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1 | 3890-3893 | 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 The torques in Eqs |
1 | 3891-3894 | 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 The torques in Eqs (4 |
1 | 3892-3895 | This
makes the forces collinear and the net force and
torque zero The torques in Eqs (4 26) and (4 |
1 | 3893-3896 | The torques in Eqs (4 26) and (4 27)
can be expressed as vector product of the magnetic
moment of the coil and the magnetic field |
1 | 3894-3897 | (4 26) and (4 27)
can be expressed as vector product of the magnetic
moment of the coil and the magnetic field We define the magnetic moment of the current
loop as,
m = I A
(4 |
1 | 3895-3898 | 26) and (4 27)
can be expressed as vector product of the magnetic
moment of the coil and the magnetic field We define the magnetic moment of the current
loop as,
m = I A
(4 28)
where the direction of the area vector A is given by
the right-hand thumb rule and is directed into
the plane of the paper in Fig |
1 | 3896-3899 | 27)
can be expressed as vector product of the magnetic
moment of the coil and the magnetic field We define the magnetic moment of the current
loop as,
m = I A
(4 28)
where the direction of the area vector A is given by
the right-hand thumb rule and is directed into
the plane of the paper in Fig 4 |
1 | 3897-3900 | We define the magnetic moment of the current
loop as,
m = I A
(4 28)
where the direction of the area vector A is given by
the right-hand thumb rule and is directed into
the plane of the paper in Fig 4 18 |
1 | 3898-3901 | 28)
where the direction of the area vector A is given by
the right-hand thumb rule and is directed into
the plane of the paper in Fig 4 18 Then as the
angle between m and B is q , Eqs |
1 | 3899-3902 | 4 18 Then as the
angle between m and B is q , Eqs (4 |
1 | 3900-3903 | 18 Then as the
angle between m and B is q , Eqs (4 26) and (4 |
1 | 3901-3904 | Then as the
angle between m and B is q , Eqs (4 26) and (4 27)
can be expressed by one expression
(4 |
1 | 3902-3905 | (4 26) and (4 27)
can be expressed by one expression
(4 29)
This is analogous to the electrostatic case
(Electric dipole of dipole moment pe in an electric
field E) |
1 | 3903-3906 | 26) and (4 27)
can be expressed by one expression
(4 29)
This is analogous to the electrostatic case
(Electric dipole of dipole moment pe in an electric
field E) ττ
= p××
E
e
As is clear from Eq |
1 | 3904-3907 | 27)
can be expressed by one expression
(4 29)
This is analogous to the electrostatic case
(Electric dipole of dipole moment pe in an electric
field E) ττ
= p××
E
e
As is clear from Eq (4 |
1 | 3905-3908 | 29)
This is analogous to the electrostatic case
(Electric dipole of dipole moment pe in an electric
field E) ττ
= p××
E
e
As is clear from Eq (4 28), the dimensions of the
magnetic moment are [A][L2] and its unit is Am2 |
1 | 3906-3909 | ττ
= p××
E
e
As is clear from Eq (4 28), the dimensions of the
magnetic moment are [A][L2] and its unit is Am2 From Eq |
1 | 3907-3910 | (4 28), the dimensions of the
magnetic moment are [A][L2] and its unit is Am2 From Eq (4 |
1 | 3908-3911 | 28), the dimensions of the
magnetic moment are [A][L2] and its unit is Am2 From Eq (4 29), we see that the torque ttttt
vanishes when m is either parallel or antiparallel
to the magnetic field B |
1 | 3909-3912 | From Eq (4 29), we see that the torque ttttt
vanishes when m is either parallel or antiparallel
to the magnetic field B This indicates a state of
equilibrium as there is no torque on the coil (this
also applies to any object with a magnetic moment
m) |
1 | 3910-3913 | (4 29), we see that the torque ttttt
vanishes when m is either parallel or antiparallel
to the magnetic field B This indicates a state of
equilibrium as there is no torque on the coil (this
also applies to any object with a magnetic moment
m) When m and B are parallel the equilibrium is
a stable one |
1 | 3911-3914 | 29), we see that the torque ttttt
vanishes when m is either parallel or antiparallel
to the magnetic field B This indicates a state of
equilibrium as there is no torque on the coil (this
also applies to any object with a magnetic moment
m) When m and B are parallel the equilibrium is
a stable one Any small rotation of the coil
produces a torque which brings it back to its original position |
1 | 3912-3915 | This indicates a state of
equilibrium as there is no torque on the coil (this
also applies to any object with a magnetic moment
m) When m and B are parallel the equilibrium is
a stable one Any small rotation of the coil
produces a torque which brings it back to its original position When
they are antiparallel, the equilibrium is unstable as any rotation produces
a torque which increases with the amount of rotation |
1 | 3913-3916 | When m and B are parallel the equilibrium is
a stable one Any small rotation of the coil
produces a torque which brings it back to its original position When
they are antiparallel, the equilibrium is unstable as any rotation produces
a torque which increases with the amount of rotation The presence of
this torque is also the reason why a small magnet or any magnetic dipole
aligns itself with the external magnetic field |
1 | 3914-3917 | Any small rotation of the coil
produces a torque which brings it back to its original position When
they are antiparallel, the equilibrium is unstable as any rotation produces
a torque which increases with the amount of rotation The presence of
this torque is also the reason why a small magnet or any magnetic dipole
aligns itself with the external magnetic field If the loop has N closely wound turns, the expression for torque, Eq |
1 | 3915-3918 | When
they are antiparallel, the equilibrium is unstable as any rotation produces
a torque which increases with the amount of rotation The presence of
this torque is also the reason why a small magnet or any magnetic dipole
aligns itself with the external magnetic field If the loop has N closely wound turns, the expression for torque, Eq (4 |
1 | 3916-3919 | The presence of
this torque is also the reason why a small magnet or any magnetic dipole
aligns itself with the external magnetic field If the loop has N closely wound turns, the expression for torque, Eq (4 29), still holds, with
m = N I A
(4 |
1 | 3917-3920 | If the loop has N closely wound turns, the expression for torque, Eq (4 29), still holds, with
m = N I A
(4 30)
Example 4 |
1 | 3918-3921 | (4 29), still holds, with
m = N I A
(4 30)
Example 4 11 A 100 turn closely wound circular coil of radius 10 cm
carries a current of 3 |
1 | 3919-3922 | 29), still holds, with
m = N I A
(4 30)
Example 4 11 A 100 turn closely wound circular coil of radius 10 cm
carries a current of 3 2 A |
1 | 3920-3923 | 30)
Example 4 11 A 100 turn closely wound circular coil of radius 10 cm
carries a current of 3 2 A (a) What is the field at the centre of the
coil |
1 | 3921-3924 | 11 A 100 turn closely wound circular coil of radius 10 cm
carries a current of 3 2 A (a) What is the field at the centre of the
coil (b) What is the magnetic moment of this coil |
1 | 3922-3925 | 2 A (a) What is the field at the centre of the
coil (b) What is the magnetic moment of this coil The coil is placed in a vertical plane and is free to rotate about a
horizontal axis which coincides with its diameter |
1 | 3923-3926 | (a) What is the field at the centre of the
coil (b) What is the magnetic moment of this coil The coil is placed in a vertical plane and is free to rotate about a
horizontal axis which coincides with its diameter A uniform magnetic
field of 2T in the horizontal direction exists such that initially the axis
of the coil is in the direction of the field |
1 | 3924-3927 | (b) What is the magnetic moment of this coil The coil is placed in a vertical plane and is free to rotate about a
horizontal axis which coincides with its diameter A uniform magnetic
field of 2T in the horizontal direction exists such that initially the axis
of the coil is in the direction of the field The coil rotates through an
angle of 90° under the influence of the magnetic field |
1 | 3925-3928 | The coil is placed in a vertical plane and is free to rotate about a
horizontal axis which coincides with its diameter A uniform magnetic
field of 2T in the horizontal direction exists such that initially the axis
of the coil is in the direction of the field The coil rotates through an
angle of 90° under the influence of the magnetic field (c) What are the
magnitudes of the torques on the coil in the initial and final position |
1 | 3926-3929 | A uniform magnetic
field of 2T in the horizontal direction exists such that initially the axis
of the coil is in the direction of the field The coil rotates through an
angle of 90° under the influence of the magnetic field (c) What are the
magnitudes of the torques on the coil in the initial and final position (d) What is the angular speed acquired by the coil when it has rotated
by 90° |
1 | 3927-3930 | The coil rotates through an
angle of 90° under the influence of the magnetic field (c) What are the
magnitudes of the torques on the coil in the initial and final position (d) What is the angular speed acquired by the coil when it has rotated
by 90° The moment of inertia of the coil is 0 |
1 | 3928-3931 | (c) What are the
magnitudes of the torques on the coil in the initial and final position (d) What is the angular speed acquired by the coil when it has rotated
by 90° The moment of inertia of the coil is 0 1 kg m2 |
1 | 3929-3932 | (d) What is the angular speed acquired by the coil when it has rotated
by 90° The moment of inertia of the coil is 0 1 kg m2 FIGURE 4 |
1 | 3930-3933 | The moment of inertia of the coil is 0 1 kg m2 FIGURE 4 19 (a) The area vector of the loop
ABCD makes an arbitrary angle q with
the magnetic field |
1 | 3931-3934 | 1 kg m2 FIGURE 4 19 (a) The area vector of the loop
ABCD makes an arbitrary angle q with
the magnetic field (b) Top view of
the loop |
1 | 3932-3935 | FIGURE 4 19 (a) The area vector of the loop
ABCD makes an arbitrary angle q with
the magnetic field (b) Top view of
the loop The forces F1 and F2 acting
on the arms AB and CD
are indicated |
1 | 3933-3936 | 19 (a) The area vector of the loop
ABCD makes an arbitrary angle q with
the magnetic field (b) Top view of
the loop The forces F1 and F2 acting
on the arms AB and CD
are indicated Rationalised 2023-24
127
Moving Charges and
Magnetism
EXAMPLE 4 |
1 | 3934-3937 | (b) Top view of
the loop The forces F1 and F2 acting
on the arms AB and CD
are indicated Rationalised 2023-24
127
Moving Charges and
Magnetism
EXAMPLE 4 12
EXAMPLE 4 |
1 | 3935-3938 | The forces F1 and F2 acting
on the arms AB and CD
are indicated Rationalised 2023-24
127
Moving Charges and
Magnetism
EXAMPLE 4 12
EXAMPLE 4 11
Solution
(a) From Eq |
1 | 3936-3939 | Rationalised 2023-24
127
Moving Charges and
Magnetism
EXAMPLE 4 12
EXAMPLE 4 11
Solution
(a) From Eq (4 |
1 | 3937-3940 | 12
EXAMPLE 4 11
Solution
(a) From Eq (4 16)
B
= µ0RNI
2
Here, N = 100; I = 3 |
1 | 3938-3941 | 11
Solution
(a) From Eq (4 16)
B
= µ0RNI
2
Here, N = 100; I = 3 2 A, and R = 0 |
1 | 3939-3942 | (4 16)
B
= µ0RNI
2
Here, N = 100; I = 3 2 A, and R = 0 1 m |
1 | 3940-3943 | 16)
B
= µ0RNI
2
Here, N = 100; I = 3 2 A, and R = 0 1 m Hence,
=
×
×
×
−
−
4
10
10
2
10
5
1
(using p ´ 3 |
1 | 3941-3944 | 2 A, and R = 0 1 m Hence,
=
×
×
×
−
−
4
10
10
2
10
5
1
(using p ´ 3 2 = 10)
= 2 × 10–3 T
The direction is given by the right-hand thumb rule |
1 | 3942-3945 | 1 m Hence,
=
×
×
×
−
−
4
10
10
2
10
5
1
(using p ´ 3 2 = 10)
= 2 × 10–3 T
The direction is given by the right-hand thumb rule (b) The magnetic moment is given by Eq |
1 | 3943-3946 | Hence,
=
×
×
×
−
−
4
10
10
2
10
5
1
(using p ´ 3 2 = 10)
= 2 × 10–3 T
The direction is given by the right-hand thumb rule (b) The magnetic moment is given by Eq (4 |
1 | 3944-3947 | 2 = 10)
= 2 × 10–3 T
The direction is given by the right-hand thumb rule (b) The magnetic moment is given by Eq (4 30),
m = N I A = N I p r2 = 100 × 3 |
1 | 3945-3948 | (b) The magnetic moment is given by Eq (4 30),
m = N I A = N I p r2 = 100 × 3 2 × 3 |
1 | 3946-3949 | (4 30),
m = N I A = N I p r2 = 100 × 3 2 × 3 14 × 10–2 = 10 A m2
The direction is once again given by the right-hand thumb rule |
1 | 3947-3950 | 30),
m = N I A = N I p r2 = 100 × 3 2 × 3 14 × 10–2 = 10 A m2
The direction is once again given by the right-hand thumb rule (c) t = m × B [from Eq |
1 | 3948-3951 | 2 × 3 14 × 10–2 = 10 A m2
The direction is once again given by the right-hand thumb rule (c) t = m × B [from Eq (4 |
1 | 3949-3952 | 14 × 10–2 = 10 A m2
The direction is once again given by the right-hand thumb rule (c) t = m × B [from Eq (4 29)]
= m B sin q
Initially, q = 0 |
1 | 3950-3953 | (c) t = m × B [from Eq (4 29)]
= m B sin q
Initially, q = 0 Thus, initial torque ti = 0 |
1 | 3951-3954 | (4 29)]
= m B sin q
Initially, q = 0 Thus, initial torque ti = 0 Finally, q = p/2 (or 90º) |
1 | 3952-3955 | 29)]
= m B sin q
Initially, q = 0 Thus, initial torque ti = 0 Finally, q = p/2 (or 90º) Thus, final torque tf = m B = 10 ´ 2 = 20 N m |
1 | 3953-3956 | Thus, initial torque ti = 0 Finally, q = p/2 (or 90º) Thus, final torque tf = m B = 10 ´ 2 = 20 N m (d) From Newton’s second law,
I
where I is the moment of inertia of the coil |
1 | 3954-3957 | Finally, q = p/2 (or 90º) Thus, final torque tf = m B = 10 ´ 2 = 20 N m (d) From Newton’s second law,
I
where I is the moment of inertia of the coil From chain rule,
d
d
d
d
d
d
d
d
t
t
Using this,
I
d
sin
d
m B
Integrating from q = 0 to q = p/2,
Example 4 |
1 | 3955-3958 | Thus, final torque tf = m B = 10 ´ 2 = 20 N m (d) From Newton’s second law,
I
where I is the moment of inertia of the coil From chain rule,
d
d
d
d
d
d
d
d
t
t
Using this,
I
d
sin
d
m B
Integrating from q = 0 to q = p/2,
Example 4 12
(a) A current-carrying circular loop lies on a smooth horizontal plane |
1 | 3956-3959 | (d) From Newton’s second law,
I
where I is the moment of inertia of the coil From chain rule,
d
d
d
d
d
d
d
d
t
t
Using this,
I
d
sin
d
m B
Integrating from q = 0 to q = p/2,
Example 4 12
(a) A current-carrying circular loop lies on a smooth horizontal plane Can a uniform magnetic field be set up in such a manner that
the loop turns around itself (i |
1 | 3957-3960 | From chain rule,
d
d
d
d
d
d
d
d
t
t
Using this,
I
d
sin
d
m B
Integrating from q = 0 to q = p/2,
Example 4 12
(a) A current-carrying circular loop lies on a smooth horizontal plane Can a uniform magnetic field be set up in such a manner that
the loop turns around itself (i e |
1 | 3958-3961 | 12
(a) A current-carrying circular loop lies on a smooth horizontal plane Can a uniform magnetic field be set up in such a manner that
the loop turns around itself (i e , turns about the vertical axis) |
1 | 3959-3962 | Can a uniform magnetic field be set up in such a manner that
the loop turns around itself (i e , turns about the vertical axis) (b) A current-carrying circular loop is located in a uniform external
magnetic field |
1 | 3960-3963 | e , turns about the vertical axis) (b) A current-carrying circular loop is located in a uniform external
magnetic field If the loop is
free to turn, what is its orientation
of stable equilibrium |
1 | 3961-3964 | , turns about the vertical axis) (b) A current-carrying circular loop is located in a uniform external
magnetic field If the loop is
free to turn, what is its orientation
of stable equilibrium Show that in this orientation, the flux of
Rationalised 2023-24
Physics
128
EXAMPLE 4 |
1 | 3962-3965 | (b) A current-carrying circular loop is located in a uniform external
magnetic field If the loop is
free to turn, what is its orientation
of stable equilibrium Show that in this orientation, the flux of
Rationalised 2023-24
Physics
128
EXAMPLE 4 12
the total field (external field + field produced by the loop) is
maximum |
1 | 3963-3966 | If the loop is
free to turn, what is its orientation
of stable equilibrium Show that in this orientation, the flux of
Rationalised 2023-24
Physics
128
EXAMPLE 4 12
the total field (external field + field produced by the loop) is
maximum (c) A loop of irregular shape carrying current is located in an external
magnetic field |
1 | 3964-3967 | Show that in this orientation, the flux of
Rationalised 2023-24
Physics
128
EXAMPLE 4 12
the total field (external field + field produced by the loop) is
maximum (c) A loop of irregular shape carrying current is located in an external
magnetic field If the wire is flexible, why does it change to a
circular shape |
1 | 3965-3968 | 12
the total field (external field + field produced by the loop) is
maximum (c) A loop of irregular shape carrying current is located in an external
magnetic field If the wire is flexible, why does it change to a
circular shape Solution
(a) No, because that would require ttttt to be in the vertical direction |
1 | 3966-3969 | (c) A loop of irregular shape carrying current is located in an external
magnetic field If the wire is flexible, why does it change to a
circular shape Solution
(a) No, because that would require ttttt to be in the vertical direction But ttttt = I A × B, and since A of the horizontal loop is in the vertical
direction, t would be in the plane of the loop for any B |
1 | 3967-3970 | If the wire is flexible, why does it change to a
circular shape Solution
(a) No, because that would require ttttt to be in the vertical direction But ttttt = I A × B, and since A of the horizontal loop is in the vertical
direction, t would be in the plane of the loop for any B (b) Orientation of stable equilibrium is one where the area vector A
of the loop is in the direction of external magnetic field |
1 | 3968-3971 | Solution
(a) No, because that would require ttttt to be in the vertical direction But ttttt = I A × B, and since A of the horizontal loop is in the vertical
direction, t would be in the plane of the loop for any B (b) Orientation of stable equilibrium is one where the area vector A
of the loop is in the direction of external magnetic field In this
orientation, the magnetic field produced by the loop is in the same
direction as external field, both normal to the plane of the loop,
thus giving rise to maximum flux of the total field |
1 | 3969-3972 | But ttttt = I A × B, and since A of the horizontal loop is in the vertical
direction, t would be in the plane of the loop for any B (b) Orientation of stable equilibrium is one where the area vector A
of the loop is in the direction of external magnetic field In this
orientation, the magnetic field produced by the loop is in the same
direction as external field, both normal to the plane of the loop,
thus giving rise to maximum flux of the total field (c) It assumes circular shape with its plane normal to the field to
maximise flux, since for a given perimeter, a circle encloses greater
area than any other shape |
1 | 3970-3973 | (b) Orientation of stable equilibrium is one where the area vector A
of the loop is in the direction of external magnetic field In this
orientation, the magnetic field produced by the loop is in the same
direction as external field, both normal to the plane of the loop,
thus giving rise to maximum flux of the total field (c) It assumes circular shape with its plane normal to the field to
maximise flux, since for a given perimeter, a circle encloses greater
area than any other shape 4 |
1 | 3971-3974 | In this
orientation, the magnetic field produced by the loop is in the same
direction as external field, both normal to the plane of the loop,
thus giving rise to maximum flux of the total field (c) It assumes circular shape with its plane normal to the field to
maximise flux, since for a given perimeter, a circle encloses greater
area than any other shape 4 9 |
1 | 3972-3975 | (c) It assumes circular shape with its plane normal to the field to
maximise flux, since for a given perimeter, a circle encloses greater
area than any other shape 4 9 2 Circular current loop as a magnetic dipole
In this section, we shall consider the elementary magnetic element: the
current loop |
1 | 3973-3976 | 4 9 2 Circular current loop as a magnetic dipole
In this section, we shall consider the elementary magnetic element: the
current loop We shall show that the magnetic field (at large distances)
due to current in a circular current loop is very similar in behaviour to
the electric field of an electric dipole |
1 | 3974-3977 | 9 2 Circular current loop as a magnetic dipole
In this section, we shall consider the elementary magnetic element: the
current loop We shall show that the magnetic field (at large distances)
due to current in a circular current loop is very similar in behaviour to
the electric field of an electric dipole In Section 4 |
1 | 3975-3978 | 2 Circular current loop as a magnetic dipole
In this section, we shall consider the elementary magnetic element: the
current loop We shall show that the magnetic field (at large distances)
due to current in a circular current loop is very similar in behaviour to
the electric field of an electric dipole In Section 4 6, we have evaluated
the magnetic field on the axis of a circular loop, of a radius R, carrying a
steady current I |
1 | 3976-3979 | We shall show that the magnetic field (at large distances)
due to current in a circular current loop is very similar in behaviour to
the electric field of an electric dipole In Section 4 6, we have evaluated
the magnetic field on the axis of a circular loop, of a radius R, carrying a
steady current I The magnitude of this field is [(Eq |
1 | 3977-3980 | In Section 4 6, we have evaluated
the magnetic field on the axis of a circular loop, of a radius R, carrying a
steady current I The magnitude of this field is [(Eq (4 |
1 | 3978-3981 | 6, we have evaluated
the magnetic field on the axis of a circular loop, of a radius R, carrying a
steady current I The magnitude of this field is [(Eq (4 15)],
(
)
2
0
3/2
2
2
2
µ
=
+
I R
B
x
R
and its direction is along the axis and given by the right-hand thumb
rule (Fig |
1 | 3979-3982 | The magnitude of this field is [(Eq (4 15)],
(
)
2
0
3/2
2
2
2
µ
=
+
I R
B
x
R
and its direction is along the axis and given by the right-hand thumb
rule (Fig 4 |
1 | 3980-3983 | (4 15)],
(
)
2
0
3/2
2
2
2
µ
=
+
I R
B
x
R
and its direction is along the axis and given by the right-hand thumb
rule (Fig 4 12) |
1 | 3981-3984 | 15)],
(
)
2
0
3/2
2
2
2
µ
=
+
I R
B
x
R
and its direction is along the axis and given by the right-hand thumb
rule (Fig 4 12) Here, x is the distance along the axis from the centre of
the loop |
1 | 3982-3985 | 4 12) Here, x is the distance along the axis from the centre of
the loop For x >> R, we may drop the R2 term in the denominator |
1 | 3983-3986 | 12) Here, x is the distance along the axis from the centre of
the loop For x >> R, we may drop the R2 term in the denominator Thus,
2
0
3
2
IR
B
x
µ
=
Note that the area of the loop A = pR2 |
1 | 3984-3987 | Here, x is the distance along the axis from the centre of
the loop For x >> R, we may drop the R2 term in the denominator Thus,
2
0
3
2
IR
B
x
µ
=
Note that the area of the loop A = pR2 Thus,
0
3
2
IA
B
x
=µ
π
As earlier, we define the magnetic moment m to have a magnitude IA,
m = I A |
1 | 3985-3988 | For x >> R, we may drop the R2 term in the denominator Thus,
2
0
3
2
IR
B
x
µ
=
Note that the area of the loop A = pR2 Thus,
0
3
2
IA
B
x
=µ
π
As earlier, we define the magnetic moment m to have a magnitude IA,
m = I A Hence,
B
≃ µ0m
3
2 πx
π
0
3
2
4
x
=µ
m
[4 |
1 | 3986-3989 | Thus,
2
0
3
2
IR
B
x
µ
=
Note that the area of the loop A = pR2 Thus,
0
3
2
IA
B
x
=µ
π
As earlier, we define the magnetic moment m to have a magnitude IA,
m = I A Hence,
B
≃ µ0m
3
2 πx
π
0
3
2
4
x
=µ
m
[4 31(a)]
The expression of Eq |
1 | 3987-3990 | Thus,
0
3
2
IA
B
x
=µ
π
As earlier, we define the magnetic moment m to have a magnitude IA,
m = I A Hence,
B
≃ µ0m
3
2 πx
π
0
3
2
4
x
=µ
m
[4 31(a)]
The expression of Eq [4 |
1 | 3988-3991 | Hence,
B
≃ µ0m
3
2 πx
π
0
3
2
4
x
=µ
m
[4 31(a)]
The expression of Eq [4 31(a)] is very similar to an expression obtained
earlier for the electric field of a dipole |
1 | 3989-3992 | 31(a)]
The expression of Eq [4 31(a)] is very similar to an expression obtained
earlier for the electric field of a dipole The similarity may be seen if we
substitute,
0
1/0
µ
ε
→
Rationalised 2023-24
129
Moving Charges and
Magnetism
e
m→
p (electrostatic dipole)
B→
E (electrostatic field)
We then obtain,
3
0
2
4
e
εx
=
π
p
E
which is precisely the field for an electric dipole at a point on its axis |
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