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1 | 5090-5093 | EXAMPLE 6 6
Rationalised 2023-24
Physics
164
EXAMPLE 6 6
FIGURE 6 11
Solution
Method I
As the rod is rotated, free electrons in the rod move towards the outer
end due to Lorentz force and get distributed over the ring |
1 | 5091-5094 | 6
Rationalised 2023-24
Physics
164
EXAMPLE 6 6
FIGURE 6 11
Solution
Method I
As the rod is rotated, free electrons in the rod move towards the outer
end due to Lorentz force and get distributed over the ring Thus, the
resulting separation of charges produces an emf across the ends of
the rod |
1 | 5092-5095 | 6
FIGURE 6 11
Solution
Method I
As the rod is rotated, free electrons in the rod move towards the outer
end due to Lorentz force and get distributed over the ring Thus, the
resulting separation of charges produces an emf across the ends of
the rod At a certain value of emf, there is no more flow of electrons
and a steady state is reached |
1 | 5093-5096 | 11
Solution
Method I
As the rod is rotated, free electrons in the rod move towards the outer
end due to Lorentz force and get distributed over the ring Thus, the
resulting separation of charges produces an emf across the ends of
the rod At a certain value of emf, there is no more flow of electrons
and a steady state is reached Using Eq |
1 | 5094-5097 | Thus, the
resulting separation of charges produces an emf across the ends of
the rod At a certain value of emf, there is no more flow of electrons
and a steady state is reached Using Eq (6 |
1 | 5095-5098 | At a certain value of emf, there is no more flow of electrons
and a steady state is reached Using Eq (6 5), the magnitude of the
emf generated across a length dr of the rod as it moves at right angles
to the magnetic field is given by
d
Bvd
r
ε = |
1 | 5096-5099 | Using Eq (6 5), the magnitude of the
emf generated across a length dr of the rod as it moves at right angles
to the magnetic field is given by
d
Bvd
r
ε = Hence,
ε
ε
=
= ∫
∫
d
Bvd
r
R
0
=
=
∫ B
r
r
B
R
R
ω
ω
d
2
0
2
Note that we have used v = w r |
1 | 5097-5100 | (6 5), the magnitude of the
emf generated across a length dr of the rod as it moves at right angles
to the magnetic field is given by
d
Bvd
r
ε = Hence,
ε
ε
=
= ∫
∫
d
Bvd
r
R
0
=
=
∫ B
r
r
B
R
R
ω
ω
d
2
0
2
Note that we have used v = w r This gives
e
2
1
1 |
1 | 5098-5101 | 5), the magnitude of the
emf generated across a length dr of the rod as it moves at right angles
to the magnetic field is given by
d
Bvd
r
ε = Hence,
ε
ε
=
= ∫
∫
d
Bvd
r
R
0
=
=
∫ B
r
r
B
R
R
ω
ω
d
2
0
2
Note that we have used v = w r This gives
e
2
1
1 0
2
50
(1 )
=2
×
×
π ×
×
= 157 V
Method II
To calculate the emf, we can imagine a closed loop OPQ in which
point O and P are connected with a resistor R and OQ is the rotating
rod |
1 | 5099-5102 | Hence,
ε
ε
=
= ∫
∫
d
Bvd
r
R
0
=
=
∫ B
r
r
B
R
R
ω
ω
d
2
0
2
Note that we have used v = w r This gives
e
2
1
1 0
2
50
(1 )
=2
×
×
π ×
×
= 157 V
Method II
To calculate the emf, we can imagine a closed loop OPQ in which
point O and P are connected with a resistor R and OQ is the rotating
rod The potential difference across the resistor is then equal to the
induced emf and equals B × (rate of change of area of loop) |
1 | 5100-5103 | This gives
e
2
1
1 0
2
50
(1 )
=2
×
×
π ×
×
= 157 V
Method II
To calculate the emf, we can imagine a closed loop OPQ in which
point O and P are connected with a resistor R and OQ is the rotating
rod The potential difference across the resistor is then equal to the
induced emf and equals B × (rate of change of area of loop) If q is the
angle between the rod and the radius of the circle at P at time t, the
area of the sector OPQ is given by
2
2
1
2
2
R
R
θ
θ
π
×
=
π
where R is the radius of the circle |
1 | 5101-5104 | 0
2
50
(1 )
=2
×
×
π ×
×
= 157 V
Method II
To calculate the emf, we can imagine a closed loop OPQ in which
point O and P are connected with a resistor R and OQ is the rotating
rod The potential difference across the resistor is then equal to the
induced emf and equals B × (rate of change of area of loop) If q is the
angle between the rod and the radius of the circle at P at time t, the
area of the sector OPQ is given by
2
2
1
2
2
R
R
θ
θ
π
×
=
π
where R is the radius of the circle Hence, the induced emf is
e = B
t
R
×
dd
1
2
2θ =
2
2
1
d
2
d
2
θ
ω
= B R
BR
t
[Note:
d
2
dt
θ
ω
ν
=
=
π
]
This expression is identical to the expression obtained by Method I
and we get the same value of e |
1 | 5102-5105 | The potential difference across the resistor is then equal to the
induced emf and equals B × (rate of change of area of loop) If q is the
angle between the rod and the radius of the circle at P at time t, the
area of the sector OPQ is given by
2
2
1
2
2
R
R
θ
θ
π
×
=
π
where R is the radius of the circle Hence, the induced emf is
e = B
t
R
×
dd
1
2
2θ =
2
2
1
d
2
d
2
θ
ω
= B R
BR
t
[Note:
d
2
dt
θ
ω
ν
=
=
π
]
This expression is identical to the expression obtained by Method I
and we get the same value of e Rationalised 2023-24
Electromagnetic
Induction
165
EXAMPLE 6 |
1 | 5103-5106 | If q is the
angle between the rod and the radius of the circle at P at time t, the
area of the sector OPQ is given by
2
2
1
2
2
R
R
θ
θ
π
×
=
π
where R is the radius of the circle Hence, the induced emf is
e = B
t
R
×
dd
1
2
2θ =
2
2
1
d
2
d
2
θ
ω
= B R
BR
t
[Note:
d
2
dt
θ
ω
ν
=
=
π
]
This expression is identical to the expression obtained by Method I
and we get the same value of e Rationalised 2023-24
Electromagnetic
Induction
165
EXAMPLE 6 7
Example 6 |
1 | 5104-5107 | Hence, the induced emf is
e = B
t
R
×
dd
1
2
2θ =
2
2
1
d
2
d
2
θ
ω
= B R
BR
t
[Note:
d
2
dt
θ
ω
ν
=
=
π
]
This expression is identical to the expression obtained by Method I
and we get the same value of e Rationalised 2023-24
Electromagnetic
Induction
165
EXAMPLE 6 7
Example 6 7
A wheel with 10 metallic spokes each 0 |
1 | 5105-5108 | Rationalised 2023-24
Electromagnetic
Induction
165
EXAMPLE 6 7
Example 6 7
A wheel with 10 metallic spokes each 0 5 m long is rotated with a
speed of 120 rev/min in a plane normal to the horizontal component
of earth’s magnetic field HE at a place |
1 | 5106-5109 | 7
Example 6 7
A wheel with 10 metallic spokes each 0 5 m long is rotated with a
speed of 120 rev/min in a plane normal to the horizontal component
of earth’s magnetic field HE at a place If HE = 0 |
1 | 5107-5110 | 7
A wheel with 10 metallic spokes each 0 5 m long is rotated with a
speed of 120 rev/min in a plane normal to the horizontal component
of earth’s magnetic field HE at a place If HE = 0 4 G at the place, what
is the induced emf between the axle and the rim of the wheel |
1 | 5108-5111 | 5 m long is rotated with a
speed of 120 rev/min in a plane normal to the horizontal component
of earth’s magnetic field HE at a place If HE = 0 4 G at the place, what
is the induced emf between the axle and the rim of the wheel Note
that 1 G = 10–4 T |
1 | 5109-5112 | If HE = 0 4 G at the place, what
is the induced emf between the axle and the rim of the wheel Note
that 1 G = 10–4 T Solution
Induced emf = (1/2) w B R2
= (1/2) × 4p × 0 |
1 | 5110-5113 | 4 G at the place, what
is the induced emf between the axle and the rim of the wheel Note
that 1 G = 10–4 T Solution
Induced emf = (1/2) w B R2
= (1/2) × 4p × 0 4 × 10–4 × (0 |
1 | 5111-5114 | Note
that 1 G = 10–4 T Solution
Induced emf = (1/2) w B R2
= (1/2) × 4p × 0 4 × 10–4 × (0 5)2
= 6 |
1 | 5112-5115 | Solution
Induced emf = (1/2) w B R2
= (1/2) × 4p × 0 4 × 10–4 × (0 5)2
= 6 28 × 10–5 V
The number of spokes is immaterial because the emf’s across the
spokes are in parallel |
1 | 5113-5116 | 4 × 10–4 × (0 5)2
= 6 28 × 10–5 V
The number of spokes is immaterial because the emf’s across the
spokes are in parallel 6 |
1 | 5114-5117 | 5)2
= 6 28 × 10–5 V
The number of spokes is immaterial because the emf’s across the
spokes are in parallel 6 7 INDUCTANCE
An electric current can be induced in a coil by flux change produced by
another coil in its vicinity or flux change produced by the same coil |
1 | 5115-5118 | 28 × 10–5 V
The number of spokes is immaterial because the emf’s across the
spokes are in parallel 6 7 INDUCTANCE
An electric current can be induced in a coil by flux change produced by
another coil in its vicinity or flux change produced by the same coil These
two situations are described separately in the next two sub-sections |
1 | 5116-5119 | 6 7 INDUCTANCE
An electric current can be induced in a coil by flux change produced by
another coil in its vicinity or flux change produced by the same coil These
two situations are described separately in the next two sub-sections However, in both the cases, the flux through a coil is proportional to the
current |
1 | 5117-5120 | 7 INDUCTANCE
An electric current can be induced in a coil by flux change produced by
another coil in its vicinity or flux change produced by the same coil These
two situations are described separately in the next two sub-sections However, in both the cases, the flux through a coil is proportional to the
current That is, FB a I |
1 | 5118-5121 | These
two situations are described separately in the next two sub-sections However, in both the cases, the flux through a coil is proportional to the
current That is, FB a I Further, if the geometry of the coil does not vary with time then,
d
d
d
d
B
I
t
t
Φ
∝
For a closely wound coil of N turns, the same magnetic flux is linked
with all the turns |
1 | 5119-5122 | However, in both the cases, the flux through a coil is proportional to the
current That is, FB a I Further, if the geometry of the coil does not vary with time then,
d
d
d
d
B
I
t
t
Φ
∝
For a closely wound coil of N turns, the same magnetic flux is linked
with all the turns When the flux FB through the coil changes, each turn
contributes to the induced emf |
1 | 5120-5123 | That is, FB a I Further, if the geometry of the coil does not vary with time then,
d
d
d
d
B
I
t
t
Φ
∝
For a closely wound coil of N turns, the same magnetic flux is linked
with all the turns When the flux FB through the coil changes, each turn
contributes to the induced emf Therefore, a term called flux linkage is
used which is equal to NFB for a closely wound coil and in such a case
NFB∝ I
The constant of proportionality, in this relation, is called inductance |
1 | 5121-5124 | Further, if the geometry of the coil does not vary with time then,
d
d
d
d
B
I
t
t
Φ
∝
For a closely wound coil of N turns, the same magnetic flux is linked
with all the turns When the flux FB through the coil changes, each turn
contributes to the induced emf Therefore, a term called flux linkage is
used which is equal to NFB for a closely wound coil and in such a case
NFB∝ I
The constant of proportionality, in this relation, is called inductance We shall see that inductance depends only on the geometry of the coil
and intrinsic material properties |
1 | 5122-5125 | When the flux FB through the coil changes, each turn
contributes to the induced emf Therefore, a term called flux linkage is
used which is equal to NFB for a closely wound coil and in such a case
NFB∝ I
The constant of proportionality, in this relation, is called inductance We shall see that inductance depends only on the geometry of the coil
and intrinsic material properties This aspect is akin to capacitance which
for a parallel plate capacitor depends on the plate area and plate separation
(geometry) and the dielectric constant K of the intervening medium
(intrinsic material property) |
1 | 5123-5126 | Therefore, a term called flux linkage is
used which is equal to NFB for a closely wound coil and in such a case
NFB∝ I
The constant of proportionality, in this relation, is called inductance We shall see that inductance depends only on the geometry of the coil
and intrinsic material properties This aspect is akin to capacitance which
for a parallel plate capacitor depends on the plate area and plate separation
(geometry) and the dielectric constant K of the intervening medium
(intrinsic material property) Inductance is a scalar quantity |
1 | 5124-5127 | We shall see that inductance depends only on the geometry of the coil
and intrinsic material properties This aspect is akin to capacitance which
for a parallel plate capacitor depends on the plate area and plate separation
(geometry) and the dielectric constant K of the intervening medium
(intrinsic material property) Inductance is a scalar quantity It has the dimensions of [M L2 T –2 A–2]
given by the dimensions of flux divided by the dimensions of current |
1 | 5125-5128 | This aspect is akin to capacitance which
for a parallel plate capacitor depends on the plate area and plate separation
(geometry) and the dielectric constant K of the intervening medium
(intrinsic material property) Inductance is a scalar quantity It has the dimensions of [M L2 T –2 A–2]
given by the dimensions of flux divided by the dimensions of current The
SI unit of inductance is henry and is denoted by H |
1 | 5126-5129 | Inductance is a scalar quantity It has the dimensions of [M L2 T –2 A–2]
given by the dimensions of flux divided by the dimensions of current The
SI unit of inductance is henry and is denoted by H It is named in honour
of Joseph Henry who discovered electromagnetic induction in USA,
independently of Faraday in England |
1 | 5127-5130 | It has the dimensions of [M L2 T –2 A–2]
given by the dimensions of flux divided by the dimensions of current The
SI unit of inductance is henry and is denoted by H It is named in honour
of Joseph Henry who discovered electromagnetic induction in USA,
independently of Faraday in England 6 |
1 | 5128-5131 | The
SI unit of inductance is henry and is denoted by H It is named in honour
of Joseph Henry who discovered electromagnetic induction in USA,
independently of Faraday in England 6 7 |
1 | 5129-5132 | It is named in honour
of Joseph Henry who discovered electromagnetic induction in USA,
independently of Faraday in England 6 7 1 Mutual inductance
Consider Fig |
1 | 5130-5133 | 6 7 1 Mutual inductance
Consider Fig 6 |
1 | 5131-5134 | 7 1 Mutual inductance
Consider Fig 6 11 which shows two long co-axial solenoids each of length
l |
1 | 5132-5135 | 1 Mutual inductance
Consider Fig 6 11 which shows two long co-axial solenoids each of length
l We denote the radius of the inner solenoid S1
by r1 and the number of
turns per unit length by n1 |
1 | 5133-5136 | 6 11 which shows two long co-axial solenoids each of length
l We denote the radius of the inner solenoid S1
by r1 and the number of
turns per unit length by n1 The corresponding quantities for the outer
solenoid S2 are r2 and n2, respectively |
1 | 5134-5137 | 11 which shows two long co-axial solenoids each of length
l We denote the radius of the inner solenoid S1
by r1 and the number of
turns per unit length by n1 The corresponding quantities for the outer
solenoid S2 are r2 and n2, respectively Let N1 and N2 be the total number
of turns of coils S1 and S2, respectively |
1 | 5135-5138 | We denote the radius of the inner solenoid S1
by r1 and the number of
turns per unit length by n1 The corresponding quantities for the outer
solenoid S2 are r2 and n2, respectively Let N1 and N2 be the total number
of turns of coils S1 and S2, respectively Rationalised 2023-24
Physics
166
When a current I2 is set up through S2, it in turn sets
up a magnetic flux through S1 |
1 | 5136-5139 | The corresponding quantities for the outer
solenoid S2 are r2 and n2, respectively Let N1 and N2 be the total number
of turns of coils S1 and S2, respectively Rationalised 2023-24
Physics
166
When a current I2 is set up through S2, it in turn sets
up a magnetic flux through S1 Let us denote it by F1 |
1 | 5137-5140 | Let N1 and N2 be the total number
of turns of coils S1 and S2, respectively Rationalised 2023-24
Physics
166
When a current I2 is set up through S2, it in turn sets
up a magnetic flux through S1 Let us denote it by F1 The corresponding flux linkage with solenoid S1 is
N1
1
M12 2
I
Φ =
(6 |
1 | 5138-5141 | Rationalised 2023-24
Physics
166
When a current I2 is set up through S2, it in turn sets
up a magnetic flux through S1 Let us denote it by F1 The corresponding flux linkage with solenoid S1 is
N1
1
M12 2
I
Φ =
(6 7)
M12 is called the mutual inductance of solenoid S1 with
respect to solenoid S2 |
1 | 5139-5142 | Let us denote it by F1 The corresponding flux linkage with solenoid S1 is
N1
1
M12 2
I
Φ =
(6 7)
M12 is called the mutual inductance of solenoid S1 with
respect to solenoid S2 It is also referred to as the
coefficient of mutual induction |
1 | 5140-5143 | The corresponding flux linkage with solenoid S1 is
N1
1
M12 2
I
Φ =
(6 7)
M12 is called the mutual inductance of solenoid S1 with
respect to solenoid S2 It is also referred to as the
coefficient of mutual induction For these simple co-axial solenoids it is possible to
calculate M12 |
1 | 5141-5144 | 7)
M12 is called the mutual inductance of solenoid S1 with
respect to solenoid S2 It is also referred to as the
coefficient of mutual induction For these simple co-axial solenoids it is possible to
calculate M12 The magnetic field due to the current I2 in
S2 is m0n2I2 |
1 | 5142-5145 | It is also referred to as the
coefficient of mutual induction For these simple co-axial solenoids it is possible to
calculate M12 The magnetic field due to the current I2 in
S2 is m0n2I2 The resulting flux linkage with coil S1 is,
(
) (
) (
)
2
1
1
1
1
0
2 2
N
n l
r
n I
Φ
µ
=
π
2
0
1
2
1
2
n n
r l I
=µ
π
(6 |
1 | 5143-5146 | For these simple co-axial solenoids it is possible to
calculate M12 The magnetic field due to the current I2 in
S2 is m0n2I2 The resulting flux linkage with coil S1 is,
(
) (
) (
)
2
1
1
1
1
0
2 2
N
n l
r
n I
Φ
µ
=
π
2
0
1
2
1
2
n n
r l I
=µ
π
(6 8)
where n1l is the total number of turns in solenoid S1 |
1 | 5144-5147 | The magnetic field due to the current I2 in
S2 is m0n2I2 The resulting flux linkage with coil S1 is,
(
) (
) (
)
2
1
1
1
1
0
2 2
N
n l
r
n I
Φ
µ
=
π
2
0
1
2
1
2
n n
r l I
=µ
π
(6 8)
where n1l is the total number of turns in solenoid S1 Thus,
from Eq |
1 | 5145-5148 | The resulting flux linkage with coil S1 is,
(
) (
) (
)
2
1
1
1
1
0
2 2
N
n l
r
n I
Φ
µ
=
π
2
0
1
2
1
2
n n
r l I
=µ
π
(6 8)
where n1l is the total number of turns in solenoid S1 Thus,
from Eq (6 |
1 | 5146-5149 | 8)
where n1l is the total number of turns in solenoid S1 Thus,
from Eq (6 9) and Eq |
1 | 5147-5150 | Thus,
from Eq (6 9) and Eq (6 |
1 | 5148-5151 | (6 9) and Eq (6 10),
M12 = m0n1n2pr 2
1l
(6 |
1 | 5149-5152 | 9) and Eq (6 10),
M12 = m0n1n2pr 2
1l
(6 9)
Note that we neglected the edge effects and considered
the magnetic field m0n2I2 to be uniform throughout the
length and width of the solenoid S2 |
1 | 5150-5153 | (6 10),
M12 = m0n1n2pr 2
1l
(6 9)
Note that we neglected the edge effects and considered
the magnetic field m0n2I2 to be uniform throughout the
length and width of the solenoid S2 This is a good approximation keeping
in mind that the solenoid is long, implying l >> r2 |
1 | 5151-5154 | 10),
M12 = m0n1n2pr 2
1l
(6 9)
Note that we neglected the edge effects and considered
the magnetic field m0n2I2 to be uniform throughout the
length and width of the solenoid S2 This is a good approximation keeping
in mind that the solenoid is long, implying l >> r2 We now consider the reverse case |
1 | 5152-5155 | 9)
Note that we neglected the edge effects and considered
the magnetic field m0n2I2 to be uniform throughout the
length and width of the solenoid S2 This is a good approximation keeping
in mind that the solenoid is long, implying l >> r2 We now consider the reverse case A current I1 is passed through the
solenoid S1 and the flux linkage with coil S2 is,
N2F2 = M21 I1
(6 |
1 | 5153-5156 | This is a good approximation keeping
in mind that the solenoid is long, implying l >> r2 We now consider the reverse case A current I1 is passed through the
solenoid S1 and the flux linkage with coil S2 is,
N2F2 = M21 I1
(6 10)
M21 is called the mutual inductance of solenoid S2 with respect to
solenoid S1 |
1 | 5154-5157 | We now consider the reverse case A current I1 is passed through the
solenoid S1 and the flux linkage with coil S2 is,
N2F2 = M21 I1
(6 10)
M21 is called the mutual inductance of solenoid S2 with respect to
solenoid S1 The flux due to the current I1 in S1 can be assumed to be confined
solely inside S1 since the solenoids are very long |
1 | 5155-5158 | A current I1 is passed through the
solenoid S1 and the flux linkage with coil S2 is,
N2F2 = M21 I1
(6 10)
M21 is called the mutual inductance of solenoid S2 with respect to
solenoid S1 The flux due to the current I1 in S1 can be assumed to be confined
solely inside S1 since the solenoids are very long Thus, flux linkage with
solenoid S2 is
(
) (
) (
)
2
2
2
2
1
0
1 1
N
n l
r
n I
Φ
µ
=
π
where n2l is the total number of turns of S2 |
1 | 5156-5159 | 10)
M21 is called the mutual inductance of solenoid S2 with respect to
solenoid S1 The flux due to the current I1 in S1 can be assumed to be confined
solely inside S1 since the solenoids are very long Thus, flux linkage with
solenoid S2 is
(
) (
) (
)
2
2
2
2
1
0
1 1
N
n l
r
n I
Φ
µ
=
π
where n2l is the total number of turns of S2 From Eq |
1 | 5157-5160 | The flux due to the current I1 in S1 can be assumed to be confined
solely inside S1 since the solenoids are very long Thus, flux linkage with
solenoid S2 is
(
) (
) (
)
2
2
2
2
1
0
1 1
N
n l
r
n I
Φ
µ
=
π
where n2l is the total number of turns of S2 From Eq (6 |
1 | 5158-5161 | Thus, flux linkage with
solenoid S2 is
(
) (
) (
)
2
2
2
2
1
0
1 1
N
n l
r
n I
Φ
µ
=
π
where n2l is the total number of turns of S2 From Eq (6 12),
M21 = m0n1n2pr 2
1l
(6 |
1 | 5159-5162 | From Eq (6 12),
M21 = m0n1n2pr 2
1l
(6 11)
Using Eq |
1 | 5160-5163 | (6 12),
M21 = m0n1n2pr 2
1l
(6 11)
Using Eq (6 |
1 | 5161-5164 | 12),
M21 = m0n1n2pr 2
1l
(6 11)
Using Eq (6 11) and Eq |
1 | 5162-5165 | 11)
Using Eq (6 11) and Eq (6 |
1 | 5163-5166 | (6 11) and Eq (6 12), we get
M12 = M21= M (say)
(6 |
1 | 5164-5167 | 11) and Eq (6 12), we get
M12 = M21= M (say)
(6 12)
We have demonstrated this equality for long co-axial solenoids |
1 | 5165-5168 | (6 12), we get
M12 = M21= M (say)
(6 12)
We have demonstrated this equality for long co-axial solenoids However, the relation is far more general |
1 | 5166-5169 | 12), we get
M12 = M21= M (say)
(6 12)
We have demonstrated this equality for long co-axial solenoids However, the relation is far more general Note that if the inner solenoid
was much shorter than (and placed well inside) the outer solenoid, then
we could still have calculated the flux linkage N1F1 because the inner
solenoid is effectively immersed in a uniform magnetic field due to the
outer solenoid |
1 | 5167-5170 | 12)
We have demonstrated this equality for long co-axial solenoids However, the relation is far more general Note that if the inner solenoid
was much shorter than (and placed well inside) the outer solenoid, then
we could still have calculated the flux linkage N1F1 because the inner
solenoid is effectively immersed in a uniform magnetic field due to the
outer solenoid In this case, the calculation of M12 would be easy |
1 | 5168-5171 | However, the relation is far more general Note that if the inner solenoid
was much shorter than (and placed well inside) the outer solenoid, then
we could still have calculated the flux linkage N1F1 because the inner
solenoid is effectively immersed in a uniform magnetic field due to the
outer solenoid In this case, the calculation of M12 would be easy However,
it would be extremely difficult to calculate the flux linkage with the outer
solenoid as the magnetic field due to the inner solenoid would vary across
the length as well as cross section of the outer solenoid |
1 | 5169-5172 | Note that if the inner solenoid
was much shorter than (and placed well inside) the outer solenoid, then
we could still have calculated the flux linkage N1F1 because the inner
solenoid is effectively immersed in a uniform magnetic field due to the
outer solenoid In this case, the calculation of M12 would be easy However,
it would be extremely difficult to calculate the flux linkage with the outer
solenoid as the magnetic field due to the inner solenoid would vary across
the length as well as cross section of the outer solenoid Therefore, the
calculation of M21 would also be extremely difficult in this case |
1 | 5170-5173 | In this case, the calculation of M12 would be easy However,
it would be extremely difficult to calculate the flux linkage with the outer
solenoid as the magnetic field due to the inner solenoid would vary across
the length as well as cross section of the outer solenoid Therefore, the
calculation of M21 would also be extremely difficult in this case The
equality M12=M21 is very useful in such situations |
1 | 5171-5174 | However,
it would be extremely difficult to calculate the flux linkage with the outer
solenoid as the magnetic field due to the inner solenoid would vary across
the length as well as cross section of the outer solenoid Therefore, the
calculation of M21 would also be extremely difficult in this case The
equality M12=M21 is very useful in such situations FIGURE 6 |
1 | 5172-5175 | Therefore, the
calculation of M21 would also be extremely difficult in this case The
equality M12=M21 is very useful in such situations FIGURE 6 12 Two long co-axial
solenoids of same
length l |
1 | 5173-5176 | The
equality M12=M21 is very useful in such situations FIGURE 6 12 Two long co-axial
solenoids of same
length l Rationalised 2023-24
Electromagnetic
Induction
167
EXAMPLE 6 |
1 | 5174-5177 | FIGURE 6 12 Two long co-axial
solenoids of same
length l Rationalised 2023-24
Electromagnetic
Induction
167
EXAMPLE 6 8
We explained the above example with air as the medium within the
solenoids |
1 | 5175-5178 | 12 Two long co-axial
solenoids of same
length l Rationalised 2023-24
Electromagnetic
Induction
167
EXAMPLE 6 8
We explained the above example with air as the medium within the
solenoids Instead, if a medium of relative permeability mr had been present,
the mutual inductance would be
M =mr m0 n1n2p r2
1 l
It is also important to know that the mutual inductance of a pair of
coils, solenoids, etc |
1 | 5176-5179 | Rationalised 2023-24
Electromagnetic
Induction
167
EXAMPLE 6 8
We explained the above example with air as the medium within the
solenoids Instead, if a medium of relative permeability mr had been present,
the mutual inductance would be
M =mr m0 n1n2p r2
1 l
It is also important to know that the mutual inductance of a pair of
coils, solenoids, etc , depends on their separation as well as their relative
orientation |
1 | 5177-5180 | 8
We explained the above example with air as the medium within the
solenoids Instead, if a medium of relative permeability mr had been present,
the mutual inductance would be
M =mr m0 n1n2p r2
1 l
It is also important to know that the mutual inductance of a pair of
coils, solenoids, etc , depends on their separation as well as their relative
orientation Example 6 |
1 | 5178-5181 | Instead, if a medium of relative permeability mr had been present,
the mutual inductance would be
M =mr m0 n1n2p r2
1 l
It is also important to know that the mutual inductance of a pair of
coils, solenoids, etc , depends on their separation as well as their relative
orientation Example 6 8 Two concentric circular coils, one of small radius r1 and
the other of large radius r2, such that r1 << r2, are placed co-axially
with centres coinciding |
1 | 5179-5182 | , depends on their separation as well as their relative
orientation Example 6 8 Two concentric circular coils, one of small radius r1 and
the other of large radius r2, such that r1 << r2, are placed co-axially
with centres coinciding Obtain the mutual inductance of the
arrangement |
1 | 5180-5183 | Example 6 8 Two concentric circular coils, one of small radius r1 and
the other of large radius r2, such that r1 << r2, are placed co-axially
with centres coinciding Obtain the mutual inductance of the
arrangement Solution Let a current I2 flow through the outer circular coil |
1 | 5181-5184 | 8 Two concentric circular coils, one of small radius r1 and
the other of large radius r2, such that r1 << r2, are placed co-axially
with centres coinciding Obtain the mutual inductance of the
arrangement Solution Let a current I2 flow through the outer circular coil The
field at the centre of the coil is B2 = m0I2 / 2r2 |
1 | 5182-5185 | Obtain the mutual inductance of the
arrangement Solution Let a current I2 flow through the outer circular coil The
field at the centre of the coil is B2 = m0I2 / 2r2 Since the other
co-axially placed coil has a very small radius, B2 may be considered
constant over its cross-sectional area |
1 | 5183-5186 | Solution Let a current I2 flow through the outer circular coil The
field at the centre of the coil is B2 = m0I2 / 2r2 Since the other
co-axially placed coil has a very small radius, B2 may be considered
constant over its cross-sectional area Hence,
F1 = pr 2
1B2
2
0
1
2
22
r
I
µ πr
=
= M12 I2
Thus,
2
0
1
12
2
2
r
M
r
µ π
=
From Eq |
1 | 5184-5187 | The
field at the centre of the coil is B2 = m0I2 / 2r2 Since the other
co-axially placed coil has a very small radius, B2 may be considered
constant over its cross-sectional area Hence,
F1 = pr 2
1B2
2
0
1
2
22
r
I
µ πr
=
= M12 I2
Thus,
2
0
1
12
2
2
r
M
r
µ π
=
From Eq (6 |
1 | 5185-5188 | Since the other
co-axially placed coil has a very small radius, B2 may be considered
constant over its cross-sectional area Hence,
F1 = pr 2
1B2
2
0
1
2
22
r
I
µ πr
=
= M12 I2
Thus,
2
0
1
12
2
2
r
M
r
µ π
=
From Eq (6 12)
2
0
1
12
21
2
2
r
M
M
r
µ π
=
=
Note that we calculated M12 from an approximate value of F1, assuming
the magnetic field B2 to be uniform over the area p r1
2 |
1 | 5186-5189 | Hence,
F1 = pr 2
1B2
2
0
1
2
22
r
I
µ πr
=
= M12 I2
Thus,
2
0
1
12
2
2
r
M
r
µ π
=
From Eq (6 12)
2
0
1
12
21
2
2
r
M
M
r
µ π
=
=
Note that we calculated M12 from an approximate value of F1, assuming
the magnetic field B2 to be uniform over the area p r1
2 However, we
can accept this value because r1 << r2 |
1 | 5187-5190 | (6 12)
2
0
1
12
21
2
2
r
M
M
r
µ π
=
=
Note that we calculated M12 from an approximate value of F1, assuming
the magnetic field B2 to be uniform over the area p r1
2 However, we
can accept this value because r1 << r2 Now, let us recollect Experiment 6 |
1 | 5188-5191 | 12)
2
0
1
12
21
2
2
r
M
M
r
µ π
=
=
Note that we calculated M12 from an approximate value of F1, assuming
the magnetic field B2 to be uniform over the area p r1
2 However, we
can accept this value because r1 << r2 Now, let us recollect Experiment 6 3 in Section 6 |
1 | 5189-5192 | However, we
can accept this value because r1 << r2 Now, let us recollect Experiment 6 3 in Section 6 2 |
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