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1 | 3490-3493 | 5
The magnitude of this field is,
0
2
d sin
d
4
I
l
r
µ
θ
=
π
B
[4 11(b)]
where we have used the property of cross-product Equation [4 11 (a)]
constitutes our basic equation for the magnetic field |
1 | 3491-3494 | 11(b)]
where we have used the property of cross-product Equation [4 11 (a)]
constitutes our basic equation for the magnetic field The proportionality
constant in SI units has the exact value,
7
0
10
Tm/A
µ4
−
π=
[4 |
1 | 3492-3495 | Equation [4 11 (a)]
constitutes our basic equation for the magnetic field The proportionality
constant in SI units has the exact value,
7
0
10
Tm/A
µ4
−
π=
[4 11(c)]
We call m0 the permeability of free space (or vacuum) |
1 | 3493-3496 | 11 (a)]
constitutes our basic equation for the magnetic field The proportionality
constant in SI units has the exact value,
7
0
10
Tm/A
µ4
−
π=
[4 11(c)]
We call m0 the permeability of free space (or vacuum) The Biot-Savart law for the magnetic field has certain similarities, as
well as, differences with the Coulomb’s law for the electrostatic field |
1 | 3494-3497 | The proportionality
constant in SI units has the exact value,
7
0
10
Tm/A
µ4
−
π=
[4 11(c)]
We call m0 the permeability of free space (or vacuum) The Biot-Savart law for the magnetic field has certain similarities, as
well as, differences with the Coulomb’s law for the electrostatic field Some
of these are:
(i)
Both are long range, since both depend inversely on the square of
distance from the source to the point of interest |
1 | 3495-3498 | 11(c)]
We call m0 the permeability of free space (or vacuum) The Biot-Savart law for the magnetic field has certain similarities, as
well as, differences with the Coulomb’s law for the electrostatic field Some
of these are:
(i)
Both are long range, since both depend inversely on the square of
distance from the source to the point of interest The principle of
superposition applies to both fields |
1 | 3496-3499 | The Biot-Savart law for the magnetic field has certain similarities, as
well as, differences with the Coulomb’s law for the electrostatic field Some
of these are:
(i)
Both are long range, since both depend inversely on the square of
distance from the source to the point of interest The principle of
superposition applies to both fields [In this connection, note that
the magnetic field is linear in the source I dl just as the electrostatic
field is linear in its source: the electric charge |
1 | 3497-3500 | Some
of these are:
(i)
Both are long range, since both depend inversely on the square of
distance from the source to the point of interest The principle of
superposition applies to both fields [In this connection, note that
the magnetic field is linear in the source I dl just as the electrostatic
field is linear in its source: the electric charge ]
(ii) The electrostatic field is produced by a scalar source, namely, the electric
charge |
1 | 3498-3501 | The principle of
superposition applies to both fields [In this connection, note that
the magnetic field is linear in the source I dl just as the electrostatic
field is linear in its source: the electric charge ]
(ii) The electrostatic field is produced by a scalar source, namely, the electric
charge The magnetic field is produced by a vector source I dl |
1 | 3499-3502 | [In this connection, note that
the magnetic field is linear in the source I dl just as the electrostatic
field is linear in its source: the electric charge ]
(ii) The electrostatic field is produced by a scalar source, namely, the electric
charge The magnetic field is produced by a vector source I dl (iii) The electrostatic field is along the displacement vector joining the
source and the field point |
1 | 3500-3503 | ]
(ii) The electrostatic field is produced by a scalar source, namely, the electric
charge The magnetic field is produced by a vector source I dl (iii) The electrostatic field is along the displacement vector joining the
source and the field point The magnetic field is perpendicular to the
plane containing the displacement vector r and the current element I dl |
1 | 3501-3504 | The magnetic field is produced by a vector source I dl (iii) The electrostatic field is along the displacement vector joining the
source and the field point The magnetic field is perpendicular to the
plane containing the displacement vector r and the current element I dl (iv) There is an angle dependence in the Biot-Savart law which is not
present in the electrostatic case |
1 | 3502-3505 | (iii) The electrostatic field is along the displacement vector joining the
source and the field point The magnetic field is perpendicular to the
plane containing the displacement vector r and the current element I dl (iv) There is an angle dependence in the Biot-Savart law which is not
present in the electrostatic case In Fig |
1 | 3503-3506 | The magnetic field is perpendicular to the
plane containing the displacement vector r and the current element I dl (iv) There is an angle dependence in the Biot-Savart law which is not
present in the electrostatic case In Fig 4 |
1 | 3504-3507 | (iv) There is an angle dependence in the Biot-Savart law which is not
present in the electrostatic case In Fig 4 7, the magnetic field at any
point in the direction of dl (the dashed line) is zero |
1 | 3505-3508 | In Fig 4 7, the magnetic field at any
point in the direction of dl (the dashed line) is zero Along this line,
q = 0, sin q = 0 and from Eq |
1 | 3506-3509 | 4 7, the magnetic field at any
point in the direction of dl (the dashed line) is zero Along this line,
q = 0, sin q = 0 and from Eq [4 |
1 | 3507-3510 | 7, the magnetic field at any
point in the direction of dl (the dashed line) is zero Along this line,
q = 0, sin q = 0 and from Eq [4 11(a)], |dB| = 0 |
1 | 3508-3511 | Along this line,
q = 0, sin q = 0 and from Eq [4 11(a)], |dB| = 0 There is an interesting relation between e0, the permittivity of free
space; m0, the permeability of free space; and c, the speed of light in vacuum:
(
)
0
0
0
0
4
µ4
ε µ
ε
=
π
π
(
)
7
9
1
10
9
10
−
=
×
8 2
2
1
1
(3
10 )
c
=
=
×
We will discuss this connection further in Chapter 8 on the
electromagnetic waves |
1 | 3509-3512 | [4 11(a)], |dB| = 0 There is an interesting relation between e0, the permittivity of free
space; m0, the permeability of free space; and c, the speed of light in vacuum:
(
)
0
0
0
0
4
µ4
ε µ
ε
=
π
π
(
)
7
9
1
10
9
10
−
=
×
8 2
2
1
1
(3
10 )
c
=
=
×
We will discuss this connection further in Chapter 8 on the
electromagnetic waves Since the speed of light in vacuum is constant,
the product m0e0 is fixed in magnitude |
1 | 3510-3513 | 11(a)], |dB| = 0 There is an interesting relation between e0, the permittivity of free
space; m0, the permeability of free space; and c, the speed of light in vacuum:
(
)
0
0
0
0
4
µ4
ε µ
ε
=
π
π
(
)
7
9
1
10
9
10
−
=
×
8 2
2
1
1
(3
10 )
c
=
=
×
We will discuss this connection further in Chapter 8 on the
electromagnetic waves Since the speed of light in vacuum is constant,
the product m0e0 is fixed in magnitude Choosing the value of either e0 or
m0, fixes the value of the other |
1 | 3511-3514 | There is an interesting relation between e0, the permittivity of free
space; m0, the permeability of free space; and c, the speed of light in vacuum:
(
)
0
0
0
0
4
µ4
ε µ
ε
=
π
π
(
)
7
9
1
10
9
10
−
=
×
8 2
2
1
1
(3
10 )
c
=
=
×
We will discuss this connection further in Chapter 8 on the
electromagnetic waves Since the speed of light in vacuum is constant,
the product m0e0 is fixed in magnitude Choosing the value of either e0 or
m0, fixes the value of the other In SI units, m0 is fixed to be equal to
4p × 10–7
in magnitude |
1 | 3512-3515 | Since the speed of light in vacuum is constant,
the product m0e0 is fixed in magnitude Choosing the value of either e0 or
m0, fixes the value of the other In SI units, m0 is fixed to be equal to
4p × 10–7
in magnitude Example 4 |
1 | 3513-3516 | Choosing the value of either e0 or
m0, fixes the value of the other In SI units, m0 is fixed to be equal to
4p × 10–7
in magnitude Example 4 5 An element
x iˆ
l
is placed at the origin and carries
a large current I = 10 A (Fig |
1 | 3514-3517 | In SI units, m0 is fixed to be equal to
4p × 10–7
in magnitude Example 4 5 An element
x iˆ
l
is placed at the origin and carries
a large current I = 10 A (Fig 4 |
1 | 3515-3518 | Example 4 5 An element
x iˆ
l
is placed at the origin and carries
a large current I = 10 A (Fig 4 8) |
1 | 3516-3519 | 5 An element
x iˆ
l
is placed at the origin and carries
a large current I = 10 A (Fig 4 8) What is the magnetic field on the y-
axis at a distance of 0 |
1 | 3517-3520 | 4 8) What is the magnetic field on the y-
axis at a distance of 0 5 m |
1 | 3518-3521 | 8) What is the magnetic field on the y-
axis at a distance of 0 5 m Dx = 1 cm |
1 | 3519-3522 | What is the magnetic field on the y-
axis at a distance of 0 5 m Dx = 1 cm FIGURE 4 |
1 | 3520-3523 | 5 m Dx = 1 cm FIGURE 4 8
Rationalised 2023-24
115
Moving Charges and
Magnetism
Solution
0
2
d sin
|d
|
4
I
l
r
µ
θ
=
π
B
[using Eq |
1 | 3521-3524 | Dx = 1 cm FIGURE 4 8
Rationalised 2023-24
115
Moving Charges and
Magnetism
Solution
0
2
d sin
|d
|
4
I
l
r
µ
θ
=
π
B
[using Eq (4 |
1 | 3522-3525 | FIGURE 4 8
Rationalised 2023-24
115
Moving Charges and
Magnetism
Solution
0
2
d sin
|d
|
4
I
l
r
µ
θ
=
π
B
[using Eq (4 11)]
2
d
10
m
l
x
−
= ∆
=
, I = 10 A, r = 0 |
1 | 3523-3526 | 8
Rationalised 2023-24
115
Moving Charges and
Magnetism
Solution
0
2
d sin
|d
|
4
I
l
r
µ
θ
=
π
B
[using Eq (4 11)]
2
d
10
m
l
x
−
= ∆
=
, I = 10 A, r = 0 5 m = y,
7
0
T m
/4
10
A
µ
−
π =
q = 90° ; sin q = 1
7
2
2
10
10
10
d
25
10
−
−
−
×
×
=
×
B
= 4 × 10–8 T
The direction of the field is in the +z-direction |
1 | 3524-3527 | (4 11)]
2
d
10
m
l
x
−
= ∆
=
, I = 10 A, r = 0 5 m = y,
7
0
T m
/4
10
A
µ
−
π =
q = 90° ; sin q = 1
7
2
2
10
10
10
d
25
10
−
−
−
×
×
=
×
B
= 4 × 10–8 T
The direction of the field is in the +z-direction This is so since,
ˆ
ˆ
d
×
i ×
j
x
y
r
l
(
)
ˆ
ˆ
y
x
=
∆
i × j
ˆ
y
x
=
∆ k
We remind you of the following cyclic property of cross-products,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
;
;
×
=
×
=
×
=
i
j
k
j
k
i k
i
j
Note that the field is small in magnitude |
1 | 3525-3528 | 11)]
2
d
10
m
l
x
−
= ∆
=
, I = 10 A, r = 0 5 m = y,
7
0
T m
/4
10
A
µ
−
π =
q = 90° ; sin q = 1
7
2
2
10
10
10
d
25
10
−
−
−
×
×
=
×
B
= 4 × 10–8 T
The direction of the field is in the +z-direction This is so since,
ˆ
ˆ
d
×
i ×
j
x
y
r
l
(
)
ˆ
ˆ
y
x
=
∆
i × j
ˆ
y
x
=
∆ k
We remind you of the following cyclic property of cross-products,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
;
;
×
=
×
=
×
=
i
j
k
j
k
i k
i
j
Note that the field is small in magnitude In the next section, we shall use the Biot-Savart law to calculate the
magnetic field due to a circular loop |
1 | 3526-3529 | 5 m = y,
7
0
T m
/4
10
A
µ
−
π =
q = 90° ; sin q = 1
7
2
2
10
10
10
d
25
10
−
−
−
×
×
=
×
B
= 4 × 10–8 T
The direction of the field is in the +z-direction This is so since,
ˆ
ˆ
d
×
i ×
j
x
y
r
l
(
)
ˆ
ˆ
y
x
=
∆
i × j
ˆ
y
x
=
∆ k
We remind you of the following cyclic property of cross-products,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
;
;
×
=
×
=
×
=
i
j
k
j
k
i k
i
j
Note that the field is small in magnitude In the next section, we shall use the Biot-Savart law to calculate the
magnetic field due to a circular loop 4 |
1 | 3527-3530 | This is so since,
ˆ
ˆ
d
×
i ×
j
x
y
r
l
(
)
ˆ
ˆ
y
x
=
∆
i × j
ˆ
y
x
=
∆ k
We remind you of the following cyclic property of cross-products,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
;
;
×
=
×
=
×
=
i
j
k
j
k
i k
i
j
Note that the field is small in magnitude In the next section, we shall use the Biot-Savart law to calculate the
magnetic field due to a circular loop 4 5 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR
CURRENT LOOP
In this section, we shall evaluate the magnetic field due
to a circular coil along its axis |
1 | 3528-3531 | In the next section, we shall use the Biot-Savart law to calculate the
magnetic field due to a circular loop 4 5 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR
CURRENT LOOP
In this section, we shall evaluate the magnetic field due
to a circular coil along its axis The evaluation entails
summing up the effect of infinitesimal current elements
(I dl) mentioned in the previous section |
1 | 3529-3532 | 4 5 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR
CURRENT LOOP
In this section, we shall evaluate the magnetic field due
to a circular coil along its axis The evaluation entails
summing up the effect of infinitesimal current elements
(I dl) mentioned in the previous section We assume
that the current I is steady and that the evaluation is
carried out in free space (i |
1 | 3530-3533 | 5 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR
CURRENT LOOP
In this section, we shall evaluate the magnetic field due
to a circular coil along its axis The evaluation entails
summing up the effect of infinitesimal current elements
(I dl) mentioned in the previous section We assume
that the current I is steady and that the evaluation is
carried out in free space (i e |
1 | 3531-3534 | The evaluation entails
summing up the effect of infinitesimal current elements
(I dl) mentioned in the previous section We assume
that the current I is steady and that the evaluation is
carried out in free space (i e , vacuum) |
1 | 3532-3535 | We assume
that the current I is steady and that the evaluation is
carried out in free space (i e , vacuum) Fig |
1 | 3533-3536 | e , vacuum) Fig 4 |
1 | 3534-3537 | , vacuum) Fig 4 9 depicts a circular loop carrying a steady
current I |
1 | 3535-3538 | Fig 4 9 depicts a circular loop carrying a steady
current I The loop is placed in the y-z plane with its
centre at the origin O and has a radius R |
1 | 3536-3539 | 4 9 depicts a circular loop carrying a steady
current I The loop is placed in the y-z plane with its
centre at the origin O and has a radius R The x-axis is
the axis of the loop |
1 | 3537-3540 | 9 depicts a circular loop carrying a steady
current I The loop is placed in the y-z plane with its
centre at the origin O and has a radius R The x-axis is
the axis of the loop We wish to calculate the magnetic
field at the point P on this axis |
1 | 3538-3541 | The loop is placed in the y-z plane with its
centre at the origin O and has a radius R The x-axis is
the axis of the loop We wish to calculate the magnetic
field at the point P on this axis Let x be the distance of
P from the centre O of the loop |
1 | 3539-3542 | The x-axis is
the axis of the loop We wish to calculate the magnetic
field at the point P on this axis Let x be the distance of
P from the centre O of the loop Consider a conducting element dl of the loop |
1 | 3540-3543 | We wish to calculate the magnetic
field at the point P on this axis Let x be the distance of
P from the centre O of the loop Consider a conducting element dl of the loop This is
shown in Fig |
1 | 3541-3544 | Let x be the distance of
P from the centre O of the loop Consider a conducting element dl of the loop This is
shown in Fig 4 |
1 | 3542-3545 | Consider a conducting element dl of the loop This is
shown in Fig 4 9 |
1 | 3543-3546 | This is
shown in Fig 4 9 The magnitude dB of the magnetic
field due to dl is given by the Biot-Savart law [Eq |
1 | 3544-3547 | 4 9 The magnitude dB of the magnetic
field due to dl is given by the Biot-Savart law [Eq 4 |
1 | 3545-3548 | 9 The magnitude dB of the magnetic
field due to dl is given by the Biot-Savart law [Eq 4 11(a)],
0
3
4
I d× r
dB
r
l
(4 |
1 | 3546-3549 | The magnitude dB of the magnetic
field due to dl is given by the Biot-Savart law [Eq 4 11(a)],
0
3
4
I d× r
dB
r
l
(4 12)
Now r 2 = x 2 + R 2 |
1 | 3547-3550 | 4 11(a)],
0
3
4
I d× r
dB
r
l
(4 12)
Now r 2 = x 2 + R 2 Further, any element of the loop
will be perpendicular to the displacement vector from
the element to the axial point |
1 | 3548-3551 | 11(a)],
0
3
4
I d× r
dB
r
l
(4 12)
Now r 2 = x 2 + R 2 Further, any element of the loop
will be perpendicular to the displacement vector from
the element to the axial point For example, the element dl in Fig |
1 | 3549-3552 | 12)
Now r 2 = x 2 + R 2 Further, any element of the loop
will be perpendicular to the displacement vector from
the element to the axial point For example, the element dl in Fig 4 |
1 | 3550-3553 | Further, any element of the loop
will be perpendicular to the displacement vector from
the element to the axial point For example, the element dl in Fig 4 9 is
in the y-z plane, whereas, the displacement vector r from dl to the axial
point P is in the x-y plane |
1 | 3551-3554 | For example, the element dl in Fig 4 9 is
in the y-z plane, whereas, the displacement vector r from dl to the axial
point P is in the x-y plane Hence |dl × r|=r dl |
1 | 3552-3555 | 4 9 is
in the y-z plane, whereas, the displacement vector r from dl to the axial
point P is in the x-y plane Hence |dl × r|=r dl Thus,
(
)
π
0
2
2
d
d
4
I l
B
x
R
=µ
+
(4 |
1 | 3553-3556 | 9 is
in the y-z plane, whereas, the displacement vector r from dl to the axial
point P is in the x-y plane Hence |dl × r|=r dl Thus,
(
)
π
0
2
2
d
d
4
I l
B
x
R
=µ
+
(4 13)
FIGURE 4 |
1 | 3554-3557 | Hence |dl × r|=r dl Thus,
(
)
π
0
2
2
d
d
4
I l
B
x
R
=µ
+
(4 13)
FIGURE 4 9 Magnetic field on the
axis of a current carrying circular
loop of radius R |
1 | 3555-3558 | Thus,
(
)
π
0
2
2
d
d
4
I l
B
x
R
=µ
+
(4 13)
FIGURE 4 9 Magnetic field on the
axis of a current carrying circular
loop of radius R Shown are the
magnetic field dB (due to a line
element dl ) and its
components along and
perpendicular to the axis |
1 | 3556-3559 | 13)
FIGURE 4 9 Magnetic field on the
axis of a current carrying circular
loop of radius R Shown are the
magnetic field dB (due to a line
element dl ) and its
components along and
perpendicular to the axis EXAMPLE 4 |
1 | 3557-3560 | 9 Magnetic field on the
axis of a current carrying circular
loop of radius R Shown are the
magnetic field dB (due to a line
element dl ) and its
components along and
perpendicular to the axis EXAMPLE 4 5
Rationalised 2023-24
Physics
116
The direction of dB is shown in Fig |
1 | 3558-3561 | Shown are the
magnetic field dB (due to a line
element dl ) and its
components along and
perpendicular to the axis EXAMPLE 4 5
Rationalised 2023-24
Physics
116
The direction of dB is shown in Fig 4 |
1 | 3559-3562 | EXAMPLE 4 5
Rationalised 2023-24
Physics
116
The direction of dB is shown in Fig 4 9 |
1 | 3560-3563 | 5
Rationalised 2023-24
Physics
116
The direction of dB is shown in Fig 4 9 It is perpendicular to the
plane formed by dl and r |
1 | 3561-3564 | 4 9 It is perpendicular to the
plane formed by dl and r It has an x-component dBx and a component
perpendicular to x-axis, dB^ |
1 | 3562-3565 | 9 It is perpendicular to the
plane formed by dl and r It has an x-component dBx and a component
perpendicular to x-axis, dB^ When the components perpendicular to
the x-axis are summed over, they cancel out and we obtain a null result |
1 | 3563-3566 | It is perpendicular to the
plane formed by dl and r It has an x-component dBx and a component
perpendicular to x-axis, dB^ When the components perpendicular to
the x-axis are summed over, they cancel out and we obtain a null result For example, the dB^ component due to dl is cancelled by the contribution
due to the diametrically opposite dl element, shown in
Fig |
1 | 3564-3567 | It has an x-component dBx and a component
perpendicular to x-axis, dB^ When the components perpendicular to
the x-axis are summed over, they cancel out and we obtain a null result For example, the dB^ component due to dl is cancelled by the contribution
due to the diametrically opposite dl element, shown in
Fig 4 |
1 | 3565-3568 | When the components perpendicular to
the x-axis are summed over, they cancel out and we obtain a null result For example, the dB^ component due to dl is cancelled by the contribution
due to the diametrically opposite dl element, shown in
Fig 4 9 |
1 | 3566-3569 | For example, the dB^ component due to dl is cancelled by the contribution
due to the diametrically opposite dl element, shown in
Fig 4 9 Thus, only the x-component survives |
1 | 3567-3570 | 4 9 Thus, only the x-component survives The net contribution along
x-direction can be obtained by integrating dBx = dB cos q over the loop |
1 | 3568-3571 | 9 Thus, only the x-component survives The net contribution along
x-direction can be obtained by integrating dBx = dB cos q over the loop For Fig |
1 | 3569-3572 | Thus, only the x-component survives The net contribution along
x-direction can be obtained by integrating dBx = dB cos q over the loop For Fig 4 |
1 | 3570-3573 | The net contribution along
x-direction can be obtained by integrating dBx = dB cos q over the loop For Fig 4 9,
2
2 1/2
cos
(
)
R
x
R
θ =
+
(4 |
1 | 3571-3574 | For Fig 4 9,
2
2 1/2
cos
(
)
R
x
R
θ =
+
(4 14)
From Eqs |
1 | 3572-3575 | 4 9,
2
2 1/2
cos
(
)
R
x
R
θ =
+
(4 14)
From Eqs (4 |
1 | 3573-3576 | 9,
2
2 1/2
cos
(
)
R
x
R
θ =
+
(4 14)
From Eqs (4 13) and (4 |
1 | 3574-3577 | 14)
From Eqs (4 13) and (4 14),
(
)
π
0
3/2
2
2
d
d
4
x
I l
R
B
x
R
=µ
+
The summation of elements dl over the loop yields 2pR, the
circumference of the loop |
1 | 3575-3578 | (4 13) and (4 14),
(
)
π
0
3/2
2
2
d
d
4
x
I l
R
B
x
R
=µ
+
The summation of elements dl over the loop yields 2pR, the
circumference of the loop Thus, the magnetic field at P due to entire
circular loop is
(
)
2
0
3/2
2
2
ˆ
ˆ
2
x
I R
B
x
R
µ
=
=
+
B
i
i
(4 |
1 | 3576-3579 | 13) and (4 14),
(
)
π
0
3/2
2
2
d
d
4
x
I l
R
B
x
R
=µ
+
The summation of elements dl over the loop yields 2pR, the
circumference of the loop Thus, the magnetic field at P due to entire
circular loop is
(
)
2
0
3/2
2
2
ˆ
ˆ
2
x
I R
B
x
R
µ
=
=
+
B
i
i
(4 15)
As a special case of the above result, we may obtain the field at the centre
of the loop |
1 | 3577-3580 | 14),
(
)
π
0
3/2
2
2
d
d
4
x
I l
R
B
x
R
=µ
+
The summation of elements dl over the loop yields 2pR, the
circumference of the loop Thus, the magnetic field at P due to entire
circular loop is
(
)
2
0
3/2
2
2
ˆ
ˆ
2
x
I R
B
x
R
µ
=
=
+
B
i
i
(4 15)
As a special case of the above result, we may obtain the field at the centre
of the loop Here x = 0, and we obtain,
0
0
ˆ
2
I
R
=µ
B
i
(4 |
1 | 3578-3581 | Thus, the magnetic field at P due to entire
circular loop is
(
)
2
0
3/2
2
2
ˆ
ˆ
2
x
I R
B
x
R
µ
=
=
+
B
i
i
(4 15)
As a special case of the above result, we may obtain the field at the centre
of the loop Here x = 0, and we obtain,
0
0
ˆ
2
I
R
=µ
B
i
(4 16)
The magnetic field lines due to a circular wire form closed loops and
are shown in Fig |
1 | 3579-3582 | 15)
As a special case of the above result, we may obtain the field at the centre
of the loop Here x = 0, and we obtain,
0
0
ˆ
2
I
R
=µ
B
i
(4 16)
The magnetic field lines due to a circular wire form closed loops and
are shown in Fig 4 |
1 | 3580-3583 | Here x = 0, and we obtain,
0
0
ˆ
2
I
R
=µ
B
i
(4 16)
The magnetic field lines due to a circular wire form closed loops and
are shown in Fig 4 10 |
1 | 3581-3584 | 16)
The magnetic field lines due to a circular wire form closed loops and
are shown in Fig 4 10 The direction of the magnetic field is given by
(another) right-hand thumb rule stated below:
Curl the palm of your right hand around the circular wire with the
fingers pointing in the direction of the current |
1 | 3582-3585 | 4 10 The direction of the magnetic field is given by
(another) right-hand thumb rule stated below:
Curl the palm of your right hand around the circular wire with the
fingers pointing in the direction of the current The right-hand thumb
gives the direction of the magnetic field |
1 | 3583-3586 | 10 The direction of the magnetic field is given by
(another) right-hand thumb rule stated below:
Curl the palm of your right hand around the circular wire with the
fingers pointing in the direction of the current The right-hand thumb
gives the direction of the magnetic field FIGURE 4 |
1 | 3584-3587 | The direction of the magnetic field is given by
(another) right-hand thumb rule stated below:
Curl the palm of your right hand around the circular wire with the
fingers pointing in the direction of the current The right-hand thumb
gives the direction of the magnetic field FIGURE 4 10 The magnetic field lines for a current loop |
1 | 3585-3588 | The right-hand thumb
gives the direction of the magnetic field FIGURE 4 10 The magnetic field lines for a current loop The direction of
the field is given by the right-hand thumb rule described in the text |
1 | 3586-3589 | FIGURE 4 10 The magnetic field lines for a current loop The direction of
the field is given by the right-hand thumb rule described in the text The
upper side of the loop may be thought of as the north pole and the lower
side as the south pole of a magnet |
1 | 3587-3590 | 10 The magnetic field lines for a current loop The direction of
the field is given by the right-hand thumb rule described in the text The
upper side of the loop may be thought of as the north pole and the lower
side as the south pole of a magnet Rationalised 2023-24
117
Moving Charges and
Magnetism
EXAMPLE 4 |
1 | 3588-3591 | The direction of
the field is given by the right-hand thumb rule described in the text The
upper side of the loop may be thought of as the north pole and the lower
side as the south pole of a magnet Rationalised 2023-24
117
Moving Charges and
Magnetism
EXAMPLE 4 6
Example 4 |
1 | 3589-3592 | The
upper side of the loop may be thought of as the north pole and the lower
side as the south pole of a magnet Rationalised 2023-24
117
Moving Charges and
Magnetism
EXAMPLE 4 6
Example 4 6 A straight wire carrying a current of 12 A is bent into a
semi-circular arc of radius 2 |
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