Download Sources of Magnetic Fields Chapter 28

Chapter 28
Lecture 12
Sources of Magnetic Fields
Announcements
— No quiz in discussion sections tomorrow
[review of midterm, HW questions, etc.]
— Ch 27 HW assigned (due Monday, March 6th)
— Grading of midterm is ongoing…
Goals for Chapter 28
• To study the magnetic field generated by a moving charge
• To consider magnetic field of a current-carrying conductor
• To examine the magnetic field of a long, straight, current-carrying
conductor
• To study the magnetic force between current-carrying conductors
• To consider the magnetic field of a current loop
• To examine and use Ampere’s Law
Magnetism &
Electricity
First evidence of relationship between magnetism and moving charges
discovered in 1820 by Hans Oersted. Current => Magnetic field.
Similar experiments by Ampere, Faraday and Henry discovered that a
moving magnet near a conducting loop can cause a current in the loop.
Ultimately, Maxwell showed that electricity and magnetism are different
manifestations of the same physical laws
•“Maxwell’s Equations” for electromagnetism
Magnetism &
Electricity
First evidence of relationship between magnetism and moving charges
discovered in 1820 by Hans Oersted. Current => Magnetic field.
Similar experiments by Ampere, Faraday and Henry discovered that a
moving magnet near a conducting loop can cause a current in the loop.
Ultimately, Maxwell showed that electricity and magnetism are different
manifestations of the same physical laws
•“Maxwell’s Equations” for electromagnetism
Magnetic field of a moving point charge
The constant µ0 is called the permeability of free space : µ0 = 4π x 10-7 T·m/A.
+
Moving charge creates B-field
B
rˆ ϕ
+q
rˆ '
P
ϕ’
At point P:
"
!" µ qv × rˆ
B= 0
4π r 2
v
P’
B
•
•
µ0: fundamental
constant
ˆ: unit vector
r
pointing from q to P
µ0 q vsinφ
B=
4π
r2
At point P, B points out of page (v×r + RHR)
At point P’, B points into page (v×r’ + RHR)
Note: B has same 1/r2 as E !
1 q
E=
4πε 0 r 2
Permeability of Free Space
The constant µ0 has an exact numerical value in SI units:
⎡ Tm⎤ ⎡ N ⎤ ⎡ Wb ⎤
µ0 = 4π x 10 ⎢ ⎥ or ⎢ 2 ⎥ or ⎢
⎣ A ⎦ ⎣ A ⎦ ⎣ Am⎥⎦
1
1
ε0 =
=
2
4π k
9 ⎡ Nm ⎤
4π × 9 ×10 ⎢ 2 ⎥
⎣ C ⎦
-7
The exactness of the value arises from the definition of the Ampere
µ0 and the corresponding constant of the electric field, ε0, are related to each other:
⎡ Tm⎤
10 -7 ⎢ ⎥
1
⎣ A⎦ =
µ0ε 0 =
2
9 ×1016
9 ⎡ Nm ⎤
9 ×10 ⎢ 2 ⎥
⎣ C ⎦
⎡ N
2⎤
mC
⎢
⎥
⎡ TmC2 ⎤ ⎢ C m s
s2
⎥
= 2
⎢ Nm2 A ⎥ = ⎢
Nm2 C s ⎥ m
⎣
⎦
⎢
⎥
⎣
⎦
⎡ TmC2 ⎤
⎢ Nm2 A ⎥
⎣
⎦
2
1
= 9 ×1016 ⎡ m 2 ⎤
⎣ s⎦
µ 0ε 0
1
= 3 ×108 ⎡⎣ m s ⎤⎦
µ0ε 0
Speed of light !
E and B fields compared
Charges create electric fields, and moving charges create magnetic fields.
Electric field properties (Chap 21):
•Proportional to the magnitude of the charge (|q|)
2
•Proportional to 1/r , where r is the distance from the source point to the field point.
1 q
E=
4πε 0 r 2
Magnetic field properties:
•
•
•
•
Proportional to the magnitude of the charge (|q|)
2
Proportional to 1/r , where r is the distance from the source point to the field point.
Proportional to the charge’s velocity, v.
Proportional to sin ϕ, where ϕ is the angle of the field point relative to the particle’s velocity
µ0 q vsinφ
B=
2
4π
r
The constant µ0 is called the permeability of free space :
µ0 = 4π x 10-7 T·m/A.
Quiz: Magnetic field
A positive point charge is moving directly toward point P.
The magnetic field that the point charge produces at point P
A.points from the charge toward point P.
B.points from point P toward the charge.
C.is perpendicular to the line from the point charge to point P.
D.is zero.
E.The answer depends on the speed of the point charge.
Quiz: Magnetic field
A positive point charge is moving directly toward point P.
The magnetic field that the point charge produces at point P
A.points from the charge toward point P.
B.points from point P toward the charge.
C.is perpendicular to the line from the point charge to point P.
D.is zero.
E.The answer depends on the speed of the point charge.
P
Example: forces between 2 moving protons
Two protons travel with parallel velocities of equal magnitude
and opposite direction, along lines separated by a distance r.
Find FE & FB (w/ direction) on the upper proton & determine
the ratio of their magnitudes.
Example: forces between 2 moving protons
Two protons travel with parallel velocities of equal magnitude
and opposite direction, along lines separated by a distance r.
Find FE & FB (w/ direction) on the upper proton & determine
the ratio of their magnitudes.
1.
Example: forces between 2 moving protons
Two protons travel with parallel velocities of equal magnitude
and opposite direction, along lines separated by a distance r.
Find FE & FB (w/ direction) on the upper proton & determine
the ratio of their magnitudes.
1.
2.
Example: forces between 2 moving protons
Two protons travel with parallel velocities of equal magnitude
and opposite direction, along lines separated by a distance r.
Find FE & FB (w/ direction) on the upper proton & determine
the ratio of their magnitudes.
1.
2.
velocity of upper proton is in
the -x direction.
Example: forces between 2 moving protons
Two protons travel with parallel velocities of equal magnitude
and opposite direction, along lines separated by a distance r.
Find FE & FB (w/ direction) on the upper proton & determine
the ratio of their magnitudes.
1.
2.
velocity of upper proton is in
the -x direction.
3.
If v<<c
then
FB<<FE
Quiz: Magnetic field due to moving point charge
Two positive point charges move side by side in the same direction with the same velocity.
What is the direction of the magnetic force that the upper point charge exerts on the lower one?
+q
A.toward the upper point charge (the force is attractive)
B.away from the upper point charge (the force is repulsive)
C.in the direction of the velocity
D.opposite to the direction of the velocity
E.none of the above
+q
Quiz: Magnetic field due to moving point charge
Two positive point charges move side by side in the same direction with the same velocity.
What is the direction of the magnetic force that the upper point charge exerts on the lower one?
+q
A.toward the upper point charge (the force is attractive)
B.away from the upper point charge (the force is repulsive)
C.in the direction of the velocity
D.opposite to the direction of the velocity
E.none of the above
Step 1: Find the direction of the B field on the lower point
generated by the upper point charge.: B is into the page
Step 2: Find the magnetic force on the lower point charge
Using RHR: FB points upward
+q
Magnetic Field of a Current
A current moving through a wire can be treated as many
individual point charges moving at the drift velocity, vd.
The moving charge in a segment of length dl is:
• dQ = nqAdl
The magnetic field dB will be proportional to dQ*v:
P
!
r
• dQv = nqAv dl
d
d
nqAvd = I
µ0
~
dB =
dQ~v ⇥ r̂
2
4⇡r
dl
€
Biot-Savart Law
Magnetic field of a
current element
For a finite size current element,
we sum up all the segments dl
which are carrying current.
Magnetic field of a straight current-carrying wire
Wire of length 2a carries a current I. Find the magnetic field at a distance x along the
perpendicular bisector of the wire.
~l ⇥ r̂
µ
Id
0
~ =
dB
4⇡ r2
Magnetic field of a straight current-carrying wire
Wire of length 2a carries a current I. Find the magnetic field at a distance x along the
perpendicular bisector of the wire.
x
r̂ = î
r
y
ŷ
r
~l ⇥ r̂
µ
Id
0
~ =
dB
4⇡ r2
Magnetic field of a straight current-carrying wire
Wire of length 2a carries a current I. Find the magnetic field at a distance x along the
perpendicular bisector of the wire.
x
r̂ = î
r
y
ŷ
r
~l ⇥ r̂
µ
Id
0
~ =
dB
4⇡ r2
d~l = (dy)ĵ
x
I(dy)
ĵ
⇥
(
î
µ
0
r
~ =
dB
4⇡
r2
y
r ĵ)
Magnetic field of a straight current-carrying wire
Wire of length 2a carries a current I. Find the magnetic field at a distance x along the
perpendicular bisector of the wire.
x
r̂ = î
r
y
ŷ
r
~l ⇥ r̂
µ
Id
0
~ =
dB
4⇡ r2
d~l = (dy)ĵ
x
I(dy)
ĵ
⇥
(
î
µ
0
r
~ =
dB
4⇡
r2
y
r ĵ)
Magnetic field of a straight current-carrying wire
Wire of length 2a carries a current I. Find the magnetic field at a distance x along the
perpendicular bisector of the wire.
x
r̂ = î
r
y
ŷ
r
~l ⇥ r̂
µ
Id
0
~ =
dB
4⇡ r2
d~l = (dy)ĵ
x
I(dy)
ĵ
⇥
(
î
µ
0
r
~ =
dB
4⇡
r2
~ =
dB
µ0 I dy x
k̂
3
4⇡ r
y
r ĵ)
Magnetic field of a straight current-carrying wire
Wire of length 2a carries a current I. Find the magnetic field at a distance x along the
perpendicular bisector of the wire.
x
r̂ = î
r
y
ŷ
r
~l ⇥ r̂
µ
Id
0
~ =
dB
4⇡ r2
d~l = (dy)ĵ
x
I(dy)
ĵ
⇥
(
î
µ
0
r
~ =
dB
4⇡
r2
y
r ĵ)
µ0 I dy x
k̂
3
4⇡ r
Z a
µ0 Ix
dy
~
B=
k̂
2
2
3/2
4⇡
a (x + y )
µ0 I
2a
~
p
B=
k̂
2
2
4⇡ x x + a
~ =
dB
Magnetic field of an infinite current-carrying wire
~ =
B
µ0 I
2a
p
k̂
2
2
4⇡ x x + a
If we let the current element become infinitely long (a!∞),
the magnitude of the field at a distance r (x = r) becomes:
µ0 I
B=
2π r
Magnetic field near a straight long wire
As always, the direction of the field is given by the righthand rule: the field lines loop around the wire in concentric
circles.
Quiz: Magnetic field at P
Two long, straight wires are oriented perpendicular to the xy-plane. They carry
currents of equal magnitude I in opposite directions as shown. At point P, the
magnetic field due to these currents is in:
A.the positive x-direction.
B.the negative x-direction.
C.the positive y-direction.
D.the negative y-direction.
E.none of the above
Quiz: Magnetic field at P
Two long, straight wires are oriented perpendicular to the xy-plane. They carry
currents of equal magnitude I in opposite directions as shown. At point P, the
magnetic field due to these currents is in:
A.the positive x-direction.
B.the negative x-direction.
C.the positive y-direction.
Btop-wire
D.the negative y-direction.
E.none of the above
Btotal
Bbottom-wire
What force will the upper wire feel?
~ = IL
~ ⇥B
~
F
What force will the upper wire feel?
~ = IL
~ ⇥B
~
F
B from lower wire
What force will the upper wire feel?
~ = IL
~ ⇥B
~
F
out of
page
Opposite current
=> wires repel.
F on upper wire
right
B from lower wire
Force between parallel wires
Parallel currents attract;
Antiparallel currents repel.
Force between parallel wires (cont.)
Magnetic field of the lower wire (at a distance r):
B=
µ0 I
2π r
Force on a wire with current I’ in that field:
Thus, the force per unit length between long
parallel current carrying wires:
F µ0 I I ʹ
=
L 2π r
Definition of the Ampere (and Coulomb)
One Ampere is defined as the current in two infinite wires, 1
meter apart, that results in a force per unit length of 2 × 10-7 N/m
between the wires.
This explains why the constant µ0 has an exact value in SI units.
The definition of the Ampere also fixes the definition of a
Coulomb (the charge transferred by 1 Ampere in 1 second), the
Volt and all other SI units that involve electric charge.
µ0 = 4π x10-7 T·m/A.
F µ0 I I ʹ
=
L 2π r