# 12.7 Magnetism in matter  (Page 5/13)

 Page 5 / 13

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

a. $1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{T}$ ; b. 0.60 T; c. $6.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}$

## Summary

• Materials are classified as paramagnetic, diamagnetic, or ferromagnetic, depending on how they behave in an applied magnetic field.
• Paramagnetic materials have partial alignment of their magnetic dipoles with an applied magnetic field. This is a positive magnetic susceptibility. Only a surface current remains, creating a solenoid-like magnetic field.
• Diamagnetic materials exhibit induced dipoles opposite to an applied magnetic field. This is a negative magnetic susceptibility.
• Ferromagnetic materials have groups of dipoles, called domains, which align with the applied magnetic field. However, when the field is removed, the ferromagnetic material remains magnetized, unlike paramagnetic materials. This magnetization of the material versus the applied field effect is called hysteresis.

## Key equations

 Permeability of free space ${\mu }_{0}=4\pi \phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\text{T}\cdot \text{m/A}$ Contribution to magnetic field from a current element $dB=\frac{{\mu }_{0}}{4\pi }\phantom{\rule{0.2em}{0ex}}\frac{I\phantom{\rule{0.2em}{0ex}}dl\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta }{{r}^{2}}$ Biot–Savart law $\stackrel{\to }{B}=\frac{{\mu }_{0}}{4\pi }\underset{\text{wire}}{\int }\frac{Id\stackrel{\to }{l}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{^}{r}}{{r}^{2}}$ Magnetic field due to a long straight wire $B=\frac{{\mu }_{0}I}{2\pi R}$ Force between two parallel currents $\frac{F}{l}=\frac{{\mu }_{0}{I}_{1}{I}_{2}}{2\pi r}$ Magnetic field of a current loop $B=\frac{{\mu }_{0}I}{2R}\phantom{\rule{0.2em}{0ex}}\text{(at center of loop)}$ Ampère’s law $\oint \stackrel{\to }{B}·d\stackrel{\to }{l}={\mu }_{0}I$ Magnetic field strength inside a solenoid $B={\mu }_{0}nI$ Magnetic field strength inside a toroid $B=\frac{{\mu }_{o}NI}{2\pi r}$ Magnetic permeability $\mu =\left(1+\chi \right){\mu }_{0}$ Magnetic field of a solenoid filled with paramagnetic material $B=\mu nI$

## Conceptual questions

A diamagnetic material is brought close to a permanent magnet. What happens to the material?

If you cut a bar magnet into two pieces, will you end up with one magnet with an isolated north pole and another magnet with an isolated south pole? Explain your answer.

The bar magnet will then become two magnets, each with their own north and south poles. There are no magnetic monopoles or single pole magnets.

## Problems

The magnetic field in the core of an air-filled solenoid is 1.50 T. By how much will this magnetic field decrease if the air is pumped out of the core while the current is held constant?

A solenoid has a ferromagnetic core, n = 1000 turns per meter, and I = 5.0 A. If B inside the solenoid is 2.0 T, what is $\chi$ for the core material?

317.31

A 20-A current flows through a solenoid with 2000 turns per meter. What is the magnetic field inside the solenoid if its core is (a) a vacuum and (b) filled with liquid oxygen at 90 K?

The magnetic dipole moment of the iron atom is about $2.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\text{A}·{\text{m}}^{2}.$ (a) Calculate the maximum magnetic dipole moment of a domain consisting of ${10}^{19}$ iron atoms. (b) What current would have to flow through a single circular loop of wire of diameter 1.0 cm to produce this magnetic dipole moment?

$2.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{A}·{\text{m}}^{2}$
$2.7\phantom{\rule{0.2em}{0ex}}\text{A}$

Suppose you wish to produce a 1.2-T magnetic field in a toroid with an iron core for which $\chi =4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}.$ The toroid has a mean radius of 15 cm and is wound with 500 turns. What current is required?

A current of 1.5 A flows through the windings of a large, thin toroid with 200 turns per meter. If the toroid is filled with iron for which $\chi =3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3},$ what is the magnetic field within it?

0.18 T

What is differential form of Gauss's law?
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
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
what is charge quantization
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
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
what is copper loss
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
what is the weight of the earth in space
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
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
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
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
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
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