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1 | 3690-3693 | The
field is directly proportional to the current and inversely
proportional to the distance from the (infinitely long) current source (iv)
There exists a simple rule to determine the direction of the magnetic
field due to a long wire This rule, called the right-hand rule*, is:
Grasp the wire in your right hand with your extended thumb pointing
in the direction of the current Your fingers will curl around in the
direction of the magnetic field |
1 | 3691-3694 | (iv)
There exists a simple rule to determine the direction of the magnetic
field due to a long wire This rule, called the right-hand rule*, is:
Grasp the wire in your right hand with your extended thumb pointing
in the direction of the current Your fingers will curl around in the
direction of the magnetic field Ampere’s circuital law is not new in content from Biot-Savart law |
1 | 3692-3695 | This rule, called the right-hand rule*, is:
Grasp the wire in your right hand with your extended thumb pointing
in the direction of the current Your fingers will curl around in the
direction of the magnetic field Ampere’s circuital law is not new in content from Biot-Savart law Both relate the magnetic field and the current, and both express the same
physical consequences of a steady electrical current |
1 | 3693-3696 | Your fingers will curl around in the
direction of the magnetic field Ampere’s circuital law is not new in content from Biot-Savart law Both relate the magnetic field and the current, and both express the same
physical consequences of a steady electrical current Ampere’s law is to
Biot-Savart law, what Gauss’s law is to Coulomb’s law |
1 | 3694-3697 | Ampere’s circuital law is not new in content from Biot-Savart law Both relate the magnetic field and the current, and both express the same
physical consequences of a steady electrical current Ampere’s law is to
Biot-Savart law, what Gauss’s law is to Coulomb’s law Both, Ampere’s
and Gauss’s law relate a physical quantity on the periphery or boundary
(magnetic or electric field) to another physical quantity, namely, the source,
in the interior (current or charge) |
1 | 3695-3698 | Both relate the magnetic field and the current, and both express the same
physical consequences of a steady electrical current Ampere’s law is to
Biot-Savart law, what Gauss’s law is to Coulomb’s law Both, Ampere’s
and Gauss’s law relate a physical quantity on the periphery or boundary
(magnetic or electric field) to another physical quantity, namely, the source,
in the interior (current or charge) We also note that Ampere’s circuital
law holds for steady currents which do not fluctuate with time |
1 | 3696-3699 | Ampere’s law is to
Biot-Savart law, what Gauss’s law is to Coulomb’s law Both, Ampere’s
and Gauss’s law relate a physical quantity on the periphery or boundary
(magnetic or electric field) to another physical quantity, namely, the source,
in the interior (current or charge) We also note that Ampere’s circuital
law holds for steady currents which do not fluctuate with time The
following example will help us understand what is meant by the term
enclosed current |
1 | 3697-3700 | Both, Ampere’s
and Gauss’s law relate a physical quantity on the periphery or boundary
(magnetic or electric field) to another physical quantity, namely, the source,
in the interior (current or charge) We also note that Ampere’s circuital
law holds for steady currents which do not fluctuate with time The
following example will help us understand what is meant by the term
enclosed current Example 4 |
1 | 3698-3701 | We also note that Ampere’s circuital
law holds for steady currents which do not fluctuate with time The
following example will help us understand what is meant by the term
enclosed current Example 4 8 Figure 4 |
1 | 3699-3702 | The
following example will help us understand what is meant by the term
enclosed current Example 4 8 Figure 4 13 shows a long straight wire of a circular
cross-section (radius a) carrying steady current I |
1 | 3700-3703 | Example 4 8 Figure 4 13 shows a long straight wire of a circular
cross-section (radius a) carrying steady current I The current I is
uniformly distributed across this cross-section |
1 | 3701-3704 | 8 Figure 4 13 shows a long straight wire of a circular
cross-section (radius a) carrying steady current I The current I is
uniformly distributed across this cross-section Calculate the
magnetic field in the region r < a and r > a |
1 | 3702-3705 | 13 shows a long straight wire of a circular
cross-section (radius a) carrying steady current I The current I is
uniformly distributed across this cross-section Calculate the
magnetic field in the region r < a and r > a FIGURE 4 |
1 | 3703-3706 | The current I is
uniformly distributed across this cross-section Calculate the
magnetic field in the region r < a and r > a FIGURE 4 13
*
Note that there are two distinct right-hand rules: One which gives the direction
of B on the axis of current-loop and the other which gives direction of B
for a straight conducting wire |
1 | 3704-3707 | Calculate the
magnetic field in the region r < a and r > a FIGURE 4 13
*
Note that there are two distinct right-hand rules: One which gives the direction
of B on the axis of current-loop and the other which gives direction of B
for a straight conducting wire Fingers and thumb play different roles in
the two |
1 | 3705-3708 | FIGURE 4 13
*
Note that there are two distinct right-hand rules: One which gives the direction
of B on the axis of current-loop and the other which gives direction of B
for a straight conducting wire Fingers and thumb play different roles in
the two Rationalised 2023-24
Physics
120
EXAMPLE 4 |
1 | 3706-3709 | 13
*
Note that there are two distinct right-hand rules: One which gives the direction
of B on the axis of current-loop and the other which gives direction of B
for a straight conducting wire Fingers and thumb play different roles in
the two Rationalised 2023-24
Physics
120
EXAMPLE 4 8
Solution (a) Consider the case r > a |
1 | 3707-3710 | Fingers and thumb play different roles in
the two Rationalised 2023-24
Physics
120
EXAMPLE 4 8
Solution (a) Consider the case r > a The Amperian loop, labelled 2,
is a circle concentric with the cross-section |
1 | 3708-3711 | Rationalised 2023-24
Physics
120
EXAMPLE 4 8
Solution (a) Consider the case r > a The Amperian loop, labelled 2,
is a circle concentric with the cross-section For this loop,
L = 2 p r
Ie = Current enclosed by the loop = I
The result is the familiar expression for a long straight wire
B (2p r) = m0I
π
0
2
I
B
r
=µ
[4 |
1 | 3709-3712 | 8
Solution (a) Consider the case r > a The Amperian loop, labelled 2,
is a circle concentric with the cross-section For this loop,
L = 2 p r
Ie = Current enclosed by the loop = I
The result is the familiar expression for a long straight wire
B (2p r) = m0I
π
0
2
I
B
r
=µ
[4 19(a)]
1
B
∝r
(r > a)
Now the current enclosed Ie is not I, but is less than this value |
1 | 3710-3713 | The Amperian loop, labelled 2,
is a circle concentric with the cross-section For this loop,
L = 2 p r
Ie = Current enclosed by the loop = I
The result is the familiar expression for a long straight wire
B (2p r) = m0I
π
0
2
I
B
r
=µ
[4 19(a)]
1
B
∝r
(r > a)
Now the current enclosed Ie is not I, but is less than this value Since the current distribution is uniform, the current enclosed is,
I
I
ar
e =
π
π
2
2
2
2
aIr
=
Using Ampere’s law,
π
2
0
2
(2
)
I r
B
r
a
µ
=
B
aI
r
=
µ0
2
2�
[4 |
1 | 3711-3714 | For this loop,
L = 2 p r
Ie = Current enclosed by the loop = I
The result is the familiar expression for a long straight wire
B (2p r) = m0I
π
0
2
I
B
r
=µ
[4 19(a)]
1
B
∝r
(r > a)
Now the current enclosed Ie is not I, but is less than this value Since the current distribution is uniform, the current enclosed is,
I
I
ar
e =
π
π
2
2
2
2
aIr
=
Using Ampere’s law,
π
2
0
2
(2
)
I r
B
r
a
µ
=
B
aI
r
=
µ0
2
2�
[4 19(b)]
B µ r (r < a)
FIGURE 4 |
1 | 3712-3715 | 19(a)]
1
B
∝r
(r > a)
Now the current enclosed Ie is not I, but is less than this value Since the current distribution is uniform, the current enclosed is,
I
I
ar
e =
π
π
2
2
2
2
aIr
=
Using Ampere’s law,
π
2
0
2
(2
)
I r
B
r
a
µ
=
B
aI
r
=
µ0
2
2�
[4 19(b)]
B µ r (r < a)
FIGURE 4 14
Figure (4 |
1 | 3713-3716 | Since the current distribution is uniform, the current enclosed is,
I
I
ar
e =
π
π
2
2
2
2
aIr
=
Using Ampere’s law,
π
2
0
2
(2
)
I r
B
r
a
µ
=
B
aI
r
=
µ0
2
2�
[4 19(b)]
B µ r (r < a)
FIGURE 4 14
Figure (4 14) shows a plot of the magnitude of B with distance r
from the centre of the wire |
1 | 3714-3717 | 19(b)]
B µ r (r < a)
FIGURE 4 14
Figure (4 14) shows a plot of the magnitude of B with distance r
from the centre of the wire The direction of the field is tangential to
the respective circular loop (1 or 2) and given by the right-hand
rule described earlier in this section |
1 | 3715-3718 | 14
Figure (4 14) shows a plot of the magnitude of B with distance r
from the centre of the wire The direction of the field is tangential to
the respective circular loop (1 or 2) and given by the right-hand
rule described earlier in this section This example possesses the required symmetry so that Ampere’s
law can be applied readily |
1 | 3716-3719 | 14) shows a plot of the magnitude of B with distance r
from the centre of the wire The direction of the field is tangential to
the respective circular loop (1 or 2) and given by the right-hand
rule described earlier in this section This example possesses the required symmetry so that Ampere’s
law can be applied readily It should be noted that while Ampere’s circuital law holds for any
loop, it may not always facilitate an evaluation of the magnetic field in
every case |
1 | 3717-3720 | The direction of the field is tangential to
the respective circular loop (1 or 2) and given by the right-hand
rule described earlier in this section This example possesses the required symmetry so that Ampere’s
law can be applied readily It should be noted that while Ampere’s circuital law holds for any
loop, it may not always facilitate an evaluation of the magnetic field in
every case For example, for the case of the circular loop discussed in
Section 4 |
1 | 3718-3721 | This example possesses the required symmetry so that Ampere’s
law can be applied readily It should be noted that while Ampere’s circuital law holds for any
loop, it may not always facilitate an evaluation of the magnetic field in
every case For example, for the case of the circular loop discussed in
Section 4 6, it cannot be applied to extract the simple expression
B = m0I/2R [Eq |
1 | 3719-3722 | It should be noted that while Ampere’s circuital law holds for any
loop, it may not always facilitate an evaluation of the magnetic field in
every case For example, for the case of the circular loop discussed in
Section 4 6, it cannot be applied to extract the simple expression
B = m0I/2R [Eq (4 |
1 | 3720-3723 | For example, for the case of the circular loop discussed in
Section 4 6, it cannot be applied to extract the simple expression
B = m0I/2R [Eq (4 16)] for the field at the centre of the loop |
1 | 3721-3724 | 6, it cannot be applied to extract the simple expression
B = m0I/2R [Eq (4 16)] for the field at the centre of the loop However,
there exists a large number of situations of high symmetry where the law
can be conveniently applied |
1 | 3722-3725 | (4 16)] for the field at the centre of the loop However,
there exists a large number of situations of high symmetry where the law
can be conveniently applied We shall use it in the next section to calculate
p
Rationalised 2023-24
121
Moving Charges and
Magnetism
the magnetic field produced by two commonly used and very useful
magnetic systems: the solenoid and the toroid |
1 | 3723-3726 | 16)] for the field at the centre of the loop However,
there exists a large number of situations of high symmetry where the law
can be conveniently applied We shall use it in the next section to calculate
p
Rationalised 2023-24
121
Moving Charges and
Magnetism
the magnetic field produced by two commonly used and very useful
magnetic systems: the solenoid and the toroid 4 |
1 | 3724-3727 | However,
there exists a large number of situations of high symmetry where the law
can be conveniently applied We shall use it in the next section to calculate
p
Rationalised 2023-24
121
Moving Charges and
Magnetism
the magnetic field produced by two commonly used and very useful
magnetic systems: the solenoid and the toroid 4 7 THE SOLENOID
We shall discuss a long solenoid |
1 | 3725-3728 | We shall use it in the next section to calculate
p
Rationalised 2023-24
121
Moving Charges and
Magnetism
the magnetic field produced by two commonly used and very useful
magnetic systems: the solenoid and the toroid 4 7 THE SOLENOID
We shall discuss a long solenoid By long solenoid we mean that the
solenoid’s length is large compared to its radius |
1 | 3726-3729 | 4 7 THE SOLENOID
We shall discuss a long solenoid By long solenoid we mean that the
solenoid’s length is large compared to its radius It consists of a long wire
wound in the form of a helix where the neighbouring turns are closely
spaced |
1 | 3727-3730 | 7 THE SOLENOID
We shall discuss a long solenoid By long solenoid we mean that the
solenoid’s length is large compared to its radius It consists of a long wire
wound in the form of a helix where the neighbouring turns are closely
spaced So each turn can be regarded as a circular loop |
1 | 3728-3731 | By long solenoid we mean that the
solenoid’s length is large compared to its radius It consists of a long wire
wound in the form of a helix where the neighbouring turns are closely
spaced So each turn can be regarded as a circular loop The net magnetic
field is the vector sum of the fields due to all the turns |
1 | 3729-3732 | It consists of a long wire
wound in the form of a helix where the neighbouring turns are closely
spaced So each turn can be regarded as a circular loop The net magnetic
field is the vector sum of the fields due to all the turns Enamelled wires
are used for winding so that turns are insulated from each other |
1 | 3730-3733 | So each turn can be regarded as a circular loop The net magnetic
field is the vector sum of the fields due to all the turns Enamelled wires
are used for winding so that turns are insulated from each other Figure 4 |
1 | 3731-3734 | The net magnetic
field is the vector sum of the fields due to all the turns Enamelled wires
are used for winding so that turns are insulated from each other Figure 4 15 displays the magnetic field lines for a finite solenoid |
1 | 3732-3735 | Enamelled wires
are used for winding so that turns are insulated from each other Figure 4 15 displays the magnetic field lines for a finite solenoid We
show a section of this solenoid in an enlarged manner in Fig |
1 | 3733-3736 | Figure 4 15 displays the magnetic field lines for a finite solenoid We
show a section of this solenoid in an enlarged manner in Fig 4 |
1 | 3734-3737 | 15 displays the magnetic field lines for a finite solenoid We
show a section of this solenoid in an enlarged manner in Fig 4 15(a) |
1 | 3735-3738 | We
show a section of this solenoid in an enlarged manner in Fig 4 15(a) Figure 4 |
1 | 3736-3739 | 4 15(a) Figure 4 15(b) shows the entire finite solenoid with its magnetic field |
1 | 3737-3740 | 15(a) Figure 4 15(b) shows the entire finite solenoid with its magnetic field In
Fig |
1 | 3738-3741 | Figure 4 15(b) shows the entire finite solenoid with its magnetic field In
Fig 4 |
1 | 3739-3742 | 15(b) shows the entire finite solenoid with its magnetic field In
Fig 4 15(a), it is clear from the circular loops that the field between two
neighbouring turns vanishes |
1 | 3740-3743 | In
Fig 4 15(a), it is clear from the circular loops that the field between two
neighbouring turns vanishes In Fig |
1 | 3741-3744 | 4 15(a), it is clear from the circular loops that the field between two
neighbouring turns vanishes In Fig 4 |
1 | 3742-3745 | 15(a), it is clear from the circular loops that the field between two
neighbouring turns vanishes In Fig 4 15(b), we see that the field at the
interior mid-point P is uniform, strong and along the axis of the solenoid |
1 | 3743-3746 | In Fig 4 15(b), we see that the field at the
interior mid-point P is uniform, strong and along the axis of the solenoid The field at the exterior mid-point Q is weak and moreover is along the
axis of the solenoid with no perpendicular or normal component |
1 | 3744-3747 | 4 15(b), we see that the field at the
interior mid-point P is uniform, strong and along the axis of the solenoid The field at the exterior mid-point Q is weak and moreover is along the
axis of the solenoid with no perpendicular or normal component As the
FIGURE 4 |
1 | 3745-3748 | 15(b), we see that the field at the
interior mid-point P is uniform, strong and along the axis of the solenoid The field at the exterior mid-point Q is weak and moreover is along the
axis of the solenoid with no perpendicular or normal component As the
FIGURE 4 15 (a) The magnetic field due to a section of the solenoid which has been
stretched out for clarity |
1 | 3746-3749 | The field at the exterior mid-point Q is weak and moreover is along the
axis of the solenoid with no perpendicular or normal component As the
FIGURE 4 15 (a) The magnetic field due to a section of the solenoid which has been
stretched out for clarity Only the exterior semi-circular part is shown |
1 | 3747-3750 | As the
FIGURE 4 15 (a) The magnetic field due to a section of the solenoid which has been
stretched out for clarity Only the exterior semi-circular part is shown Notice
how the circular loops between neighbouring turns tend to cancel |
1 | 3748-3751 | 15 (a) The magnetic field due to a section of the solenoid which has been
stretched out for clarity Only the exterior semi-circular part is shown Notice
how the circular loops between neighbouring turns tend to cancel (b) The magnetic field of a finite solenoid |
1 | 3749-3752 | Only the exterior semi-circular part is shown Notice
how the circular loops between neighbouring turns tend to cancel (b) The magnetic field of a finite solenoid FIGURE 4 |
1 | 3750-3753 | Notice
how the circular loops between neighbouring turns tend to cancel (b) The magnetic field of a finite solenoid FIGURE 4 16 The magnetic field of a very long solenoid |
1 | 3751-3754 | (b) The magnetic field of a finite solenoid FIGURE 4 16 The magnetic field of a very long solenoid We consider a
rectangular Amperian loop abcd to determine the field |
1 | 3752-3755 | FIGURE 4 16 The magnetic field of a very long solenoid We consider a
rectangular Amperian loop abcd to determine the field Rationalised 2023-24
Physics
122
EXAMPLE 4 |
1 | 3753-3756 | 16 The magnetic field of a very long solenoid We consider a
rectangular Amperian loop abcd to determine the field Rationalised 2023-24
Physics
122
EXAMPLE 4 9
solenoid is made longer it appears like a long cylindrical metal sheet |
1 | 3754-3757 | We consider a
rectangular Amperian loop abcd to determine the field Rationalised 2023-24
Physics
122
EXAMPLE 4 9
solenoid is made longer it appears like a long cylindrical metal sheet Figure 4 |
1 | 3755-3758 | Rationalised 2023-24
Physics
122
EXAMPLE 4 9
solenoid is made longer it appears like a long cylindrical metal sheet Figure 4 16 represents this idealised picture |
1 | 3756-3759 | 9
solenoid is made longer it appears like a long cylindrical metal sheet Figure 4 16 represents this idealised picture The field outside the solenoid
approaches zero |
1 | 3757-3760 | Figure 4 16 represents this idealised picture The field outside the solenoid
approaches zero We shall assume that the field outside is zero |
1 | 3758-3761 | 16 represents this idealised picture The field outside the solenoid
approaches zero We shall assume that the field outside is zero The field
inside becomes everywhere parallel to the axis |
1 | 3759-3762 | The field outside the solenoid
approaches zero We shall assume that the field outside is zero The field
inside becomes everywhere parallel to the axis Consider a rectangular Amperian loop abcd |
1 | 3760-3763 | We shall assume that the field outside is zero The field
inside becomes everywhere parallel to the axis Consider a rectangular Amperian loop abcd Along cd the field is zero
as argued above |
1 | 3761-3764 | The field
inside becomes everywhere parallel to the axis Consider a rectangular Amperian loop abcd Along cd the field is zero
as argued above Along transverse sections bc and ad, the field component
is zero |
1 | 3762-3765 | Consider a rectangular Amperian loop abcd Along cd the field is zero
as argued above Along transverse sections bc and ad, the field component
is zero Thus, these two sections make no contribution |
1 | 3763-3766 | Along cd the field is zero
as argued above Along transverse sections bc and ad, the field component
is zero Thus, these two sections make no contribution Let the field along
ab be B |
1 | 3764-3767 | Along transverse sections bc and ad, the field component
is zero Thus, these two sections make no contribution Let the field along
ab be B Thus, the relevant length of the Amperian loop is, L = h |
1 | 3765-3768 | Thus, these two sections make no contribution Let the field along
ab be B Thus, the relevant length of the Amperian loop is, L = h Let n be the number of turns per unit length, then the total number
of turns is nh |
1 | 3766-3769 | Let the field along
ab be B Thus, the relevant length of the Amperian loop is, L = h Let n be the number of turns per unit length, then the total number
of turns is nh The enclosed current is, Ie = I (n h), where I is the current
in the solenoid |
1 | 3767-3770 | Thus, the relevant length of the Amperian loop is, L = h Let n be the number of turns per unit length, then the total number
of turns is nh The enclosed current is, Ie = I (n h), where I is the current
in the solenoid From Ampere’s circuital law [Eq |
1 | 3768-3771 | Let n be the number of turns per unit length, then the total number
of turns is nh The enclosed current is, Ie = I (n h), where I is the current
in the solenoid From Ampere’s circuital law [Eq 4 |
1 | 3769-3772 | The enclosed current is, Ie = I (n h), where I is the current
in the solenoid From Ampere’s circuital law [Eq 4 17 (b)]
BL = m0Ie, B h = m0I (n h)
B = m0 n I
(4 |
1 | 3770-3773 | From Ampere’s circuital law [Eq 4 17 (b)]
BL = m0Ie, B h = m0I (n h)
B = m0 n I
(4 20)
The direction of the field is given by the right-hand rule |
1 | 3771-3774 | 4 17 (b)]
BL = m0Ie, B h = m0I (n h)
B = m0 n I
(4 20)
The direction of the field is given by the right-hand rule The solenoid
is commonly used to obtain a uniform magnetic field |
1 | 3772-3775 | 17 (b)]
BL = m0Ie, B h = m0I (n h)
B = m0 n I
(4 20)
The direction of the field is given by the right-hand rule The solenoid
is commonly used to obtain a uniform magnetic field We shall see in the
next chapter that a large field is possible by inserting a soft iron core
inside the solenoid |
1 | 3773-3776 | 20)
The direction of the field is given by the right-hand rule The solenoid
is commonly used to obtain a uniform magnetic field We shall see in the
next chapter that a large field is possible by inserting a soft iron core
inside the solenoid Example 4 |
1 | 3774-3777 | The solenoid
is commonly used to obtain a uniform magnetic field We shall see in the
next chapter that a large field is possible by inserting a soft iron core
inside the solenoid Example 4 9 A solenoid of length 0 |
1 | 3775-3778 | We shall see in the
next chapter that a large field is possible by inserting a soft iron core
inside the solenoid Example 4 9 A solenoid of length 0 5 m has a radius of 1 cm and is
made up of 500 turns |
1 | 3776-3779 | Example 4 9 A solenoid of length 0 5 m has a radius of 1 cm and is
made up of 500 turns It carries a current of 5 A |
1 | 3777-3780 | 9 A solenoid of length 0 5 m has a radius of 1 cm and is
made up of 500 turns It carries a current of 5 A What is the
magnitude of the magnetic field inside the solenoid |
1 | 3778-3781 | 5 m has a radius of 1 cm and is
made up of 500 turns It carries a current of 5 A What is the
magnitude of the magnetic field inside the solenoid Solution The number of turns per unit length is,
500
1000
n =0 |
1 | 3779-3782 | It carries a current of 5 A What is the
magnitude of the magnetic field inside the solenoid Solution The number of turns per unit length is,
500
1000
n =0 5
=
turns/m
The length l = 0 |
1 | 3780-3783 | What is the
magnitude of the magnetic field inside the solenoid Solution The number of turns per unit length is,
500
1000
n =0 5
=
turns/m
The length l = 0 5 m and radius r = 0 |
1 | 3781-3784 | Solution The number of turns per unit length is,
500
1000
n =0 5
=
turns/m
The length l = 0 5 m and radius r = 0 01 m |
1 | 3782-3785 | 5
=
turns/m
The length l = 0 5 m and radius r = 0 01 m Thus, l/a = 50 i |
1 | 3783-3786 | 5 m and radius r = 0 01 m Thus, l/a = 50 i e |
1 | 3784-3787 | 01 m Thus, l/a = 50 i e , l >> a |
1 | 3785-3788 | Thus, l/a = 50 i e , l >> a Hence, we can use the long solenoid formula, namely, Eq |
1 | 3786-3789 | e , l >> a Hence, we can use the long solenoid formula, namely, Eq (4 |
1 | 3787-3790 | , l >> a Hence, we can use the long solenoid formula, namely, Eq (4 20)
B = m0n I
= 4p × 10–7 × 103 × 5
= 6 |
1 | 3788-3791 | Hence, we can use the long solenoid formula, namely, Eq (4 20)
B = m0n I
= 4p × 10–7 × 103 × 5
= 6 28 × 10–3 T
FIGURE 4 |
1 | 3789-3792 | (4 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 |
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