CBSE Class 12 > Class 12 Videos > L29 : Numericals - Matter & Magnetism, Physics, Class 12
L29 Numericals - Matter & Magnetism, Physics, Class 12 Video Lecture
FAQs on L29 : Numericals - Matter & Magnetism, Physics, Class 12
| 1. What is matter and magnetism? |  |
Ans. Matter refers to anything that occupies space and has mass. It can exist in three states - solid, liquid, or gas. Magnetism, on the other hand, is a property of certain materials to attract or repel other materials. It is caused by the alignment of magnetic moments in the material.
| 2. What are the different types of magnets? | |
Ans. There are three main types of magnets - permanent magnets, temporary magnets, and electromagnets. Permanent magnets retain their magnetism even after the external magnetic field is removed. Temporary magnets become magnetized in the presence of a magnetic field but lose their magnetism quickly. Electromagnets are created by passing an electric current through a coil, generating a magnetic field.
| 3. How does magnetism affect matter? | |
Ans. Magnetism can affect matter in various ways. When a magnetic material comes in contact with a magnet, it can be attracted or repelled depending on the orientation of the magnetic fields. Magnetism can also induce electrical currents in conductive materials, known as electromagnetic induction. This phenomenon is the basis for many electrical devices like generators and transformers.
| 4. What are some practical applications of magnetism in daily life? | |
Ans. Magnetism has several practical applications in our daily lives. Some examples include: - Magnetic compasses use Earth's magnetic field to determine direction. - Magnetic resonance imaging (MRI) machines use powerful magnets to create detailed images of the inside of the body. - Credit cards and magnetic strips on ID cards use magnetism for data storage. - Electric motors and generators utilize magnetism to convert electrical energy into mechanical energy and vice versa.
| 5. How can matter be magnetized? | |
Ans. Matter can be magnetized in several ways. Ferromagnetic materials, such as iron, can be easily magnetized by placing them in a strong magnetic field. This aligns the magnetic domains within the material, resulting in a net magnetic field. Additionally, matter can be magnetized by inducing an electric current in a coil of wire and passing it through the material. This creates a magnetic field that magnetizes the material.
Text Transcript from Video
hello friends this video on matter and
magnetism part 29 is brought to you by
exam feel calm no more fear from exam
please make sure that you have watched
all the videos till part 28 before going
ahead with pad 29 so let us look at the
first problem it says that a toroidal
solenoid has 3000 turns and a mean
radius of 10 centimeters so that means
let me first draw a toroidal solenoid so
a toroidal solenoid would look somewhat
like this so it says that it has 3000
turns that is number of turns is 3000
and it has a mean radius of 10
centimeters that is 10 into 10 to the
power minus 2 meters it has a soft iron
core of relative permeability mm so
relative permeability is 2005 the
magnitude of the magnetic fields in the
core when a current of 1 ampere is
passed to the solenoid so we have to
find out the magnitude of magnetic field
in the core so let us suppose this is
the soft iron core inside the solenoid
so this is the soft iron core inside the
solenoid so how do we find the magnitude
of the magnetic field inside the
solenoid we already know that the
magnitude of magnetic field inside the
solenoid is given by mu naught and I
when we were talking about free space
like mu naught and I then when no
material was inserted here right so this
was for free space but now that a
material is inserted inside and that to
the material has a relative permeability
of 2000 that means in this case here
this will be equal to the magnetic
we'll be equal to mu into n I so what is
mu mu is nothing but mu I into nu not
into n what is n that is number of turns
per unit length so this will be total
number of turns divided by the length of
the solenoid into current right so what
would be the length of the solenoid
length is nothing but the circumference
of the wire so that would be equal to vu
naught mu R n divided by 2 pi I into I
so the magnetic field comes out to be mu
I as 2002 mu naught is 4 pi into 10 to
the power minus 7 into n is 3000 divided
by 2 pi into R is 10 into 10 to the
power minus 2 and 2 I is 1 ampere so
this PI and Phi will get cancelled and
this will come out to be 12 Tesla so
this is the value of the magnetic field
inside the core of the solenoid right so
now let us go ahead and look at the next
problem it says that a magnetizing field
of 1,600 ampere per meter produces a
magnetic flux of 2.4 indepen to the
power minus 5 Weber in a bag of iron of
cross-section zero point 2 centimeter
square so magnetizing field that as H is
given as 1600 ampere per meter it
produces a magnetic flux
what is magnetic flux magnetic flux is
nothing but integration of B dot DS
right that is the number of magnetic
lines of force crossing a particular
area so this is defined as magnetic flux
which we often denote as Phi B so the
magnetic flux is given as 2 point 4 into
10 to the power minus 5 Weber and the
cross sectional area of the bar is given
as zero point 2 centimeter square at a
zero point two
10 to the power minus 4 meter square so
we have to calculate the permeability
and the susceptibility so we have to
calculate the permeability and the
magnetic susceptibility of the back now
with the help of whatever we studied so
far we know that nu is equal to D by H
we already know this and what is B from
this definition we can say that D is
nothing but Phi B divided by a where a
is the area because integration of D s
is nothing but area so instead of B we
can write it as Phi B divided by area
this divided by H so this would be the
value of MU that is the permeability now
Phi B is 2 point 4 into 10 to the power
minus 5 divided by area and 0.2 into 10
to the power minus 4 this divided by H
is 1600 so this value comes out to be 7
point 5 into 10 to the power minus 4
Tesla
per ampere meter so this would be the
value of permeability now susceptibility
we know that relation between relative
permeability and susceptibility
susceptibility is mu is equal to 1 plus
Chi M now what is mu i mu is nothing but
mu divided by mu naught is equal to 1
plus Chi M therefore Chi M that the
susceptibility will be equal to MU
divided by mu naught minus 1 right so
what is mu mu we have calculated just
now that is 7 point 5 into 10 to the
power minus 4 divided by mu naught is 4
pi into 10 to the power minus 7 minus 1
so this comes out to be 5 9 6 point 1 so
this is the value of magnetic
susceptibility so basically you will
have you would have to there is you have
to there is no other go you
need to remember all these relationships
the relations between permeability and
susceptibility relations between
magnetizing field and intensity of
magnetization so all the things these
terminologies are very very important
and you will have to remember all those
things let us now look at the last
problem of this lesson it says that a
sample of paramagnetic salt contains 2
into 10 to the power 24 atomic dipoles
each of dipole moment 1.5 into 10 to the
power minus 23 Julieta so what is the
number of atomic dipoles 2 into 10 to
the power 24 and the magnetic dipole of
each of them is 1.5 into 10 to the power
minus 23 June 5 Tesla the sample is
placed under a homogeneous magnetic
field of 0.6 4 Tesla and cooled to a
temperature of 4.2 Kelvin okay the
degree of magnetic saturation saturation
achieved is equal to 15% that means you
have a paramagnetic salt so that means
you have a paramagnetic substance what
did you do you place it in a magnetic
field so what will happen when you place
a paramagnetic substance in magnetic
field the substance will get magnetized
right so it says that the magnetic
saturation is achieved is equal to 15%
so what does the total dipole moment of
the sample for a magnetic field of 0.99
Tesla and a temperature of 2.8 kelvins
so what is will happen if you reduce the
magnet in if you increase the magnetic
field and reduce the temperature further
she first we even have to calculate the
total dipole moment of the substance as
a whole right so what would be the total
dipole moment the total dipole moment
would be equal to the number of atomic
dipoles into the
dipole moment of each atom so total
number of atoms is n so these many atoms
and each atom has a dipole moment of M
so you may multiply both of them so 2
into 10 to the power 24 into 1.5 into 10
to the power minus 23
so that is 30 so this 30 is the total
dipole moment now it says that the
magnetic saturation that is achieved is
equal to 15% that means the dipole
moment that is induced due to the
application of the magnetic field is
only 15% right so therefore the
effective dipole moment or the effective
magnetic moment will be equal to 15% of
30 right because it seems that the
magnetic saturation is only 30 it should
have been the total magnetic moment
should have been 30 but since the
saturation achieved is only 15 percent
that means only 15 percent of this value
is the actual magnetic moment that is
induced right so because whatever is the
magnetic moment that is induced that is
what will reflect in the magnetization
so this comes out to be 4 point 5 Joule
per Tesla so this is the effective
magnetic moment of the substance on the
application application of magnetic
field so what do we see now we see that
the initial magnetic moment in the first
case we see that the magnetic field that
is applied is 0.6 for Tesla this is the
magnetic field applied in the first case
and the temperature in the first cases
4.2 Kelvin and the magnetic moment in
the first case is equal to 4 point 5
Joule per Tesla now after some time they
increase the magnetic
0.98 Tesla they decrease the temperature
to 2.8 Kelvin and they are asking us to
calculate the magnetic the total dipole
moment of the sample under this
situation right so we have to calculate
m2 right now we will apply
Curie's law so as per Quirino we know
that the susceptibility is equal to it
is directly proportional to 1 by T or we
can say susceptibility is equal to C by
P where C is a Curie constant right this
is Curie constant now how do we define
susceptibility susceptibility is the
ratio of magnetization to magnetic
intensity so this is equal to C by P so
the C is of constant so we can say that
M is equal to H C divided by thing right
now we also know that this M is directly
Probot is this M this M is nothing but
magnetization so magnetization is
actually dependent on the magnetic
moment now whenever magnetic moment gets
induced whenever there is a net magnetic
moment the substance is magnetized
whenever the net magnetic moment is 0
the substances demagnetized state so
that means this magnetization depends on
magnetic moment so magnetization is
dependent on magnetic moment and also
this H that is magnetic intensity is
dependent on magnetic field right so
this is very logical so we can write
this expression as M is proportional to
B C divided by P so from this we can say
that B empty divided by P is constant
because this is a constant value right
so now using this we can write that
therefore M 1 T 1 by B 1 should be equal
to M 2 T 2 by B 2 right so from this we
can say that M 2 is equal to M 1 T 1 T 2
divided by B 1 T 2 so now it will
quickly put the values that you already
have here we have noted out all the
values right so we will put those values
so M 1 is 4.5 into T 1 is 4 point 2
Kelvin B 2 is 0.98 Tesla B 1 is 0.64 and
T 2 is 2 point 8 so when you calculate
this you get the result as 7 point 8 7 5
2 / Tesla so this would be the value of
the net dipole moment when the
temperature is further reduced at a
magnetic field of value 0.98 Tesla so
this is how you generally approach
problems so I hope that the concepts of
magnetism are now clear to you so
practice more problems so that you can
key a transfer your concepts further
because in physics Eisenmann I just
cannot so on each and every problem for
you so it is like you should understand
the basic concepts once the basics I
clear then it is up to your practice the
more you practice the concepts get
clearer over and again so see you on in
the next lesson
thank you please visit exam fee or comm
to watch free educational videos try 3
online tests get the best quality study
materials study from the best tutors and
mentors and much more thank you once
again
magnetism part 29 is brought to you by
exam feel calm no more fear from exam
please make sure that you have watched
all the videos till part 28 before going
ahead with pad 29 so let us look at the
first problem it says that a toroidal
solenoid has 3000 turns and a mean
radius of 10 centimeters so that means
let me first draw a toroidal solenoid so
a toroidal solenoid would look somewhat
like this so it says that it has 3000
turns that is number of turns is 3000
and it has a mean radius of 10
centimeters that is 10 into 10 to the
power minus 2 meters it has a soft iron
core of relative permeability mm so
relative permeability is 2005 the
magnitude of the magnetic fields in the
core when a current of 1 ampere is
passed to the solenoid so we have to
find out the magnitude of magnetic field
in the core so let us suppose this is
the soft iron core inside the solenoid
so this is the soft iron core inside the
solenoid so how do we find the magnitude
of the magnetic field inside the
solenoid we already know that the
magnitude of magnetic field inside the
solenoid is given by mu naught and I
when we were talking about free space
like mu naught and I then when no
material was inserted here right so this
was for free space but now that a
material is inserted inside and that to
the material has a relative permeability
of 2000 that means in this case here
this will be equal to the magnetic
we'll be equal to mu into n I so what is
mu mu is nothing but mu I into nu not
into n what is n that is number of turns
per unit length so this will be total
number of turns divided by the length of
the solenoid into current right so what
would be the length of the solenoid
length is nothing but the circumference
of the wire so that would be equal to vu
naught mu R n divided by 2 pi I into I
so the magnetic field comes out to be mu
I as 2002 mu naught is 4 pi into 10 to
the power minus 7 into n is 3000 divided
by 2 pi into R is 10 into 10 to the
power minus 2 and 2 I is 1 ampere so
this PI and Phi will get cancelled and
this will come out to be 12 Tesla so
this is the value of the magnetic field
inside the core of the solenoid right so
now let us go ahead and look at the next
problem it says that a magnetizing field
of 1,600 ampere per meter produces a
magnetic flux of 2.4 indepen to the
power minus 5 Weber in a bag of iron of
cross-section zero point 2 centimeter
square so magnetizing field that as H is
given as 1600 ampere per meter it
produces a magnetic flux
what is magnetic flux magnetic flux is
nothing but integration of B dot DS
right that is the number of magnetic
lines of force crossing a particular
area so this is defined as magnetic flux
which we often denote as Phi B so the
magnetic flux is given as 2 point 4 into
10 to the power minus 5 Weber and the
cross sectional area of the bar is given
as zero point 2 centimeter square at a
zero point two
10 to the power minus 4 meter square so
we have to calculate the permeability
and the susceptibility so we have to
calculate the permeability and the
magnetic susceptibility of the back now
with the help of whatever we studied so
far we know that nu is equal to D by H
we already know this and what is B from
this definition we can say that D is
nothing but Phi B divided by a where a
is the area because integration of D s
is nothing but area so instead of B we
can write it as Phi B divided by area
this divided by H so this would be the
value of MU that is the permeability now
Phi B is 2 point 4 into 10 to the power
minus 5 divided by area and 0.2 into 10
to the power minus 4 this divided by H
is 1600 so this value comes out to be 7
point 5 into 10 to the power minus 4
Tesla
per ampere meter so this would be the
value of permeability now susceptibility
we know that relation between relative
permeability and susceptibility
susceptibility is mu is equal to 1 plus
Chi M now what is mu i mu is nothing but
mu divided by mu naught is equal to 1
plus Chi M therefore Chi M that the
susceptibility will be equal to MU
divided by mu naught minus 1 right so
what is mu mu we have calculated just
now that is 7 point 5 into 10 to the
power minus 4 divided by mu naught is 4
pi into 10 to the power minus 7 minus 1
so this comes out to be 5 9 6 point 1 so
this is the value of magnetic
susceptibility so basically you will
have you would have to there is you have
to there is no other go you
need to remember all these relationships
the relations between permeability and
susceptibility relations between
magnetizing field and intensity of
magnetization so all the things these
terminologies are very very important
and you will have to remember all those
things let us now look at the last
problem of this lesson it says that a
sample of paramagnetic salt contains 2
into 10 to the power 24 atomic dipoles
each of dipole moment 1.5 into 10 to the
power minus 23 Julieta so what is the
number of atomic dipoles 2 into 10 to
the power 24 and the magnetic dipole of
each of them is 1.5 into 10 to the power
minus 23 June 5 Tesla the sample is
placed under a homogeneous magnetic
field of 0.6 4 Tesla and cooled to a
temperature of 4.2 Kelvin okay the
degree of magnetic saturation saturation
achieved is equal to 15% that means you
have a paramagnetic salt so that means
you have a paramagnetic substance what
did you do you place it in a magnetic
field so what will happen when you place
a paramagnetic substance in magnetic
field the substance will get magnetized
right so it says that the magnetic
saturation is achieved is equal to 15%
so what does the total dipole moment of
the sample for a magnetic field of 0.99
Tesla and a temperature of 2.8 kelvins
so what is will happen if you reduce the
magnet in if you increase the magnetic
field and reduce the temperature further
she first we even have to calculate the
total dipole moment of the substance as
a whole right so what would be the total
dipole moment the total dipole moment
would be equal to the number of atomic
dipoles into the
dipole moment of each atom so total
number of atoms is n so these many atoms
and each atom has a dipole moment of M
so you may multiply both of them so 2
into 10 to the power 24 into 1.5 into 10
to the power minus 23
so that is 30 so this 30 is the total
dipole moment now it says that the
magnetic saturation that is achieved is
equal to 15% that means the dipole
moment that is induced due to the
application of the magnetic field is
only 15% right so therefore the
effective dipole moment or the effective
magnetic moment will be equal to 15% of
30 right because it seems that the
magnetic saturation is only 30 it should
have been the total magnetic moment
should have been 30 but since the
saturation achieved is only 15 percent
that means only 15 percent of this value
is the actual magnetic moment that is
induced right so because whatever is the
magnetic moment that is induced that is
what will reflect in the magnetization
so this comes out to be 4 point 5 Joule
per Tesla so this is the effective
magnetic moment of the substance on the
application application of magnetic
field so what do we see now we see that
the initial magnetic moment in the first
case we see that the magnetic field that
is applied is 0.6 for Tesla this is the
magnetic field applied in the first case
and the temperature in the first cases
4.2 Kelvin and the magnetic moment in
the first case is equal to 4 point 5
Joule per Tesla now after some time they
increase the magnetic
0.98 Tesla they decrease the temperature
to 2.8 Kelvin and they are asking us to
calculate the magnetic the total dipole
moment of the sample under this
situation right so we have to calculate
m2 right now we will apply
Curie's law so as per Quirino we know
that the susceptibility is equal to it
is directly proportional to 1 by T or we
can say susceptibility is equal to C by
P where C is a Curie constant right this
is Curie constant now how do we define
susceptibility susceptibility is the
ratio of magnetization to magnetic
intensity so this is equal to C by P so
the C is of constant so we can say that
M is equal to H C divided by thing right
now we also know that this M is directly
Probot is this M this M is nothing but
magnetization so magnetization is
actually dependent on the magnetic
moment now whenever magnetic moment gets
induced whenever there is a net magnetic
moment the substance is magnetized
whenever the net magnetic moment is 0
the substances demagnetized state so
that means this magnetization depends on
magnetic moment so magnetization is
dependent on magnetic moment and also
this H that is magnetic intensity is
dependent on magnetic field right so
this is very logical so we can write
this expression as M is proportional to
B C divided by P so from this we can say
that B empty divided by P is constant
because this is a constant value right
so now using this we can write that
therefore M 1 T 1 by B 1 should be equal
to M 2 T 2 by B 2 right so from this we
can say that M 2 is equal to M 1 T 1 T 2
divided by B 1 T 2 so now it will
quickly put the values that you already
have here we have noted out all the
values right so we will put those values
so M 1 is 4.5 into T 1 is 4 point 2
Kelvin B 2 is 0.98 Tesla B 1 is 0.64 and
T 2 is 2 point 8 so when you calculate
this you get the result as 7 point 8 7 5
2 / Tesla so this would be the value of
the net dipole moment when the
temperature is further reduced at a
magnetic field of value 0.98 Tesla so
this is how you generally approach
problems so I hope that the concepts of
magnetism are now clear to you so
practice more problems so that you can
key a transfer your concepts further
because in physics Eisenmann I just
cannot so on each and every problem for
you so it is like you should understand
the basic concepts once the basics I
clear then it is up to your practice the
more you practice the concepts get
clearer over and again so see you on in
the next lesson
thank you please visit exam fee or comm
to watch free educational videos try 3
online tests get the best quality study
materials study from the best tutors and
mentors and much more thank you once
again
About this Video
Apr 20, 2026 Last updated
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