CBSE Class 12 > Class 12 Videos > L10 : Numericals - Matter & Magnetism, Physics, Class 12
L10 Numericals - Matter & Magnetism, Physics, Class 12 Video Lecture
FAQs on L10 : Numericals - Matter & Magnetism, Physics, Class 12
| 1. What is matter? |  |
Ans. Matter is anything that has mass and occupies space. It can exist in various forms such as solid, liquid, or gas.
| 2. How can matter be classified? | |
Ans. Matter can be classified into three main categories: elements, compounds, and mixtures. Elements are pure substances that cannot be broken down into simpler substances. Compounds are substances that are composed of two or more elements chemically combined in a fixed ratio. Mixtures are combinations of two or more substances that are physically combined and can be separated by physical means.
| 3. What is magnetism? | |
Ans. Magnetism is a phenomenon by which materials exert attractive or repulsive forces on other materials. It is caused by the motion of electric charges within the material. Magnets have two poles, north and south, and opposite poles attract while similar poles repel each other.
| 4. How does magnetism relate to matter? | |
Ans. Magnetism is a property of matter. Some materials, known as ferromagnetic materials, have the ability to become magnetized and exhibit magnetic properties. These materials have unpaired electrons that align their spins in the same direction, creating a magnetic field. Other materials, such as paramagnetic and diamagnetic materials, also exhibit weak magnetic properties but to a lesser extent.
| 5. How can matter and magnetism be studied? | |
Ans. Matter and magnetism can be studied through various experiments and observations. Scientists use techniques such as magnetic field mapping, magnetic force measurements, and magnetization studies to understand the behavior and properties of different materials. Physics principles and mathematical models are also used to analyze and explain the relationship between matter and magnetism.
Text Transcript from Video
hello friends this video on matter and
magnetism part 10 is brought to you by
exam feel calm please make sure that you
have watched all the videos to the part
9 before going ahead with part 10 that
this here comes problem 1 it states that
a planar loop of irregular shape
encloses an area of 7 point 5 into 10 to
the power minus 4 meter square okay and
carries a current of 12 amperes so if we
have a planar loop of any irregular
shape it is it has an area of 7 point 5
into 10 to the power minus 4 meter
square and it carries a current of 12
amperes the sense of flow of current
appears to be clockwise to an observer
so whoever was the observer he finds
that the current flows in the clockwise
direction now what is the magnitude and
direction of magnetic moment associated
with the current loop so we have to
calculate the magnitude and direction of
magnetic moment now how do we define
magnetic moment magnetic moment is
defined as product of current and area
so calculating the magnitude is simple
it will be 12 into 7 point 5 into 10 to
the power minus 4 which comes out to be
9 into 10 to the power minus 3 ampere
meter square so this is the magnitude of
the magnetic moment and what would be
the direction when I talk about
direction again obviously we will have
to apply the right-hand thumb rule in
this case when your current is flowing
in the clockwise direction
where does your thumb point your thumb
points somewhere inside the plane of the
paper I mean right now I am sitting in
front of a computer so when I curl my
fingers in the direction of current I
see that my thumb goes away from me
right so the thumb is so the direction
of the thumb gives me the direction of
magnetic moment so the direction of math
big moment is perpendicular and away
from the observer because right now I am
considering that I am the observer so
considering myself as the observer I
considered the direction of current as
clockwise so again by considering myself
as the observer when I curl my fingers
in the direction of the current I see
that my thumb goes away from me so that
we can say that the direction is
perpendicular away from the observer
right so let us look at the next problem
it says that a bar magnet when suspended
horizontally and perpendicular to
Earth's feet experiences a torque of 3
into 10 to the power minus 4 Newton
meter what is the magnetic moment of the
magnet that is let us suppose we have a
bar magnet which is suspended
horizontally and perpendicular to the
Earth's magnetic field let us suppose
this is the direction of the Earth's
magnetic field ok and I have suspended a
bar magnet here which is suspended
horizontally and perpendicular to the
Earth's magnetic field so let us suppose
that the bar magnet is suspended
somewhat like this which is
perpendicular to the magnetic field of
the earth I mean I cannot draw it like
this it is like something a three
dimensional figure which you will have
to imagine let us suppose this is that
the plane is in the magnetic field and
you have a bar magnet which is hanging
such that it is perpendicular to the
direction of the magnetic field okay
shall we say that the magnet is
perpendicular to magnetic field so that
means you are Teta is equal to 90
degrees we have to calculate the
magnetic moment of the magnet so
magnetic moment we have to calculate and
what is given it is given that it
experiences a torque of 3 into 10 to the
power minus 4 Newton meter now we know
that toward experienced by a magnet when
suspended in magnetic field is given by
M cross B right
the standard it enhance MD sine-theta
right so what does torque torque is
given as 3 into 10 to the power minus 4
this is equal to n which we have to
calculate what does V be is the Earth's
magnetic field and it is given that the
horizontal component of Earth's magnetic
field is given as zero point 4 into 10
to the power minus 4 into sine theta
that the sine 90 degrees which is equal
to 1 so from this we can say that the
magnetic moment will be equal to 3 into
10 to the power minus 4 divided by 0.4
into 10 to the power minus 4 so this
comes out to be 7 point 5 Joule per
Aslam so this would be the magnetic
moment of the magnet right so whenever
we talk about magnetic moment of a bar
magnet and if you want to know about the
direction the direction would always be
along the axis of the bar magnet right
okay so let us look at the third problem
which says that a bar magnet of magnetic
moment 1.5 Joule per Tesla lies aligned
with the direction of a uniform magnetic
field of 0.2 to Tesla so that means you
have a uniform magnetic field maybe
somewhat like this and you have a bar
magnet
which is aligned with the direction of
the magnetic field somewhat like this so
what is the amount of work required by
an external torque to turn the magnet so
as to align its magnetic moment so that
means it is asking us to calculate the
work that needs to be done to align the
magnet to turn the magnet so as to align
its magnetic moment so as to align its
magnetic moment normal to the field
direction so let us consider the first
part so in the first part it sees that
initial what is the initial scenario
initially that theta is equal to 0
degree because the magnetic
is along the direction of the magnetic
field so theta initial is zero the
question asks you to calculate the
amount of work to be done so that theta
becomes normal to the field direction so
that means theta final should be equal
to ninety degrees so how do you define
work done work done is nothing but the
magnetic potential energy which we
talked about so which is which is again
nothing but minus m dot B which is equal
to minus MD cos theta
right so we can write it as minus MV cos
theta
final minus cos theta initial this would
be the work done right so now what is mm
is the magnetic moment which is given as
1 point 5 Joule per Tesla B is the
magnetic field which is given as zero
point 2 2 cos theta final is cos 90
degree minus cos 0 degree so this comes
out to be zero point three three twos so
this would be the amount of work which
is required to be done to turn the
magnet such that the magnetic moment
becomes normal to the field direction in
the second part of asks or closer to the
field direction that means in the second
part
Teta initial is again the initial
scenario where it is along the field
direction that is zero degree but now it
wants it to be rotated in such a way
that the magnetic moment is opposite to
the field direction so theta final
should be 180 degrees so in this case
again the work done will be equal to
minus M B cos theta final minus cos
theta initial so this can be written as
minus 1.5 into 0.22 cos 180 degrees
minus cos 0 degrees so this comes out to
be zero point six six joules so this
would be the amount of work which is to
be done here now in the second part
asks what is the torque on the magnet in
both the pieces so let us now calculate
the torque associated in each of them so
let us first consider the first case we
know that the torque is given by torque
acting on a magnet when placed in a
uniform magnetic field is given by M
cross B which is equal to M D sine theta
right so in the first case it needs to
be rotated by 90 degrees because initial
angle was zero and it is to be rotated
to 90 degree so it has to be rotated by
90 degrees therefore theta will be equal
to 90 degree
therefore the torque associated in the
with the first case is M into B into
sine 90 degrees which comes out to be
0.33 Newton meters whereas in the second
case it has to be rotated by 180 degrees
therefore theta is equal to 180 degrees
and the torque associated with the
scheme's would be one point five into
zero point twenty two into sine 180
degrees now sine 180 degree is equal to
zero so this dot would be equal to zero
okay so you understood how we calculated
the torque and the work done on the
magnetic dipole when it is placed in a
uniform magnetic field right thank you
please visit examview dot-com to watch
free educational videos try3 online
tests get the best quality study
materials study from the best tutors and
mentors and much more thank you once
again
magnetism part 10 is brought to you by
exam feel calm please make sure that you
have watched all the videos to the part
9 before going ahead with part 10 that
this here comes problem 1 it states that
a planar loop of irregular shape
encloses an area of 7 point 5 into 10 to
the power minus 4 meter square okay and
carries a current of 12 amperes so if we
have a planar loop of any irregular
shape it is it has an area of 7 point 5
into 10 to the power minus 4 meter
square and it carries a current of 12
amperes the sense of flow of current
appears to be clockwise to an observer
so whoever was the observer he finds
that the current flows in the clockwise
direction now what is the magnitude and
direction of magnetic moment associated
with the current loop so we have to
calculate the magnitude and direction of
magnetic moment now how do we define
magnetic moment magnetic moment is
defined as product of current and area
so calculating the magnitude is simple
it will be 12 into 7 point 5 into 10 to
the power minus 4 which comes out to be
9 into 10 to the power minus 3 ampere
meter square so this is the magnitude of
the magnetic moment and what would be
the direction when I talk about
direction again obviously we will have
to apply the right-hand thumb rule in
this case when your current is flowing
in the clockwise direction
where does your thumb point your thumb
points somewhere inside the plane of the
paper I mean right now I am sitting in
front of a computer so when I curl my
fingers in the direction of current I
see that my thumb goes away from me
right so the thumb is so the direction
of the thumb gives me the direction of
magnetic moment so the direction of math
big moment is perpendicular and away
from the observer because right now I am
considering that I am the observer so
considering myself as the observer I
considered the direction of current as
clockwise so again by considering myself
as the observer when I curl my fingers
in the direction of the current I see
that my thumb goes away from me so that
we can say that the direction is
perpendicular away from the observer
right so let us look at the next problem
it says that a bar magnet when suspended
horizontally and perpendicular to
Earth's feet experiences a torque of 3
into 10 to the power minus 4 Newton
meter what is the magnetic moment of the
magnet that is let us suppose we have a
bar magnet which is suspended
horizontally and perpendicular to the
Earth's magnetic field let us suppose
this is the direction of the Earth's
magnetic field ok and I have suspended a
bar magnet here which is suspended
horizontally and perpendicular to the
Earth's magnetic field so let us suppose
that the bar magnet is suspended
somewhat like this which is
perpendicular to the magnetic field of
the earth I mean I cannot draw it like
this it is like something a three
dimensional figure which you will have
to imagine let us suppose this is that
the plane is in the magnetic field and
you have a bar magnet which is hanging
such that it is perpendicular to the
direction of the magnetic field okay
shall we say that the magnet is
perpendicular to magnetic field so that
means you are Teta is equal to 90
degrees we have to calculate the
magnetic moment of the magnet so
magnetic moment we have to calculate and
what is given it is given that it
experiences a torque of 3 into 10 to the
power minus 4 Newton meter now we know
that toward experienced by a magnet when
suspended in magnetic field is given by
M cross B right
the standard it enhance MD sine-theta
right so what does torque torque is
given as 3 into 10 to the power minus 4
this is equal to n which we have to
calculate what does V be is the Earth's
magnetic field and it is given that the
horizontal component of Earth's magnetic
field is given as zero point 4 into 10
to the power minus 4 into sine theta
that the sine 90 degrees which is equal
to 1 so from this we can say that the
magnetic moment will be equal to 3 into
10 to the power minus 4 divided by 0.4
into 10 to the power minus 4 so this
comes out to be 7 point 5 Joule per
Aslam so this would be the magnetic
moment of the magnet right so whenever
we talk about magnetic moment of a bar
magnet and if you want to know about the
direction the direction would always be
along the axis of the bar magnet right
okay so let us look at the third problem
which says that a bar magnet of magnetic
moment 1.5 Joule per Tesla lies aligned
with the direction of a uniform magnetic
field of 0.2 to Tesla so that means you
have a uniform magnetic field maybe
somewhat like this and you have a bar
magnet
which is aligned with the direction of
the magnetic field somewhat like this so
what is the amount of work required by
an external torque to turn the magnet so
as to align its magnetic moment so that
means it is asking us to calculate the
work that needs to be done to align the
magnet to turn the magnet so as to align
its magnetic moment so as to align its
magnetic moment normal to the field
direction so let us consider the first
part so in the first part it sees that
initial what is the initial scenario
initially that theta is equal to 0
degree because the magnetic
is along the direction of the magnetic
field so theta initial is zero the
question asks you to calculate the
amount of work to be done so that theta
becomes normal to the field direction so
that means theta final should be equal
to ninety degrees so how do you define
work done work done is nothing but the
magnetic potential energy which we
talked about so which is which is again
nothing but minus m dot B which is equal
to minus MD cos theta
right so we can write it as minus MV cos
theta
final minus cos theta initial this would
be the work done right so now what is mm
is the magnetic moment which is given as
1 point 5 Joule per Tesla B is the
magnetic field which is given as zero
point 2 2 cos theta final is cos 90
degree minus cos 0 degree so this comes
out to be zero point three three twos so
this would be the amount of work which
is required to be done to turn the
magnet such that the magnetic moment
becomes normal to the field direction in
the second part of asks or closer to the
field direction that means in the second
part
Teta initial is again the initial
scenario where it is along the field
direction that is zero degree but now it
wants it to be rotated in such a way
that the magnetic moment is opposite to
the field direction so theta final
should be 180 degrees so in this case
again the work done will be equal to
minus M B cos theta final minus cos
theta initial so this can be written as
minus 1.5 into 0.22 cos 180 degrees
minus cos 0 degrees so this comes out to
be zero point six six joules so this
would be the amount of work which is to
be done here now in the second part
asks what is the torque on the magnet in
both the pieces so let us now calculate
the torque associated in each of them so
let us first consider the first case we
know that the torque is given by torque
acting on a magnet when placed in a
uniform magnetic field is given by M
cross B which is equal to M D sine theta
right so in the first case it needs to
be rotated by 90 degrees because initial
angle was zero and it is to be rotated
to 90 degree so it has to be rotated by
90 degrees therefore theta will be equal
to 90 degree
therefore the torque associated in the
with the first case is M into B into
sine 90 degrees which comes out to be
0.33 Newton meters whereas in the second
case it has to be rotated by 180 degrees
therefore theta is equal to 180 degrees
and the torque associated with the
scheme's would be one point five into
zero point twenty two into sine 180
degrees now sine 180 degree is equal to
zero so this dot would be equal to zero
okay so you understood how we calculated
the torque and the work done on the
magnetic dipole when it is placed in a
uniform magnetic field right thank you
please visit examview dot-com to watch
free educational videos try3 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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