Chapter 20 The Production and Properties of Magnetic Fields.
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Transcript of Chapter 20 The Production and Properties of Magnetic Fields.
![Page 1: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/1.jpg)
Chapter 20 The Production and Properties of Magnetic Fields
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Main points
•Biot–Savart law:
•Ampère’s law:
•Gauss’ law for magnetism:
![Page 3: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/3.jpg)
What is the source of magnetic fields, if there is no magnetic charge?
Answer: electric charge in motion!
bar magnet
current in wire
moving point charge
Therefore, understanding source of field generated by bar magnet lies in understanding currents at atomic level within bulk matter.
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20-1 The Biot-Savart Law
The electric field:The magnetic field :
20
ˆ
4 r
rlIB
d
d
Permeability of Free Space 27
0 AN104
?
the magnetic field of a straight section of wire is given by the Biot–Savart law:
It is a defined value
![Page 5: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/5.jpg)
The magnitude of the field :
The direction of the field is given by the cross product.
B
P
lI
dr
Example :
lI
d
lI
d
P
r B
r B
r
B lI
d r
0B
20
ˆ
4 r
rlIB
d
d
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Using the Biot-Savart Law
lI
d r B
a
IFind the magnetic field at a point
P located a distance a from a
straight wire carrying a current
I. P
Field around a current-carrying wire:
we integrated dl over the whole wire
20
ˆ
4 r
rlIB
d
d
?
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I
alI
d r
P
l
1
2
O
Magnitude:
Direction: determined by a right-hand rule
![Page 8: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/8.jpg)
Field around a current-carrying wire
The magnetic field lines are circles around the wire
a right-hand rule
![Page 9: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/9.jpg)
(1) For an infinite long wire
)cos(cos4 21
0
aI
B
aI
B
2
0B
(2) A very long wire carrying a current I is bent into the shape shown in figure
P
a
I
1
2
B
r
Discussion
I
1
2
PThe direction of the field is given by the right-hand rule
![Page 10: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/10.jpg)
I 1
2
3R
O
Example Consider the wire shown in the figure below , find the magnetic field at a point O .
Solution
The magnetic field at point O is the sum of three fields
They are directed out of page
lId
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ACTA current I flows in the +y direction in an infinite wire; a current I also flows in the loop as shown in the diagram.
– What is Fx, the net force on the loop in the x-direction?
(a) Fx < 0 (b) Fx = 0 (c) Fx > 0
I
I
o
y
x
You may have remembered that the net force on a current loop in a uniform B-field is zero, but the B-field produced by an infinite wire is not uniform!
![Page 12: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/12.jpg)
Example Two parallel, infinitely long wires, separated by a distance a, carry parallel currents I1 and I2 , find the force exerted by one long, straight current on another.
a
2I1I
1B21f
Solution :
The magnetic field of wire 1 at the position of wire 2 is
12f
The force on per unit length of wire 2 is
Similarly, the force on per unit length of wire 1 is
attract
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(1)The force between two currents is in accordance with Newton’s third law
Discussion
Currents in the same direction attractCurrents in the opposite directions repel
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(2) Defining the ampere
One ampere is the current which, following through two infinitely long, thin, straight wires placed one meter apart in a vacuum, produces a force of newtons on each meter length of wire
This definition gives
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(3) The force between two current element
22dlI
11dlI
12r12f
21f
Similarly,
The force between two current element doesn’t abide by Newton’s third law.
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(4) Two infinite line charges of linear charge density 1 and 2 move along the length direction with velocity of v1 and v2. Find the forces between the two wires. 1v
2v
1
2
a The current of the two wires is
The electrical force on per unit length of the wire is
00
1
c
Commonly, the magnetic force is much less than the electrical force.
The magnetic force on per unit length of the wire is
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Example The figure below shows an infinite straight wire and a current loop. Find the torque on the loop.
x
Solution 1
Use
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x
Solution 2
Use
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I
b
P
r
x
y
O
xd 1
Field around a long current-carrying flat :
Break the flat into strips of width dx
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(1) (2) (3)
y
b
b
IBp 2
tan 10
(1)
(2)
i021
magnetic shield
i i
Discussion
Infinite Straight Wire
Infinite Current Sheet
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P
x
R
xO
Find the magnetic field at a point P located a distance x from the center along the axis of a circular current loop of radius R and current I.
lI
dB
d
B
d
B
r
Px
I
ixR
IRB
2/322
20
)(2
The direction of the field is given by the right-hand rule.
Field due to a current loop :
![Page 22: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/22.jpg)
(1) At the center of the circular loop
If the circular loop has N turns
Discussion
(2) The magnetic field at the center of the arc segment
I
ixR
IRB
2/322
20
)(2
B
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I
(3)
2/322
20
)(2 xRIR
B
A
nnIA
Define
30
2 xB
Magnetic dipole moment
Compare with electric dipole
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The magnetic field of a magnetic dipole
The electric field of an electric dipole
![Page 25: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/25.jpg)
The magnetic field lines of a circular loop of a current and a bar magnet
![Page 26: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/26.jpg)
Solution
Example: A nonconducting disk of radius R has charge q and rotates with angular velocity about its axis, find the magnetic field at a point on the axis of the rotating disk.
x
OR
q
P
r
B
d
At the center of the disk
![Page 27: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/27.jpg)
Example A hydrogen atom may be described as consisting of an electron that moves in a circular orbit around a proton. The force that gives rise to the motion is the Coulomb attraction between the proton and the electron, which have charges respectively, where .The motion is further constrained by the requirement that the angular momentum has the value where n is an integer and , Planck’s constant. Calculate the magnitude and direction of the magnetic field at the location of the proton. What is the magnetic moment of the current loop?
![Page 28: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/28.jpg)
SolutionThe Coulomb attraction provides the centripetal force
The constraint on the angular momentum
The magnetic field at the location of the proton
Perpendical to the orbit
The magnetic moment
Perpendical to the orbit
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I
PR
A solenoid of radius R has n turns in per unit length, with each turn carrying a current I .
ldl
B
d
r
Magnetic field in a solenoid :
A solenoid is a length of wire coiled into a cylinder
![Page 30: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/30.jpg)
PR
l
B
d
r 21
(1) For a long solenoid
nIB 0
Discussion
(2) If the origin is at one end of the solenoid
nIB 02
1
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20-2 Gauss’ Law for Magnetism
Electrostatic field :Magnetic field :
Magnetic flux
In the Magnetic field, the number of Magnetic field lines crossing a curved surface S is defined as magnetic flux.
ABB
dd
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For a curved surface
If the magnetic field lines enter the surface
For a closed surface
If the magnetic field lines leave the surface
SI: weber (Wb)
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• No magnetic monopoles (single magnetic charge) have ever been observed
• Magnetic field lines must be continuous
• The magnetic “charge” inside any closed surface must always be zero.
Gauss’s Law For Magnetism
0d SB AB
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The magnetic field lines must be continuous
![Page 35: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/35.jpg)
20-3 Ampere’s Law
Ampere’s Law
Electrostatic field : conservative field
Magnetic field:
Take a line integral over the magnetic field along a circular path C
Path C circles a current-carrying wire
ld
Current I is through the surface whose edge is defined by the path
![Page 36: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/36.jpg)
Path C’ consists of C1 and C2
Take a line integral over the magnetic field along a closed path C’(C1+C2)
In this case, there is no current through the surface whose edge is defined by the path
![Page 37: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/37.jpg)
In the static magnetic field, the path integral of the total magnetic field around a closed path is equal to times the algebraic sum of the currents threading the path.
Ampere’s law enclosedLIμlB 0d
Integral around a closed path Current “enclosed” by that path
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(2) The enclosed current can be positive or negative contribution.
Discussion
Curl your right hand around the closed path, with the fingers pointing in the direction of integration. A current through the path in the general direction of your outstretched thumb is assigned a plus sign, and a current generally in the opposite direction is assigned a minus sign
(1)Ampere’s Law applies to an arbitrary closed path The direction of the closed path is arbitrary
![Page 39: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/39.jpg)
(3) Ampere’s law is valid only if the currents are steady.
aI
L
A finite current segment, if
1
2
For example
The integral of the magnetic field around the closed path L is
![Page 40: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/40.jpg)
Using Ampere’s law to find the magnetic field
Ampere's Law can simplify the calculation if there is symmetry of the current! (and the magnetic field )
• Choose a closed path with direction
• Evaluate line integral in Ampere’s Law (The integral is simple)
• Calculate the current enclosed by the path
• Apply Ampere’s Law
![Page 41: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/41.jpg)
Example: A long, hollow, cylindrical surface of radius R carries a current I that is uniformly distributed over the surface. Find the magnetic field both outside and inside the cylindrical surface.
R
I
rP
LSolution Because of the high degree of symmetry, we know that is tangent to the circle of radius r about the axis and constant in magnitude everywhere on the circle.
PId'dI
B
d 'dB
B
rI
B
2
0
0B
![Page 42: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/42.jpg)
ACT For a long cylinder that current is uniformly distributed over the cross-sectional area, find the magnetic field.
R
I
![Page 43: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/43.jpg)
Example Find the magnetic field of an infinite sheet of current, the current per unit length (along the direction which is perpendicular to the current) is i .
Solution i
B
'B
P ab
c d
![Page 44: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/44.jpg)
If the thickness of the sheet is dj is the current density
jd• outside the sheet
• inside the sheet
x
![Page 45: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/45.jpg)
Choose the rectangle abcd as the integral path.
Example Find the the magnetic field of a long ideal solenoid carrying a current i.
Solution
The field inside is parallel to the cylinder axis
The field outside is insignificant
1l
2l
S
![Page 46: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/46.jpg)
The Magnetic field lines for a real solenoid of finite length. The field is strong and uniform at interior points such as P1 but relatively weak at external points such as P2.
![Page 47: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/47.jpg)
o
I
N
Solution
• If the cross-sectional width is much less than the radius of the toroidal solenoid ,
r
• outside the torus
Example Find the magnetic field of a toroidal solenoid
The Magnetic field within the torus must be parallel to the walls
![Page 48: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/48.jpg)
o
I
N
h
1R
2R A
rd
If the cross-sectional surface is rectangle with width h.Find the magnetic flux in it.
![Page 49: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/49.jpg)
Example Figure below shows a cross section of a long cylindrical conductor containing a long cylindrical hole .The axes of the cylinder and the hole are parallel and are a distance d apart. The current is uniformly distributed in the conductor. The current density is Find the magnetic field in the hole.
O1 O2
d
Solution
Regard the cylindrical hole as resulting from the superposition of a complete cylinder (no hole) carrying a current in one direction and a small cylinder carrying a current in the opposite direction, both cylinders having the same current density .
![Page 50: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/50.jpg)
OPr
The magnetic field in a cylinder
O1 O2
d1r
2rUse superposition
![Page 51: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/51.jpg)
0d SB AB
0 B
Ampere’s law
enclosedLIμlB 0d
Gauss’s Law For Magnetism
Bj
0
1
Gauss equation
Stokes equation
![Page 52: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/52.jpg)
20-4 The Magnetic Field of Moving Point Charges
lI
d
P
r
lI
d+q
A
—— The magnetic field of a moving charges
30
4 r
rqB
v
![Page 53: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/53.jpg)
The Electric and Magnetic Field of Moving Point Charges
•q
v
E
Br
Ec
EB
vv 200
1
•
![Page 54: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/54.jpg)
Oa
b
Example A closed circuit consists of two semicircles of radii a and b that are connected by straight segments as shown in the figure, the linear charge density is (>0) , and rotates with angular velocity about O. Find the magnetic field at point O .
1
2
34
qd
d
Segment 1 :
v
Solution
Segment 2 :
![Page 55: Chapter 20 The Production and Properties of Magnetic Fields.](https://reader035.fdocuments.in/reader035/viewer/2022062217/56649e175503460f94b03190/html5/thumbnails/55.jpg)
Oa
b
1
2
34
qd
dvv
B
d
Segment 3 :
Segment 4 :
qd