12.7 Magnetism in matter  (Page 2/13)

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

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.

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 differential form of Gauss's law?
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i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
kya ye b.sc ka hai agar haa to konsa part
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it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
determine absolute zero
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
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Henry
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Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
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Mehmet
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
Mayowa
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Mehmet
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
explain the lack of symmetry in the field of the parallel capacitor
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe