Class 12 Exam > Class 12 Videos > Torque on current loop :part 1 - Moving Charges & Magnetism
Torque on current loop :part 1 - Moving Charges & Magnetism Video Lecture - Class 12
FAQs on Torque on current loop :part 1 - Moving Charges & Magnetism Video Lecture - Class 12
| 1. What is torque on a current loop? |  |
Ans. Torque on a current loop refers to the turning effect produced on the loop when it is placed in a magnetic field and carries an electric current. It is a vector quantity and is given by the product of the magnetic moment of the loop and the magnetic field strength.
| 2. How is torque on a current loop calculated? | |
Ans. The torque on a current loop can be calculated using the formula: torque = magnetic moment * magnetic field strength * sin(theta), where theta is the angle between the magnetic moment vector and the magnetic field vector.
| 3. What factors affect the torque on a current loop? | |
Ans. The torque on a current loop is affected by several factors, including the strength of the magnetic field, the current flowing through the loop, the area of the loop, and the orientation of the loop with respect to the magnetic field.
| 4. How does the torque on a current loop change with the orientation of the loop? | |
Ans. The torque on a current loop is maximum when the loop is aligned perpendicular to the magnetic field. As the angle between the loop and the magnetic field decreases, the torque decreases. When the loop is parallel or anti-parallel to the magnetic field, the torque becomes zero.
| 5. What is the significance of torque on a current loop? | |
Ans. The torque on a current loop plays a crucial role in various applications, such as electric motors and generators. It is responsible for the rotation of the loop, enabling the conversion of electrical energy into mechanical energy or vice versa. Understanding the torque on a current loop is essential in designing and optimizing these devices.
Text Transcript from Video
hello friends this video on matter and
magnetism part 30 is brought to you by
exam feel calm no more fear from exam
please make sure that you have watched
all the videos till 529 before going
ahead with part 13 so now let us
consider a loop which is carrying some
current that is a current carrying loop
and we will place it in a magnetic field
and we will see that at all facts on
that current carrying loop so it says
that a rectangular new carrying a steady
current and placed in a uniform magnetic
field experiences torque so this is what
we will see this is a fact which we came
to know just now and now we will move it
and we will see how this torque comes
into picture let us suppose this is a
rectangular car and this is a
rectangular dune carrying current let us
say ABCD is a rectangular loop which is
a which is carrying a current I and it
is placed in a magnetic field so what
happens at or tax on this current
carrying bag so now this basically
resembles something right I mean you
must have studied in your previous
classes about electric motor drive it
looks somewhat like this so that also
has a current carrying rectangular loop
and here you have some handle with the
help of which this entire loop it keeps
rotating so the axis of rotation is this
axis which you see here this is the axis
of rotation right so about this axis
this loop will keep rotating somewhat
like this if you look at this figure you
can see let us suppose this is the
magnetic this is the magnetic field this
is the magnetic field lines and this is
a current carrying loop and it rotates
in this way if you look at it carefully
just observe it once again so it rotates
like this so are you getting it how is
it rotating okay so similarly this loop
will also rotate like that so here let
us try to see what are the courses that
will
turn this current carrying loop so the
force which act on this side ad and BC
will be equal to will be equal and
opposite right for example in this case
now we imagine that the loop is placed
in such a way the loop is placed such
that the magnetic field B is in plane of
the loop that means the magnetic field
if this is their claim of the magnetic
field the loop is also lying in the same
plane you can see this is the direction
of magnetic field and this is how the
loop displaced so the magnetic field and
the loop they are in the same plane so
now let us try to calculate the force
which will act on each arm of this loop
now the force which will act on the arm
ad so the force that will act on an ad
will be equal to 0 by so if you look at
a B a D and B they are along the same
direction so that means the angle
because the force is given by I L cross
B force acting on a current-carrying
conductor will be given by I L cross B
right so here if you look at the side ad
the length vector as well as the
magnetic field vector they are in the
same direction therefore sine theta is
equal to zero therefore the force acting
on the an ad will be equal to zero
similarly the force acting on an BC will
also be equal to 0 because BC is also
along the same direction so there will
be no force acting on ad and BC now what
will be the force acting on and a B so
the force acting on a B will be I L
cross B right so this will be equal to
AI
L into B sine theta right so what is
this L L is nothing but the length of
this loop length of one side of this
loop length of a B and length of a B is
given as B so we say that this is B into
B sine theta what is Theta in this case
this is your length and this is your
magnetic field so it is a
three-dimensional figure right so
magnetic field is in the horizontal
plane and your loop is lying in the same
plane but it is perpendicular to the
magnetic field right you imagine the
figure you can look at this diagram so
that it becomes clear to use magnetic
field is already there on top of that
you place a loop so that load is in a
direction perpendicular to the direction
of magnetic field so this theta will be
90 degree so this will be the force that
will act on a B right and now let us try
to calculate what will be the direction
of this force direction will be given by
L cross B so apply right hand rule so
take all your four fingers towards the
length that is towards this direction
and then fold them towards right because
right is the direction of magnetic field
so I got thumb points in the downward
direction therefore the force will act
in the downward direction now let us
calculate the force that will act on the
arm CD so that force will again be given
by I L cross B which is equal to I L B
sine theta so even in this case theta
will be equal to 90 degree because the
length vector of this arm is
perpendicular to the magnetic field so
this will also become equal to I into B
into B and what would be the direction
in this case take all your four fingers
along this direction because this is the
length vector and then it goes towards
the magnetic field so take all your arm
along this direction and then it goes
towards magnetic fish
therefore your thumb points towards up
so the net force acts in the upward
direction so what would be the net force
so use if you look at it you will see
that f1 is equal to f2 but they are in
the opposite direction so that means the
net force is equal to 0 because F 1 and
F 2 are equal and opposite right so we
see that the net force acting on this
rectangular current carrying loop in
uniform magnetic field is equal to 0 so
the net force is equal to 0 but another
interesting thing which we observed is
that if you look at the arm a B the
force acts in the downward direction if
you look at the arm C D the force F 2
acts in the upward direction and these
two forces do not act along the same
line because when you when this object
is rotating like this let us suppose the
object is rotating like this you apply
some force at this point and you apply
some force at this point so will the net
force be equal to zero
however the net force will be equal to
zero but because the line of these
opposite forces are not same therefore a
torque will start acting on it so the
same thing is happening here in this
case also since the axis of rotation is
this one so and a B and C D are not
connected by the any axis right so equal
and opposite forces are acting on a B
and C D but they are not acting along
the same lines so if you look at it in
this way so it is somewhat like this on
this there is a force which acts in the
downward direction so f1 try once to
what is the aim of this f1 this f1 wants
to bring this a B down it wants to push
a B downwards
and there is a force f2 which wants to
pull the side BC upward right so
basically this is how the force will
look f1 is trying to bring the side a B
down and f2 is trying to bring the side
C D up right so it is trying to
basically rotate the loop if you think
of it very carefully let us suppose
right now it is stable there is one
force f1 which is acting on this side
and it is trying to push a B down and
this f2 is trying to push pull and C D
up in that is even though f1 and f2 are
equal and opposite but still there is a
torque which is acting on the loop
making it to rotate because f1 will make
a B to go down and this will make C D to
come up so gradually CD will start
coming up ad will start coming down and
as I decided this entire loop will start
rotating right so therefore at all acts
right so this is the scenario if you
look at it from this this side so this
is f1 this is f2 right so if you join it
with from the same tag you will get a
figure like this that is this diagram
and let us suppose if this entire length
is a so the centre of mass will be at
the center so torque will act and what
will be the magnitude of torque that
will be equal to f1 multiply it by the
moment of the arm that is a by 2 plus F
2 multiplied by a by 2 so this will be
equal to f1 is equal to f2 is equal to I
be B this into a now what is a into B a
into B is nothing but the area of the
loop a is the area of the loop so this
is the net torque which will act on this
loop when it is present in an external
magnetic field so I hope you understood
how this extern
thought will act on and do now when I
look at the same scenario but there we
will assume here what did we assume here
we had assumed that the magnetic field
is in the plane of the loop right that
is what we had assumed at the very
beginning but in the next scenario we
will assume that initially itself the
loop is placed in such a way that it
makes some angle theta with the
direction of the magnetic fields right
so that is the second scenario which we
will consider so that we get a general
expression for torque so I hope you
understood how this rotation takes place
so for the sake of that I have this
diagram which will explain you more
clearly how this rotation is taking
place right thank you please visit exam
comm to watch free educational videos
try free online tests get the best
quality study materials study from the
best tutors and mentors and much more
thank you once again
magnetism part 30 is brought to you by
exam feel calm no more fear from exam
please make sure that you have watched
all the videos till 529 before going
ahead with part 13 so now let us
consider a loop which is carrying some
current that is a current carrying loop
and we will place it in a magnetic field
and we will see that at all facts on
that current carrying loop so it says
that a rectangular new carrying a steady
current and placed in a uniform magnetic
field experiences torque so this is what
we will see this is a fact which we came
to know just now and now we will move it
and we will see how this torque comes
into picture let us suppose this is a
rectangular car and this is a
rectangular dune carrying current let us
say ABCD is a rectangular loop which is
a which is carrying a current I and it
is placed in a magnetic field so what
happens at or tax on this current
carrying bag so now this basically
resembles something right I mean you
must have studied in your previous
classes about electric motor drive it
looks somewhat like this so that also
has a current carrying rectangular loop
and here you have some handle with the
help of which this entire loop it keeps
rotating so the axis of rotation is this
axis which you see here this is the axis
of rotation right so about this axis
this loop will keep rotating somewhat
like this if you look at this figure you
can see let us suppose this is the
magnetic this is the magnetic field this
is the magnetic field lines and this is
a current carrying loop and it rotates
in this way if you look at it carefully
just observe it once again so it rotates
like this so are you getting it how is
it rotating okay so similarly this loop
will also rotate like that so here let
us try to see what are the courses that
will
turn this current carrying loop so the
force which act on this side ad and BC
will be equal to will be equal and
opposite right for example in this case
now we imagine that the loop is placed
in such a way the loop is placed such
that the magnetic field B is in plane of
the loop that means the magnetic field
if this is their claim of the magnetic
field the loop is also lying in the same
plane you can see this is the direction
of magnetic field and this is how the
loop displaced so the magnetic field and
the loop they are in the same plane so
now let us try to calculate the force
which will act on each arm of this loop
now the force which will act on the arm
ad so the force that will act on an ad
will be equal to 0 by so if you look at
a B a D and B they are along the same
direction so that means the angle
because the force is given by I L cross
B force acting on a current-carrying
conductor will be given by I L cross B
right so here if you look at the side ad
the length vector as well as the
magnetic field vector they are in the
same direction therefore sine theta is
equal to zero therefore the force acting
on the an ad will be equal to zero
similarly the force acting on an BC will
also be equal to 0 because BC is also
along the same direction so there will
be no force acting on ad and BC now what
will be the force acting on and a B so
the force acting on a B will be I L
cross B right so this will be equal to
AI
L into B sine theta right so what is
this L L is nothing but the length of
this loop length of one side of this
loop length of a B and length of a B is
given as B so we say that this is B into
B sine theta what is Theta in this case
this is your length and this is your
magnetic field so it is a
three-dimensional figure right so
magnetic field is in the horizontal
plane and your loop is lying in the same
plane but it is perpendicular to the
magnetic field right you imagine the
figure you can look at this diagram so
that it becomes clear to use magnetic
field is already there on top of that
you place a loop so that load is in a
direction perpendicular to the direction
of magnetic field so this theta will be
90 degree so this will be the force that
will act on a B right and now let us try
to calculate what will be the direction
of this force direction will be given by
L cross B so apply right hand rule so
take all your four fingers towards the
length that is towards this direction
and then fold them towards right because
right is the direction of magnetic field
so I got thumb points in the downward
direction therefore the force will act
in the downward direction now let us
calculate the force that will act on the
arm CD so that force will again be given
by I L cross B which is equal to I L B
sine theta so even in this case theta
will be equal to 90 degree because the
length vector of this arm is
perpendicular to the magnetic field so
this will also become equal to I into B
into B and what would be the direction
in this case take all your four fingers
along this direction because this is the
length vector and then it goes towards
the magnetic field so take all your arm
along this direction and then it goes
towards magnetic fish
therefore your thumb points towards up
so the net force acts in the upward
direction so what would be the net force
so use if you look at it you will see
that f1 is equal to f2 but they are in
the opposite direction so that means the
net force is equal to 0 because F 1 and
F 2 are equal and opposite right so we
see that the net force acting on this
rectangular current carrying loop in
uniform magnetic field is equal to 0 so
the net force is equal to 0 but another
interesting thing which we observed is
that if you look at the arm a B the
force acts in the downward direction if
you look at the arm C D the force F 2
acts in the upward direction and these
two forces do not act along the same
line because when you when this object
is rotating like this let us suppose the
object is rotating like this you apply
some force at this point and you apply
some force at this point so will the net
force be equal to zero
however the net force will be equal to
zero but because the line of these
opposite forces are not same therefore a
torque will start acting on it so the
same thing is happening here in this
case also since the axis of rotation is
this one so and a B and C D are not
connected by the any axis right so equal
and opposite forces are acting on a B
and C D but they are not acting along
the same lines so if you look at it in
this way so it is somewhat like this on
this there is a force which acts in the
downward direction so f1 try once to
what is the aim of this f1 this f1 wants
to bring this a B down it wants to push
a B downwards
and there is a force f2 which wants to
pull the side BC upward right so
basically this is how the force will
look f1 is trying to bring the side a B
down and f2 is trying to bring the side
C D up right so it is trying to
basically rotate the loop if you think
of it very carefully let us suppose
right now it is stable there is one
force f1 which is acting on this side
and it is trying to push a B down and
this f2 is trying to push pull and C D
up in that is even though f1 and f2 are
equal and opposite but still there is a
torque which is acting on the loop
making it to rotate because f1 will make
a B to go down and this will make C D to
come up so gradually CD will start
coming up ad will start coming down and
as I decided this entire loop will start
rotating right so therefore at all acts
right so this is the scenario if you
look at it from this this side so this
is f1 this is f2 right so if you join it
with from the same tag you will get a
figure like this that is this diagram
and let us suppose if this entire length
is a so the centre of mass will be at
the center so torque will act and what
will be the magnitude of torque that
will be equal to f1 multiply it by the
moment of the arm that is a by 2 plus F
2 multiplied by a by 2 so this will be
equal to f1 is equal to f2 is equal to I
be B this into a now what is a into B a
into B is nothing but the area of the
loop a is the area of the loop so this
is the net torque which will act on this
loop when it is present in an external
magnetic field so I hope you understood
how this extern
thought will act on and do now when I
look at the same scenario but there we
will assume here what did we assume here
we had assumed that the magnetic field
is in the plane of the loop right that
is what we had assumed at the very
beginning but in the next scenario we
will assume that initially itself the
loop is placed in such a way that it
makes some angle theta with the
direction of the magnetic fields right
so that is the second scenario which we
will consider so that we get a general
expression for torque so I hope you
understood how this rotation takes place
so for the sake of that I have this
diagram which will explain you more
clearly how this rotation is taking
place right thank you please visit exam
comm to watch free educational videos
try free online tests get the best
quality study materials study from the
best tutors and mentors and much more
thank you once again
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Dec 13, 2024 Last updated |
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