# 12.7 Magnetism in matter  (Page 2/13)

 Page 2 / 13
${U}_{B}=1.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\text{J}.$

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

${U}_{T}\approx kT=\left(1.38\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\text{J/K}\right)\left(300\phantom{\rule{0.2em}{0ex}}\text{K}\right)=4.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-21}\text{J},$

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 [link] 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. The alignment process in a paramagnetic material filling a solenoid (not shown). (a) Without an applied field, the magnetic dipoles are randomly oriented. (b) With a field, partial alignment occurs. (c) An equivalent representation of part (b). (d) The internal currents cancel, leaving an effective surface current that produces a magnetic field similar to that of a finite solenoid.

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

$\stackrel{\to }{B}={\stackrel{\to }{B}}_{0}+{\stackrel{\to }{B}}_{m},$

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

${\stackrel{\to }{B}}_{m}=\chi {\stackrel{\to }{B}}_{0},$

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

$\stackrel{\to }{B}={\stackrel{\to }{B}}_{0}+\chi {\stackrel{\to }{B}}_{0}=\left(1+\chi \right){\stackrel{\to }{B}}_{0}.$

For a sample within an infinite solenoid, this becomes

$B=\left(1+\chi \right){\mu }_{0}nI.$

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

The quantity

$\mu =\left(1+\chi \right){\mu }_{0}.$

is called the magnetic permeability of a material. In terms of $\mu ,$ [link] can be written as

$B=\mu nI$

for the filled solenoid.

Magnetic susceptibilities
Paramagnetic Materials $\chi$ Diamagnetic Materials $\chi$
Aluminum $2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$ Bismuth $-1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$
Calcium $1.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$ Carbon (diamond) $-2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$
Chromium $3.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}$ Copper $-9.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}$
Magnesium $1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$ Lead $-1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$
Oxygen gas (1 atm) $1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}$ Mercury $-2.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$
Oxygen liquid (90 K) $3.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}$ Hydrogen gas (1 atm) $-2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}$
Tungsten $6.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$ Nitrogen gas (1 atm) $-6.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}$
Air (1 atm) $3.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}$ Water $-9.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}$

## 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.

what is principle of superposition
what are questions that are likely to come out during exam
what is electricity
watt is electricity.
electricity ka full definition with formula
Jyoti
If a point charge is released from rest in a uniform electric field will it follow a field line? Will it do so if the electric field is not uniform?
Maxwell's stress tensor is
Yes
doris
neither vector nor scalar
Anil
if 6.0×10^13 electrons are placed on a metal sphere of charge 9.0micro Coulombs, what is the net charge on the sphere
18.51micro Coulombs
ASHOK
Is it possible to find the magnetic field of a circular loop at the centre by using ampere's law?
Is it possible to find the magnetic field of a circular loop at it's centre?
yes
Brother
The density of a gas of relative molecular mass 28 at a certain temperature is 0.90 K kgmcube.The root mean square speed of the gas molecules at that temperature is 602ms.Assuming that the rate of diffusion of a gas in inversely proportional to the square root of its density,calculate the density of
A hot liquid at 80degree Celsius is added to 600g of the same liquid originally at 10 degree Celsius. when the mixture reaches 30 degree Celsius, what will be the total mass of the liquid?
Under which topic
doris
what is electrostatics
Study of charges which are at rest
himanshu
Explain Kinematics
Two equal positive charges are repelling each other. The force on the charge on the left is 3.0 Newtons. Using your notes on Coulomb's law, and the forces acting on each of the charges, what is the force on the charge on the right?
Using the same two positive charges, the left positive charge is increased so that its charge is 4 times LARGER than the charge on the right. Using your notes on Coulomb's law and changes to the charge, once the charge is increased, what is the new force of repulsion between the two positive charges?
Nya
A mass 'm' is attached to a spring oscillates every 5 second. If the mass is increased by a 5 kg, the period increases by 3 second. Find its initial mass 'm'
a hot water tank containing 50,000g of water is heated by an electric immersion heater rated at 3kilowatt,240volt, calculate the current
what is charge
product of current and time
Jaffar By Richley Crapo By Stephen Voron By Sarah Warren By Cath Yu By Stephen Voron By Madison Christian By Jams Kalo By Rhodes By OpenStax By OpenStax