Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 3790-3793 | 20)
B = m0n I
= 4p × 10–7 × 103 × 5
= 6 28 × 10–3 T
FIGURE 4 17 Two long straight
parallel conductors carrying steady
currents Ia and Ib and separated by a
distance d Ba is the magnetic field set
up by conductor ‘a’ at conductor ‘b’ |
1 | 3791-3794 | 28 × 10–3 T
FIGURE 4 17 Two long straight
parallel conductors carrying steady
currents Ia and Ib and separated by a
distance d Ba is the magnetic field set
up by conductor ‘a’ at conductor ‘b’ 4 |
1 | 3792-3795 | 17 Two long straight
parallel conductors carrying steady
currents Ia and Ib and separated by a
distance d Ba is the magnetic field set
up by conductor ‘a’ at conductor ‘b’ 4 8 FORCE BETWEEN TWO PARALLEL
CURRENTS, THE AMPERE
We have learnt that there exists a magnetic field due to a
conductor carrying a current which obeys the Biot-Savart
law |
1 | 3793-3796 | Ba is the magnetic field set
up by conductor ‘a’ at conductor ‘b’ 4 8 FORCE BETWEEN TWO PARALLEL
CURRENTS, THE AMPERE
We have learnt that there exists a magnetic field due to a
conductor carrying a current which obeys the Biot-Savart
law Further, we have learnt that an external magnetic field
will exert a force on a current-carrying conductor |
1 | 3794-3797 | 4 8 FORCE BETWEEN TWO PARALLEL
CURRENTS, THE AMPERE
We have learnt that there exists a magnetic field due to a
conductor carrying a current which obeys the Biot-Savart
law Further, we have learnt that an external magnetic field
will exert a force on a current-carrying conductor This
follows from the Lorentz force formula |
1 | 3795-3798 | 8 FORCE BETWEEN TWO PARALLEL
CURRENTS, THE AMPERE
We have learnt that there exists a magnetic field due to a
conductor carrying a current which obeys the Biot-Savart
law Further, we have learnt that an external magnetic field
will exert a force on a current-carrying conductor This
follows from the Lorentz force formula Thus, it is logical
to expect that two current-carrying conductors placed near
each other will exert (magnetic) forces on each other |
1 | 3796-3799 | Further, we have learnt that an external magnetic field
will exert a force on a current-carrying conductor This
follows from the Lorentz force formula Thus, it is logical
to expect that two current-carrying conductors placed near
each other will exert (magnetic) forces on each other In
the period 1820-25, Ampere studied the nature of this
magnetic force and its dependence on the magnitude of
the current, on the shape and size of the conductors, as
well as, the distances between the conductors |
1 | 3797-3800 | This
follows from the Lorentz force formula Thus, it is logical
to expect that two current-carrying conductors placed near
each other will exert (magnetic) forces on each other In
the period 1820-25, Ampere studied the nature of this
magnetic force and its dependence on the magnitude of
the current, on the shape and size of the conductors, as
well as, the distances between the conductors In this
section, we shall take the simple example of two parallel
current- carrying conductors, which will perhaps help us
to appreciate Ampere’s painstaking work |
1 | 3798-3801 | Thus, it is logical
to expect that two current-carrying conductors placed near
each other will exert (magnetic) forces on each other In
the period 1820-25, Ampere studied the nature of this
magnetic force and its dependence on the magnitude of
the current, on the shape and size of the conductors, as
well as, the distances between the conductors In this
section, we shall take the simple example of two parallel
current- carrying conductors, which will perhaps help us
to appreciate Ampere’s painstaking work Rationalised 2023-24
123
Moving Charges and
Magnetism
Figure 4 |
1 | 3799-3802 | In
the period 1820-25, Ampere studied the nature of this
magnetic force and its dependence on the magnitude of
the current, on the shape and size of the conductors, as
well as, the distances between the conductors In this
section, we shall take the simple example of two parallel
current- carrying conductors, which will perhaps help us
to appreciate Ampere’s painstaking work Rationalised 2023-24
123
Moving Charges and
Magnetism
Figure 4 17 shows two long parallel conductors a and b separated
by a distance d and carrying (parallel) currents Ia and Ib, respectively |
1 | 3800-3803 | In this
section, we shall take the simple example of two parallel
current- carrying conductors, which will perhaps help us
to appreciate Ampere’s painstaking work Rationalised 2023-24
123
Moving Charges and
Magnetism
Figure 4 17 shows two long parallel conductors a and b separated
by a distance d and carrying (parallel) currents Ia and Ib, respectively The conductor ‘a’ produces, the same magnetic field Ba at all points
along the conductor ‘b’ |
1 | 3801-3804 | Rationalised 2023-24
123
Moving Charges and
Magnetism
Figure 4 17 shows two long parallel conductors a and b separated
by a distance d and carrying (parallel) currents Ia and Ib, respectively The conductor ‘a’ produces, the same magnetic field Ba at all points
along the conductor ‘b’ The right-hand rule tells us that the direction of
this field is downwards (when the conductors are placed horizontally) |
1 | 3802-3805 | 17 shows two long parallel conductors a and b separated
by a distance d and carrying (parallel) currents Ia and Ib, respectively The conductor ‘a’ produces, the same magnetic field Ba at all points
along the conductor ‘b’ The right-hand rule tells us that the direction of
this field is downwards (when the conductors are placed horizontally) Its magnitude is given by Eq |
1 | 3803-3806 | The conductor ‘a’ produces, the same magnetic field Ba at all points
along the conductor ‘b’ The right-hand rule tells us that the direction of
this field is downwards (when the conductors are placed horizontally) Its magnitude is given by Eq [4 |
1 | 3804-3807 | The right-hand rule tells us that the direction of
this field is downwards (when the conductors are placed horizontally) Its magnitude is given by Eq [4 19(a)] or from Ampere’s circuital law,
0
2
a
a
I
B
d
=µ
π
The conductor ‘b’ carrying a current Ib will experience a sideways
force due to the field Ba |
1 | 3805-3808 | Its magnitude is given by Eq [4 19(a)] or from Ampere’s circuital law,
0
2
a
a
I
B
d
=µ
π
The conductor ‘b’ carrying a current Ib will experience a sideways
force due to the field Ba The direction of this force is towards the
conductor ‘a’ (Verify this) |
1 | 3806-3809 | [4 19(a)] or from Ampere’s circuital law,
0
2
a
a
I
B
d
=µ
π
The conductor ‘b’ carrying a current Ib will experience a sideways
force due to the field Ba The direction of this force is towards the
conductor ‘a’ (Verify this) We label this force as Fba, the force on a
segment L of ‘b’ due to ‘a’ |
1 | 3807-3810 | 19(a)] or from Ampere’s circuital law,
0
2
a
a
I
B
d
=µ
π
The conductor ‘b’ carrying a current Ib will experience a sideways
force due to the field Ba The direction of this force is towards the
conductor ‘a’ (Verify this) We label this force as Fba, the force on a
segment L of ‘b’ due to ‘a’ The magnitude of this force is given by
Eq |
1 | 3808-3811 | The direction of this force is towards the
conductor ‘a’ (Verify this) We label this force as Fba, the force on a
segment L of ‘b’ due to ‘a’ The magnitude of this force is given by
Eq (4 |
1 | 3809-3812 | We label this force as Fba, the force on a
segment L of ‘b’ due to ‘a’ The magnitude of this force is given by
Eq (4 4),
Fba = Ib L Ba
0
2
a
I Ib
L
d
=µ
π
(4 |
1 | 3810-3813 | The magnitude of this force is given by
Eq (4 4),
Fba = Ib L Ba
0
2
a
I Ib
L
d
=µ
π
(4 23)
It is of course possible to compute the force on ‘a’ due to ‘b’ |
1 | 3811-3814 | (4 4),
Fba = Ib L Ba
0
2
a
I Ib
L
d
=µ
π
(4 23)
It is of course possible to compute the force on ‘a’ due to ‘b’ From
considerations similar to above we can find the force Fab, on a segment of
length L of ‘a’ due to the current in ‘b’ |
1 | 3812-3815 | 4),
Fba = Ib L Ba
0
2
a
I Ib
L
d
=µ
π
(4 23)
It is of course possible to compute the force on ‘a’ due to ‘b’ From
considerations similar to above we can find the force Fab, on a segment of
length L of ‘a’ due to the current in ‘b’ It is equal in magnitude to Fba,
and directed towards ‘b’ |
1 | 3813-3816 | 23)
It is of course possible to compute the force on ‘a’ due to ‘b’ From
considerations similar to above we can find the force Fab, on a segment of
length L of ‘a’ due to the current in ‘b’ It is equal in magnitude to Fba,
and directed towards ‘b’ Thus,
Fba = –Fab
(4 |
1 | 3814-3817 | From
considerations similar to above we can find the force Fab, on a segment of
length L of ‘a’ due to the current in ‘b’ It is equal in magnitude to Fba,
and directed towards ‘b’ Thus,
Fba = –Fab
(4 24)
Note that this is consistent with Newton’s third Law |
1 | 3815-3818 | It is equal in magnitude to Fba,
and directed towards ‘b’ Thus,
Fba = –Fab
(4 24)
Note that this is consistent with Newton’s third Law Thus, at least for
parallel conductors and steady currents, we have shown that the
Biot-Savart law and the Lorentz force yield results in accordance with
Newton’s third Law* |
1 | 3816-3819 | Thus,
Fba = –Fab
(4 24)
Note that this is consistent with Newton’s third Law Thus, at least for
parallel conductors and steady currents, we have shown that the
Biot-Savart law and the Lorentz force yield results in accordance with
Newton’s third Law* We have seen from above that currents flowing in the same direction
attract each other |
1 | 3817-3820 | 24)
Note that this is consistent with Newton’s third Law Thus, at least for
parallel conductors and steady currents, we have shown that the
Biot-Savart law and the Lorentz force yield results in accordance with
Newton’s third Law* We have seen from above that currents flowing in the same direction
attract each other One can show that oppositely directed currents repel
each other |
1 | 3818-3821 | Thus, at least for
parallel conductors and steady currents, we have shown that the
Biot-Savart law and the Lorentz force yield results in accordance with
Newton’s third Law* We have seen from above that currents flowing in the same direction
attract each other One can show that oppositely directed currents repel
each other Thus,
Parallel currents attract, and antiparallel currents repel |
1 | 3819-3822 | We have seen from above that currents flowing in the same direction
attract each other One can show that oppositely directed currents repel
each other Thus,
Parallel currents attract, and antiparallel currents repel This rule is the opposite of what we find in electrostatics |
1 | 3820-3823 | One can show that oppositely directed currents repel
each other Thus,
Parallel currents attract, and antiparallel currents repel This rule is the opposite of what we find in electrostatics Like (same
sign) charges repel each other, but like (parallel) currents attract each
other |
1 | 3821-3824 | Thus,
Parallel currents attract, and antiparallel currents repel This rule is the opposite of what we find in electrostatics Like (same
sign) charges repel each other, but like (parallel) currents attract each
other Let fba represent the magnitude of the force Fba per unit length |
1 | 3822-3825 | This rule is the opposite of what we find in electrostatics Like (same
sign) charges repel each other, but like (parallel) currents attract each
other Let fba represent the magnitude of the force Fba per unit length Then,
from Eq |
1 | 3823-3826 | Like (same
sign) charges repel each other, but like (parallel) currents attract each
other Let fba represent the magnitude of the force Fba per unit length Then,
from Eq (4 |
1 | 3824-3827 | Let fba represent the magnitude of the force Fba per unit length Then,
from Eq (4 23),
π
0
2
a
b
ba
I I
f
d
=µ
(4 |
1 | 3825-3828 | Then,
from Eq (4 23),
π
0
2
a
b
ba
I I
f
d
=µ
(4 25)
The above expression is used to define the ampere (A), which is one of
the seven SI base units |
1 | 3826-3829 | (4 23),
π
0
2
a
b
ba
I I
f
d
=µ
(4 25)
The above expression is used to define the ampere (A), which is one of
the seven SI base units *
It turns out that when we have time-dependent currents and/or charges in
motion, Newton’s third law may not hold for forces between charges and/or
conductors |
1 | 3827-3830 | 23),
π
0
2
a
b
ba
I I
f
d
=µ
(4 25)
The above expression is used to define the ampere (A), which is one of
the seven SI base units *
It turns out that when we have time-dependent currents and/or charges in
motion, Newton’s third law may not hold for forces between charges and/or
conductors An essential consequence of the Newton’s third law in mechanics
is conservation of momentum of an isolated system |
1 | 3828-3831 | 25)
The above expression is used to define the ampere (A), which is one of
the seven SI base units *
It turns out that when we have time-dependent currents and/or charges in
motion, Newton’s third law may not hold for forces between charges and/or
conductors An essential consequence of the Newton’s third law in mechanics
is conservation of momentum of an isolated system This, however, holds even
for the case of time-dependent situations with electromagnetic fields, provided
the momentum carried by fields is also taken into account |
1 | 3829-3832 | *
It turns out that when we have time-dependent currents and/or charges in
motion, Newton’s third law may not hold for forces between charges and/or
conductors An essential consequence of the Newton’s third law in mechanics
is conservation of momentum of an isolated system This, however, holds even
for the case of time-dependent situations with electromagnetic fields, provided
the momentum carried by fields is also taken into account Rationalised 2023-24
Physics
124
EXAMPLE 4 |
1 | 3830-3833 | An essential consequence of the Newton’s third law in mechanics
is conservation of momentum of an isolated system This, however, holds even
for the case of time-dependent situations with electromagnetic fields, provided
the momentum carried by fields is also taken into account Rationalised 2023-24
Physics
124
EXAMPLE 4 10
The ampere is the value of that steady current which, when maintained
in each of the two very long, straight, parallel conductors of negligible
cross-section, and placed one metre apart in vacuum, would produce
on each of these conductors a force equal to 2 × 10–7 newtons per metre
of length |
1 | 3831-3834 | This, however, holds even
for the case of time-dependent situations with electromagnetic fields, provided
the momentum carried by fields is also taken into account Rationalised 2023-24
Physics
124
EXAMPLE 4 10
The ampere is the value of that steady current which, when maintained
in each of the two very long, straight, parallel conductors of negligible
cross-section, and placed one metre apart in vacuum, would produce
on each of these conductors a force equal to 2 × 10–7 newtons per metre
of length This definition of the ampere was adopted in 1946 |
1 | 3832-3835 | Rationalised 2023-24
Physics
124
EXAMPLE 4 10
The ampere is the value of that steady current which, when maintained
in each of the two very long, straight, parallel conductors of negligible
cross-section, and placed one metre apart in vacuum, would produce
on each of these conductors a force equal to 2 × 10–7 newtons per metre
of length This definition of the ampere was adopted in 1946 It is a theoretical
definition |
1 | 3833-3836 | 10
The ampere is the value of that steady current which, when maintained
in each of the two very long, straight, parallel conductors of negligible
cross-section, and placed one metre apart in vacuum, would produce
on each of these conductors a force equal to 2 × 10–7 newtons per metre
of length This definition of the ampere was adopted in 1946 It is a theoretical
definition In practice, one must eliminate the effect of the earth’s magnetic
field and substitute very long wires by multiturn coils of appropriate
geometries |
1 | 3834-3837 | This definition of the ampere was adopted in 1946 It is a theoretical
definition In practice, one must eliminate the effect of the earth’s magnetic
field and substitute very long wires by multiturn coils of appropriate
geometries An instrument called the current balance is used to measure
this mechanical force |
1 | 3835-3838 | It is a theoretical
definition In practice, one must eliminate the effect of the earth’s magnetic
field and substitute very long wires by multiturn coils of appropriate
geometries An instrument called the current balance is used to measure
this mechanical force The SI unit of charge, namely, the coulomb, can now be defined in
terms of the ampere |
1 | 3836-3839 | In practice, one must eliminate the effect of the earth’s magnetic
field and substitute very long wires by multiturn coils of appropriate
geometries An instrument called the current balance is used to measure
this mechanical force The SI unit of charge, namely, the coulomb, can now be defined in
terms of the ampere When a steady current of 1A is set up in a conductor, the quantity of
charge that flows through its cross-section in 1s is one coulomb (1C) |
1 | 3837-3840 | An instrument called the current balance is used to measure
this mechanical force The SI unit of charge, namely, the coulomb, can now be defined in
terms of the ampere When a steady current of 1A is set up in a conductor, the quantity of
charge that flows through its cross-section in 1s is one coulomb (1C) Example 4 |
1 | 3838-3841 | The SI unit of charge, namely, the coulomb, can now be defined in
terms of the ampere When a steady current of 1A is set up in a conductor, the quantity of
charge that flows through its cross-section in 1s is one coulomb (1C) Example 4 10 The horizontal component of the earth’s magnetic field
at a certain place is 3 |
1 | 3839-3842 | When a steady current of 1A is set up in a conductor, the quantity of
charge that flows through its cross-section in 1s is one coulomb (1C) Example 4 10 The horizontal component of the earth’s magnetic field
at a certain place is 3 0 ×10–5 T and the direction of the field is from
the geographic south to the geographic north |
1 | 3840-3843 | Example 4 10 The horizontal component of the earth’s magnetic field
at a certain place is 3 0 ×10–5 T and the direction of the field is from
the geographic south to the geographic north A very long straight
conductor is carrying a steady current of 1A |
1 | 3841-3844 | 10 The horizontal component of the earth’s magnetic field
at a certain place is 3 0 ×10–5 T and the direction of the field is from
the geographic south to the geographic north A very long straight
conductor is carrying a steady current of 1A What is the force per
unit length on it when it is placed on a horizontal table and the
direction of the current is (a) east to west; (b) south to north |
1 | 3842-3845 | 0 ×10–5 T and the direction of the field is from
the geographic south to the geographic north A very long straight
conductor is carrying a steady current of 1A What is the force per
unit length on it when it is placed on a horizontal table and the
direction of the current is (a) east to west; (b) south to north Solution F = Il × B
F = IlB sinq
The force per unit length is
f = F/l = I B sinq
(a) When the current is flowing from east to west,
q = 90°
Hence,
f = I B
= 1 × 3 × 10–5 = 3 × 10–5 N m–1
This is larger than the value 2×10–7 Nm–1 quoted in the definition
of the ampere |
1 | 3843-3846 | A very long straight
conductor is carrying a steady current of 1A What is the force per
unit length on it when it is placed on a horizontal table and the
direction of the current is (a) east to west; (b) south to north Solution F = Il × B
F = IlB sinq
The force per unit length is
f = F/l = I B sinq
(a) When the current is flowing from east to west,
q = 90°
Hence,
f = I B
= 1 × 3 × 10–5 = 3 × 10–5 N m–1
This is larger than the value 2×10–7 Nm–1 quoted in the definition
of the ampere Hence it is important to eliminate the effect of the
earth’s magnetic field and other stray fields while standardising
the ampere |
1 | 3844-3847 | What is the force per
unit length on it when it is placed on a horizontal table and the
direction of the current is (a) east to west; (b) south to north Solution F = Il × B
F = IlB sinq
The force per unit length is
f = F/l = I B sinq
(a) When the current is flowing from east to west,
q = 90°
Hence,
f = I B
= 1 × 3 × 10–5 = 3 × 10–5 N m–1
This is larger than the value 2×10–7 Nm–1 quoted in the definition
of the ampere Hence it is important to eliminate the effect of the
earth’s magnetic field and other stray fields while standardising
the ampere The direction of the force is downwards |
1 | 3845-3848 | Solution F = Il × B
F = IlB sinq
The force per unit length is
f = F/l = I B sinq
(a) When the current is flowing from east to west,
q = 90°
Hence,
f = I B
= 1 × 3 × 10–5 = 3 × 10–5 N m–1
This is larger than the value 2×10–7 Nm–1 quoted in the definition
of the ampere Hence it is important to eliminate the effect of the
earth’s magnetic field and other stray fields while standardising
the ampere The direction of the force is downwards This direction may be
obtained by the directional property of cross product of vectors |
1 | 3846-3849 | Hence it is important to eliminate the effect of the
earth’s magnetic field and other stray fields while standardising
the ampere The direction of the force is downwards This direction may be
obtained by the directional property of cross product of vectors (b) When the current is flowing from south to north,
q = 0o
f = 0
Hence there is no force on the conductor |
1 | 3847-3850 | The direction of the force is downwards This direction may be
obtained by the directional property of cross product of vectors (b) When the current is flowing from south to north,
q = 0o
f = 0
Hence there is no force on the conductor 4 |
1 | 3848-3851 | This direction may be
obtained by the directional property of cross product of vectors (b) When the current is flowing from south to north,
q = 0o
f = 0
Hence there is no force on the conductor 4 9 TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE
4 |
1 | 3849-3852 | (b) When the current is flowing from south to north,
q = 0o
f = 0
Hence there is no force on the conductor 4 9 TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE
4 9 |
1 | 3850-3853 | 4 9 TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE
4 9 1
Torque on a rectangular current loop in a uniform
magnetic field
We now show that a rectangular loop carrying a steady current I and
placed in a uniform magnetic field experiences a torque |
1 | 3851-3854 | 9 TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE
4 9 1
Torque on a rectangular current loop in a uniform
magnetic field
We now show that a rectangular loop carrying a steady current I and
placed in a uniform magnetic field experiences a torque It does not
experience a net force |
1 | 3852-3855 | 9 1
Torque on a rectangular current loop in a uniform
magnetic field
We now show that a rectangular loop carrying a steady current I and
placed in a uniform magnetic field experiences a torque It does not
experience a net force This behaviour is analogous to that of electric
dipole in a uniform electric field (Section 1 |
1 | 3853-3856 | 1
Torque on a rectangular current loop in a uniform
magnetic field
We now show that a rectangular loop carrying a steady current I and
placed in a uniform magnetic field experiences a torque It does not
experience a net force This behaviour is analogous to that of electric
dipole in a uniform electric field (Section 1 12) |
1 | 3854-3857 | It does not
experience a net force This behaviour is analogous to that of electric
dipole in a uniform electric field (Section 1 12) Rationalised 2023-24
125
Moving Charges and
Magnetism
We first consider the simple case when the
rectangular loop is placed such that the uniform
magnetic field B is in the plane of the loop |
1 | 3855-3858 | This behaviour is analogous to that of electric
dipole in a uniform electric field (Section 1 12) Rationalised 2023-24
125
Moving Charges and
Magnetism
We first consider the simple case when the
rectangular loop is placed such that the uniform
magnetic field B is in the plane of the loop This is
illustrated in Fig |
1 | 3856-3859 | 12) Rationalised 2023-24
125
Moving Charges and
Magnetism
We first consider the simple case when the
rectangular loop is placed such that the uniform
magnetic field B is in the plane of the loop This is
illustrated in Fig 4 |
1 | 3857-3860 | Rationalised 2023-24
125
Moving Charges and
Magnetism
We first consider the simple case when the
rectangular loop is placed such that the uniform
magnetic field B is in the plane of the loop This is
illustrated in Fig 4 18(a) |
1 | 3858-3861 | This is
illustrated in Fig 4 18(a) The field exerts no force on the two arms AD and BC
of the loop |
1 | 3859-3862 | 4 18(a) The field exerts no force on the two arms AD and BC
of the loop It is perpendicular to the arm AB of the loop
and exerts a force F1 on it which is directed into the
plane of the loop |
1 | 3860-3863 | 18(a) The field exerts no force on the two arms AD and BC
of the loop It is perpendicular to the arm AB of the loop
and exerts a force F1 on it which is directed into the
plane of the loop Its magnitude is,
F1 = I b B
Similarly, it exerts a force F2 on the arm CD and F2
is directed out of the plane of the paper |
1 | 3861-3864 | The field exerts no force on the two arms AD and BC
of the loop It is perpendicular to the arm AB of the loop
and exerts a force F1 on it which is directed into the
plane of the loop Its magnitude is,
F1 = I b B
Similarly, it exerts a force F2 on the arm CD and F2
is directed out of the plane of the paper F2 = I b B = F1
Thus, the net force on the loop is zero |
1 | 3862-3865 | It is perpendicular to the arm AB of the loop
and exerts a force F1 on it which is directed into the
plane of the loop Its magnitude is,
F1 = I b B
Similarly, it exerts a force F2 on the arm CD and F2
is directed out of the plane of the paper F2 = I b B = F1
Thus, the net force on the loop is zero There is a
torque on the loop due to the pair of forces F1 and F2 |
1 | 3863-3866 | Its magnitude is,
F1 = I b B
Similarly, it exerts a force F2 on the arm CD and F2
is directed out of the plane of the paper F2 = I b B = F1
Thus, the net force on the loop is zero There is a
torque on the loop due to the pair of forces F1 and F2 Figure 4 |
1 | 3864-3867 | F2 = I b B = F1
Thus, the net force on the loop is zero There is a
torque on the loop due to the pair of forces F1 and F2 Figure 4 18(b) shows a view of the loop from the AD
end |
1 | 3865-3868 | There is a
torque on the loop due to the pair of forces F1 and F2 Figure 4 18(b) shows a view of the loop from the AD
end It shows that the torque on the loop tends to rotate
it anticlockwise |
1 | 3866-3869 | Figure 4 18(b) shows a view of the loop from the AD
end It shows that the torque on the loop tends to rotate
it anticlockwise This torque is (in magnitude),
1
2
2
2
a
a
F
F
τ =
+
(
)
2
2
a
a
IbB
IbB
I ab B
=
+
=
= I A B
(4 |
1 | 3867-3870 | 18(b) shows a view of the loop from the AD
end It shows that the torque on the loop tends to rotate
it anticlockwise This torque is (in magnitude),
1
2
2
2
a
a
F
F
τ =
+
(
)
2
2
a
a
IbB
IbB
I ab B
=
+
=
= I A B
(4 26)
where A = ab is the area of the rectangle |
1 | 3868-3871 | It shows that the torque on the loop tends to rotate
it anticlockwise This torque is (in magnitude),
1
2
2
2
a
a
F
F
τ =
+
(
)
2
2
a
a
IbB
IbB
I ab B
=
+
=
= I A B
(4 26)
where A = ab is the area of the rectangle We next consider the case when the plane of the loop,
is not along the magnetic field, but makes an angle with
it |
1 | 3869-3872 | This torque is (in magnitude),
1
2
2
2
a
a
F
F
τ =
+
(
)
2
2
a
a
IbB
IbB
I ab B
=
+
=
= I A B
(4 26)
where A = ab is the area of the rectangle We next consider the case when the plane of the loop,
is not along the magnetic field, but makes an angle with
it We take the angle between the field and the normal to
the coil to be angle q (The previous case corresponds to
q = p/2) |
1 | 3870-3873 | 26)
where A = ab is the area of the rectangle We next consider the case when the plane of the loop,
is not along the magnetic field, but makes an angle with
it We take the angle between the field and the normal to
the coil to be angle q (The previous case corresponds to
q = p/2) Figure 4 |
1 | 3871-3874 | We next consider the case when the plane of the loop,
is not along the magnetic field, but makes an angle with
it We take the angle between the field and the normal to
the coil to be angle q (The previous case corresponds to
q = p/2) Figure 4 19 illustrates this general case |
1 | 3872-3875 | We take the angle between the field and the normal to
the coil to be angle q (The previous case corresponds to
q = p/2) Figure 4 19 illustrates this general case The forces on the arms BC and DA are equal, opposite, and act along
the axis of the coil, which connects the centres of mass of BC and DA |
1 | 3873-3876 | Figure 4 19 illustrates this general case The forces on the arms BC and DA are equal, opposite, and act along
the axis of the coil, which connects the centres of mass of BC and DA Being collinear along the axis they cancel each other, resulting in no net
force or torque |
1 | 3874-3877 | 19 illustrates this general case The forces on the arms BC and DA are equal, opposite, and act along
the axis of the coil, which connects the centres of mass of BC and DA Being collinear along the axis they cancel each other, resulting in no net
force or torque The forces on arms AB and CD are F1 and F2 |
1 | 3875-3878 | The forces on the arms BC and DA are equal, opposite, and act along
the axis of the coil, which connects the centres of mass of BC and DA Being collinear along the axis they cancel each other, resulting in no net
force or torque The forces on arms AB and CD are F1 and F2 They too
are equal and opposite, with magnitude,
F1 = F2 = I b B
But they are not collinear |
1 | 3876-3879 | Being collinear along the axis they cancel each other, resulting in no net
force or torque The forces on arms AB and CD are F1 and F2 They too
are equal and opposite, with magnitude,
F1 = F2 = I b B
But they are not collinear This results in a couple as before |
1 | 3877-3880 | The forces on arms AB and CD are F1 and F2 They too
are equal and opposite, with magnitude,
F1 = F2 = I b B
But they are not collinear This results in a couple as before The
torque is, however, less than the earlier case when plane of loop was
along the magnetic field |
1 | 3878-3881 | They too
are equal and opposite, with magnitude,
F1 = F2 = I b B
But they are not collinear This results in a couple as before The
torque is, however, less than the earlier case when plane of loop was
along the magnetic field This is because the perpendicular distance
between the forces of the couple has decreased |
1 | 3879-3882 | This results in a couple as before The
torque is, however, less than the earlier case when plane of loop was
along the magnetic field This is because the perpendicular distance
between the forces of the couple has decreased Figure 4 |
1 | 3880-3883 | The
torque is, however, less than the earlier case when plane of loop was
along the magnetic field This is because the perpendicular distance
between the forces of the couple has decreased Figure 4 19(b) is a view
of the arrangement from the AD end and it illustrates these two forces
constituting a couple |
1 | 3881-3884 | This is because the perpendicular distance
between the forces of the couple has decreased Figure 4 19(b) is a view
of the arrangement from the AD end and it illustrates these two forces
constituting a couple The magnitude of the torque on the loop is,
1
2
sin
sin
2
2
a
a
F
F
τ
θ
θ
=
+
= I ab B sin q
= I A B sin q
(4 |
1 | 3882-3885 | Figure 4 19(b) is a view
of the arrangement from the AD end and it illustrates these two forces
constituting a couple The magnitude of the torque on the loop is,
1
2
sin
sin
2
2
a
a
F
F
τ
θ
θ
=
+
= I ab B sin q
= I A B sin q
(4 27)
FIGURE 4 |
1 | 3883-3886 | 19(b) is a view
of the arrangement from the AD end and it illustrates these two forces
constituting a couple The magnitude of the torque on the loop is,
1
2
sin
sin
2
2
a
a
F
F
τ
θ
θ
=
+
= I ab B sin q
= I A B sin q
(4 27)
FIGURE 4 18 (a) A rectangular
current-carrying coil in uniform
magnetic field |
1 | 3884-3887 | The magnitude of the torque on the loop is,
1
2
sin
sin
2
2
a
a
F
F
τ
θ
θ
=
+
= I ab B sin q
= I A B sin q
(4 27)
FIGURE 4 18 (a) A rectangular
current-carrying coil in uniform
magnetic field The magnetic moment
m points downwards |
1 | 3885-3888 | 27)
FIGURE 4 18 (a) A rectangular
current-carrying coil in uniform
magnetic field The magnetic moment
m points downwards The torque ttttt is
along the axis and tends to rotate the
coil anticlockwise |
1 | 3886-3889 | 18 (a) A rectangular
current-carrying coil in uniform
magnetic field The magnetic moment
m points downwards The torque ttttt is
along the axis and tends to rotate the
coil anticlockwise (b) The couple
acting on the coil |
1 | 3887-3890 | The magnetic moment
m points downwards The torque ttttt is
along the axis and tends to rotate the
coil anticlockwise (b) The couple
acting on the coil Rationalised 2023-24
Physics
126
EXAMPLE 4 |
1 | 3888-3891 | The torque ttttt is
along the axis and tends to rotate the
coil anticlockwise (b) The couple
acting on the coil Rationalised 2023-24
Physics
126
EXAMPLE 4 11
As q à 0, the perpendicular distance between
the forces of the couple also approaches zero |
1 | 3889-3892 | (b) The couple
acting on the coil Rationalised 2023-24
Physics
126
EXAMPLE 4 11
As q à 0, the perpendicular distance between
the forces of the couple also approaches zero This
makes the forces collinear and the net force and
torque zero |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.