Class 12 Exam > Class 12 Videos > Mutual inductance between long solenoids - Electromagnetic Induction
Mutual inductance between long solenoids - Electromagnetic Induction Video Lecture - Class 12
FAQs on Mutual inductance between long solenoids - Electromagnetic Induction Video Lecture - Class 12
| 1. What is mutual inductance between long solenoids? |  |
| 2. How is mutual inductance calculated between long solenoids? | |
Ans. The mutual inductance (M) between long solenoids can be calculated using the formula:
M = μ₀N₁N₂A/l
where μ₀ is the permeability of free space, N₁ and N₂ are the number of turns in the two solenoids, A is the area of cross-section of the solenoids, and l is the distance between them.
| 3. What factors affect the mutual inductance between long solenoids? | |
Ans. The mutual inductance between long solenoids is influenced by several factors. These factors include the number of turns in the solenoids, the area of cross-section of the solenoids, the distance between the solenoids, and the permeability of free space. Increasing the number of turns, the area of cross-section, or the distance between the solenoids will increase the mutual inductance.
| 4. How does mutual inductance affect the behavior of long solenoids? | |
Ans. Mutual inductance plays a crucial role in the behavior of long solenoids. When the current in one solenoid changes, it induces an EMF in the other solenoid through mutual inductance. This induced EMF can cause a current to flow in the second solenoid, leading to the transfer of energy between the solenoids. This phenomenon is utilized in devices like transformers to transfer electrical energy efficiently.
| 5. Can mutual inductance be negative between long solenoids? | |
Ans. No, mutual inductance cannot be negative between long solenoids. It is always a positive value, indicating the coupling between the magnetic fields of the solenoids. Negative values of mutual inductance are not physically meaningful in this context.
Text Transcript from Video
hello friends this video on
electromagnetic induction part 18
is brought to you by exam feel calm no
more fear from exam please make sure
that you have watched all the videos to
the part 17 before going ahead with part
18 so now let us try to derive an
expression for mutual inductance between
long coaxial solenoids now let us try to
calculate the mutual inductance between
two long co-axial solenoids what do I
mean when I say coaxial solenoids that
means the axis of the Centers of the
both the solenoids would be the same for
example if this is one solenoid the
other solenoid would be such something
like this that is the axis of both the
solenoids would be the same so that is
the case here let us suppose this is the
axis of the coaxial solenoids let us say
that the radius of the bigger solenoid
is r2 and the radius of the smaller
solenoid is r1 right so let us call the
smaller coil smaller coil let us
represent it with radius r1 let us
represent the smaller coil with radius
r1 and number of turns n1 similarly let
us represent the larger coil with radius
r2 and total number of turns as n2 now
let us assume case one in case one we
assume that some current I 2 flows in
coil 2 that means in the outer coil
there is a current i2 which is flowing
so what will happen because of current
i2 in coil 2 there will be a magnetic
flux change in coil 1 right so this i2
in coil 2 will result in a flux Phi 1 in
coil 1 right therefore we can see that
the total flux in coil 1 or flux linkage
of coil 1 is proportional
to the current flowing in coil to so we
can write it as n 1 Phi 1 is equal to M
1 2 into I 2 where m 1 2 is mutual
inductance this is mutual inductance of
1 with respect to 2 like that is M 1 2
so this is my first equation like so
what we can write now we know that
magnetic field now what we will do is so
here we have found one expression for
total flux in point 1 now we will find
out another expression for total flux in
coil 1 but in terms of magnetic field
and area so what would be the magnetic
field due to I 2 in coil 2 now due to
the current I 2 flowing in the coil 2
what would be the magnetic field in coil
1 the magnetic field would be equal to
MU naught n I so Mew naught into n 2
into I 2 so this will be equal to MU
naught into n 2 I 2 divided by L length
so we can say that the total flux N 1
Phi 1 will be equal to n 1 Phi 1 is V
into a so what is BB is nothing but the
magnetic field due to the current in I 2
so the magnetic field due to the outer
coil is influencing the inner coil right
so we can also write another expression
for total flux that is N 1 into instead
of Phi 1 we can write this will be equal
to B into what is area area will be even
we are writing it for coil 1 but the
magnetic field on coil 1 is nothing but
the magnetic field inside the outer coil
it is somewhat like this imagine the
situation in this way this is the larger
coil this is the smaller coil so the
magnetic field inside larger coin we
covered the entire smaller coil
so if I say what is the magnetic field
inside the smaller coil the magnetic
field inside the smaller coil is nothing
but the magnetic field which is produced
by this outer foil right so the flux in
the inner coil would be the value of
this magnetic field into this small area
right that will be the flux so this
total flux will be equal to n1 into mu
not n - I - divided by L into a1 right
so this is my expression - so now from
Equation 1 and 2 now if I compare
equation 1 and equation 2 we can equate
the right-hand sides so we can say that
M 1 2 I 2 is equal to n1 into mu naught
n 2 I 2 divided by L into a 1 so this I
2 and I 2 will get cancelled therefore
we can say that M 1 2 is equal to MU
naught n 1 n 2 into a 1 divided by n so
this is the mutual inductance of coil 1
with respect to coil 2 right so this was
the scenario when the LG that current is
flowing through the outer coil and we
calculated the mutual inductance of the
inner coil with respect to the outer
curve now let us consider the reverse
scenario in the next scenario I assume
that some current i1 is flowing through
coil 1 so i1 is flowing through coil 1
so this will produce flux this will
result in flux in coil 2 okay so if
current is flowing in coil 1 so this
will produce flux in coil 2 right so we
can see that the total flux in coil 2
we'll be n 2 Phi 2 is equal to M 2 1
into I 1 right this will be the total
flux let me call this equation as
equation 3 so this will be the total
flux now let me write another expression
for total flux through coil 2 we can
write it as n 2 Phi 2 will be equal to n
2 into the magnetic field what is this
flux this flux is nothing but magnetic
field into area so what would be the
magnetic field in this case the magnetic
field will be B right into what is this
B this B is nothing but a magnetic field
due to the smaller coil that is that in
inside coil right into what would be the
area now this is an important concept
here let us this means my bigger coil
this is my smaller pipe right now that
in that case the bigger or in the
previous means the bigger coil was
producing the magnetic field in this is
the smaller coil is producing the
magnetic field so if I ask you what is
the magnetic field magnetic flux for the
bigger point what is this it is this
much magnetic field this magnetic field
stick into this much area because the
outer area is of no use because the
entire flux which is produced by the
inner coil is confined only inside the
inner coil so there is no point
considering this bigger area because
effectively the flux for the bigger poil
also is mac this magnetic field into
this area right whereas in my previous
case in my previous case it was the
reverse because the magnetic field was
produced by the outer coil so here we
can write it as n 2 into B 1 into a 1
okay so this can be written as n 2 into
what is magnetic field by the inner foil
that will be mu naught so what is the
magnetic field by the inner coil that is
mu not n I that is N 1 I
one this can be written as mu not n 1i 1
divided by L so we can write it as mu
not n 1 I 1 divided by n into a 1 so now
if you compare this equation 3 so now if
you compare equation 3 and 4 so
comparing equation 3 and 4 you can
equate the right hand side that is M 2 1
into I 1 is equal to n 2 into mu naught
n 1 I 1 divided by L into a 1 so I 1 I 1
will get cancelled therefore you get M 2
1 is equal to MU naught n 1 n 2 into a 1
divided by L now if you compare the
value of this mutual inductance of 2
with respect to 1 with 1 with respect to
you with respect to 2 in the previous
three if you compare both of this you
find that M 1 2 is equal to M 2 1 so
this is an important result because many
times it so happens that if you have too
long coaxial solenoids it is easier for
you to calculate the mutual inductance
of the smaller solenoid with respect to
the outer solenoid but calculating the
reverse is little more rigorous so what
we can do we can just calculate the
mutual inductance of one solenoid with
respect to the other and then we know
that other way around also the value
comes out to be the same now what is the
reason that both the values are same the
reason is that this basic fact was the
reason the 1 the thing which I explained
here let me explain it once again if you
have let us suppose this is the larger
solenoid this is the smaller solenoid so
the two cases which we discussed is in
the first case current flows through
this solenoid so the magnetic flux this
solenoid produces is like this in the
second case current flows through this
solenoid so the magnetic flux it
produces
like this so in the first case the
magnetic flux on the second solenoid the
induced magnetic flux on the second
solenoid is equal to the magnetic field
this magnetic field multiplied by the
small area in the second case what is
the magnetic flux induced on the larger
solenoid it is this magnetic field into
this small area so in both the cases the
value of the magnetic flux which is
induced is the same because in both the
cases we are considering the smaller
area that is why it comes out that the
mutual inductance of a 1 with respect to
2 is equal to the mutual inductance of 2
with respect to 1 thank you please visit
exam few 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
electromagnetic induction part 18
is brought to you by exam feel calm no
more fear from exam please make sure
that you have watched all the videos to
the part 17 before going ahead with part
18 so now let us try to derive an
expression for mutual inductance between
long coaxial solenoids now let us try to
calculate the mutual inductance between
two long co-axial solenoids what do I
mean when I say coaxial solenoids that
means the axis of the Centers of the
both the solenoids would be the same for
example if this is one solenoid the
other solenoid would be such something
like this that is the axis of both the
solenoids would be the same so that is
the case here let us suppose this is the
axis of the coaxial solenoids let us say
that the radius of the bigger solenoid
is r2 and the radius of the smaller
solenoid is r1 right so let us call the
smaller coil smaller coil let us
represent it with radius r1 let us
represent the smaller coil with radius
r1 and number of turns n1 similarly let
us represent the larger coil with radius
r2 and total number of turns as n2 now
let us assume case one in case one we
assume that some current I 2 flows in
coil 2 that means in the outer coil
there is a current i2 which is flowing
so what will happen because of current
i2 in coil 2 there will be a magnetic
flux change in coil 1 right so this i2
in coil 2 will result in a flux Phi 1 in
coil 1 right therefore we can see that
the total flux in coil 1 or flux linkage
of coil 1 is proportional
to the current flowing in coil to so we
can write it as n 1 Phi 1 is equal to M
1 2 into I 2 where m 1 2 is mutual
inductance this is mutual inductance of
1 with respect to 2 like that is M 1 2
so this is my first equation like so
what we can write now we know that
magnetic field now what we will do is so
here we have found one expression for
total flux in point 1 now we will find
out another expression for total flux in
coil 1 but in terms of magnetic field
and area so what would be the magnetic
field due to I 2 in coil 2 now due to
the current I 2 flowing in the coil 2
what would be the magnetic field in coil
1 the magnetic field would be equal to
MU naught n I so Mew naught into n 2
into I 2 so this will be equal to MU
naught into n 2 I 2 divided by L length
so we can say that the total flux N 1
Phi 1 will be equal to n 1 Phi 1 is V
into a so what is BB is nothing but the
magnetic field due to the current in I 2
so the magnetic field due to the outer
coil is influencing the inner coil right
so we can also write another expression
for total flux that is N 1 into instead
of Phi 1 we can write this will be equal
to B into what is area area will be even
we are writing it for coil 1 but the
magnetic field on coil 1 is nothing but
the magnetic field inside the outer coil
it is somewhat like this imagine the
situation in this way this is the larger
coil this is the smaller coil so the
magnetic field inside larger coin we
covered the entire smaller coil
so if I say what is the magnetic field
inside the smaller coil the magnetic
field inside the smaller coil is nothing
but the magnetic field which is produced
by this outer foil right so the flux in
the inner coil would be the value of
this magnetic field into this small area
right that will be the flux so this
total flux will be equal to n1 into mu
not n - I - divided by L into a1 right
so this is my expression - so now from
Equation 1 and 2 now if I compare
equation 1 and equation 2 we can equate
the right-hand sides so we can say that
M 1 2 I 2 is equal to n1 into mu naught
n 2 I 2 divided by L into a 1 so this I
2 and I 2 will get cancelled therefore
we can say that M 1 2 is equal to MU
naught n 1 n 2 into a 1 divided by n so
this is the mutual inductance of coil 1
with respect to coil 2 right so this was
the scenario when the LG that current is
flowing through the outer coil and we
calculated the mutual inductance of the
inner coil with respect to the outer
curve now let us consider the reverse
scenario in the next scenario I assume
that some current i1 is flowing through
coil 1 so i1 is flowing through coil 1
so this will produce flux this will
result in flux in coil 2 okay so if
current is flowing in coil 1 so this
will produce flux in coil 2 right so we
can see that the total flux in coil 2
we'll be n 2 Phi 2 is equal to M 2 1
into I 1 right this will be the total
flux let me call this equation as
equation 3 so this will be the total
flux now let me write another expression
for total flux through coil 2 we can
write it as n 2 Phi 2 will be equal to n
2 into the magnetic field what is this
flux this flux is nothing but magnetic
field into area so what would be the
magnetic field in this case the magnetic
field will be B right into what is this
B this B is nothing but a magnetic field
due to the smaller coil that is that in
inside coil right into what would be the
area now this is an important concept
here let us this means my bigger coil
this is my smaller pipe right now that
in that case the bigger or in the
previous means the bigger coil was
producing the magnetic field in this is
the smaller coil is producing the
magnetic field so if I ask you what is
the magnetic field magnetic flux for the
bigger point what is this it is this
much magnetic field this magnetic field
stick into this much area because the
outer area is of no use because the
entire flux which is produced by the
inner coil is confined only inside the
inner coil so there is no point
considering this bigger area because
effectively the flux for the bigger poil
also is mac this magnetic field into
this area right whereas in my previous
case in my previous case it was the
reverse because the magnetic field was
produced by the outer coil so here we
can write it as n 2 into B 1 into a 1
okay so this can be written as n 2 into
what is magnetic field by the inner foil
that will be mu naught so what is the
magnetic field by the inner coil that is
mu not n I that is N 1 I
one this can be written as mu not n 1i 1
divided by L so we can write it as mu
not n 1 I 1 divided by n into a 1 so now
if you compare this equation 3 so now if
you compare equation 3 and 4 so
comparing equation 3 and 4 you can
equate the right hand side that is M 2 1
into I 1 is equal to n 2 into mu naught
n 1 I 1 divided by L into a 1 so I 1 I 1
will get cancelled therefore you get M 2
1 is equal to MU naught n 1 n 2 into a 1
divided by L now if you compare the
value of this mutual inductance of 2
with respect to 1 with 1 with respect to
you with respect to 2 in the previous
three if you compare both of this you
find that M 1 2 is equal to M 2 1 so
this is an important result because many
times it so happens that if you have too
long coaxial solenoids it is easier for
you to calculate the mutual inductance
of the smaller solenoid with respect to
the outer solenoid but calculating the
reverse is little more rigorous so what
we can do we can just calculate the
mutual inductance of one solenoid with
respect to the other and then we know
that other way around also the value
comes out to be the same now what is the
reason that both the values are same the
reason is that this basic fact was the
reason the 1 the thing which I explained
here let me explain it once again if you
have let us suppose this is the larger
solenoid this is the smaller solenoid so
the two cases which we discussed is in
the first case current flows through
this solenoid so the magnetic flux this
solenoid produces is like this in the
second case current flows through this
solenoid so the magnetic flux it
produces
like this so in the first case the
magnetic flux on the second solenoid the
induced magnetic flux on the second
solenoid is equal to the magnetic field
this magnetic field multiplied by the
small area in the second case what is
the magnetic flux induced on the larger
solenoid it is this magnetic field into
this small area so in both the cases the
value of the magnetic flux which is
induced is the same because in both the
cases we are considering the smaller
area that is why it comes out that the
mutual inductance of a 1 with respect to
2 is equal to the mutual inductance of 2
with respect to 1 thank you please visit
exam few 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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Mar 10, 2025
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