# 12.7 Magnetism in matter  (Page 4/13)

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When ${B}_{0}$ is varied over a range of positive and negative values, B is found to behave as shown in [link] . 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 [link] 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 [link] , 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 [link] 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 ( [link] ). 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.

## Iron core in a coil

A long coil is tightly wound around an iron cylinder whose magnetization curve is shown in [link] . (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 [link] . (b) The graph is read to determine the net magnetic field for this same current. (c) The magnetic susceptibility is calculated using [link] .

## Solution

1. The applied field ${B}_{0}$ of the coil is
$\begin{array}{ccc}\hfill {B}_{0}& =\hfill & {\mu }_{0}n{I}_{0}=\left(4\pi \phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\text{T}·\text{m/A}\right)\left(2000\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\text{m}\right)\left(0.20\phantom{\rule{0.2em}{0ex}}\text{A}\right)\hfill \\ \hfill {B}_{0}& =\hfill & 5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{T}.\hfill \end{array}$
2. From inspection of the magnetization curve of [link] , 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.
3. The magnetic susceptibility is calculated to be
$\chi =\frac{B}{{B}_{0}}-1=\frac{1.4\phantom{\rule{0.2em}{0ex}}\text{T}}{5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{T}}\text{−1}=2.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}.$

## 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.

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