\text{T}.\hfill \end{array}$$
- From inspection of the magnetization curve of Figure 12.25, we see that, for this value of ${B}_{0},$ $B=1.4\phantom{\rule{0.2em}{0ex}}\text{T}.$ Notice that the internal field of the aligned atoms is much larger than the externally applied field.
- The magnetic susceptibility is calculated to be
$$\chi =\frac{B}{{B}_{0}}-1=\frac{1.4\phantom{\rule{0.2em}{0ex}}\text{T}}{5.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\text{T}}\text{\u22121}=2.8\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}.$$

#### Significance

Ferromagnetic materials have susceptibilities in the range of ${10}^{3}$ which compares well to our results here. Paramagnetic materials have fractional susceptibilities, so their applied field of the coil is much greater than the magnetic field generated by the material.### Check Your Understanding 12.8

**Check Your Understanding** Repeat the calculations from the previous example for ${I}_{0}=0.040\phantom{\rule{0.2em}{0ex}}\text{A}.$