Download Lecture 15. Magnetic Fields of Moving Charges and Currents

Lecture 15. Magnetic Fields of Moving Charges and Currents
Outline:
Magnetic Field of a Moving Charge.
Magnetic Field of Currents.
Interaction between Two Wires with Current.
Lecture 14:
Hall Effect.
Magnetic Force on a Wire Segment.
Torque on a Current-Carrying Loop.
1
Force vs. Torque
𝑑2 𝑥
𝐹⃗ = 𝐹𝑥� = 𝑚𝑎⃗ = 𝑚 2 𝑥�
𝑑𝑡
𝑑𝑣⃗
𝑎⃗ =
𝑑𝑡
𝜏⃗
𝜏⃗ = 𝜏𝜑� = 𝐿𝑎⃗𝑎𝑎𝑎
𝑎⃗𝑎𝑎𝑎
𝜑� = 𝑟̂ × 𝑣�
𝑟⃗
𝑣⃗
𝑑2𝜑
= 𝐿 2 𝜑�
𝑑𝑡
𝑑𝜔
𝑑2 𝜑
≡
= 𝜔̇ 𝜑� = 2 𝜑�
𝑑𝑡
𝑑𝑑
𝑟⃗
𝑑𝜑
𝜔 ≡ 𝜔𝜑� =
𝜑�
𝑑𝑑
𝑣⃗
𝑟⃗ × 𝑣⃗
𝜔= 2
𝑟
2
Iclicker Question
Imagine that you place a magnet in a uniform magnetic field.
Will the magnetic field exert a net force on the magnet? If so,
what is the direction of the force (Hint: use the electric dipole
analogy).
A. yes, in the direction of the magnetic field
B. yes, opposite to the direction of the magnetic field
C. yes, but the direction depends on the magnet’s orientation
D. no, the net force on the magnet is zero.
3
Iclicker Question
Imagine that you place a magnet in a uniform magnetic field.
Will the magnetic field exert a net force on the magnet? If so,
what is the direction of the force (Hint: use the electric dipole
analogy).
A. yes, in the direction of the magnetic field
B. yes, opposite to the direction of the magnetic field
C. yes, but the direction depends on the magnet’s orientation
D. no, the net force on the magnet is zero.
4
Iclicker Question
An elliptical wire loop (3
turns) with current is in
the plane of the board.
What is the torque on the
current loop?
3 turns
𝐴 = 1𝑚2
𝐵 = 1𝑇
𝐼 = 1𝐴
A. 1 Nm into board
B. 1 Nm out of board
C. 3 Nm up
D. 3 Nm down
E. Zero
𝐴
𝜏⃗ = 𝜇⃗ × 𝐵
5
Iclicker Question
An elliptical wire loop (3
turns) with current is in
the plane of the board.
What is the torque on the
current loop?
3 turns
𝐴 = 1𝑚2
𝐵 = 1𝑇
𝐼 = 1𝐴
A. 1 Nm into board
B. 1 Nm out of board
C. 3 Nm up
D. 3 Nm down
E. Zero
𝐴
𝜏⃗ = 𝜇⃗ × 𝐵
6
Iclicker Question
7
Iclicker Question
8
Magnetic Field of a Moving Charge
Positive charge moving with a constant velocity 𝒗 ≪ 𝒄:
𝑐 – the speed of light
𝜇0 𝑣⃗ × 𝑟̂ 𝜇0 𝑣⃗ × 𝑟⃗
𝐵 𝑟 =
𝑞 2 =
𝑞 3
4𝜋
𝑟
4𝜋
𝑟
- the charge is at the origin
at this moment
2
𝑁𝑠
𝜇0 = 4𝜋 ∙ 10−7 2
𝐶
velocity directed
into the plane of
the page
𝐵
𝑐2 =
1
𝜖0 𝜇 0
𝜇0
𝑁𝑠 2
−7
= 1 ∙ 10
4𝜋
𝐶2
𝜇0 - the vacuum permeability
-
a glimpse of a deep sub-structure
connecting E and B
9
Iclicker Question
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.
𝜇0 𝑣⃗ × 𝑟̂
𝐵 𝑟 =
𝑞 2
4𝜋
𝑟
10
Iclicker Question
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.
𝜇0 𝑣⃗ × 𝑟̂
𝐵 𝑟 =
𝑞 2
4𝜋
𝑟
11
Iclicker Question
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

v
+q

v
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
𝜇0 𝑣⃗ × 𝑟⃗
𝐵 𝑟 =
𝑞 3
4𝜋
𝑟
𝐹⃗ = 𝑞𝑣⃗ × 𝐵
12
Iclicker Question
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

v
+q

v
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
𝜇0 𝑣⃗ × 𝑟⃗
𝐵 𝑟 =
𝑞 3
4𝜋
𝑟
𝐹⃗ = 𝑞𝑣⃗ × 𝐵
13
Magnetic Field of a Moving Charge (cont’d)
Example: a charge of 10 nC is moving in a straight line at 30 m/s. What is the
max magnitude of B produced by the charge at a point 10 cm from its path?
What is the max magnitude of E?
𝐵𝑚𝑚𝑚
30
𝜇0 𝑞𝑣
−7
−8
−12 𝑇
=
=
10
∙
10
=
3
∙
10
0.12
4𝜋 𝑟 2
𝐸𝑚𝑚𝑚
10−8
1 𝑞
10
4 𝑁/𝐶
=
≅
10
=
10
0.12
4𝜋𝜖0 𝑟 2
𝐵𝑚𝑚𝑚
𝑣
= 𝜇0 𝜖0 𝑣 = 2
𝐸𝑚𝑚𝑚
𝑐
Magnetic force can be regarded as a relativistic correction to
the electrical force.
14
Magnetic Field of Currents
Superposition:
𝑣⃗𝑑 × 𝑟̂
𝜇0
𝐵 𝑟 =
𝑞𝑠𝑠𝑠
4𝜋
𝑟⃗ 2
1820 – first observation of the
magnetic field due to currents
The charge of carriers in a wire segment 𝐴𝐴𝐴:
the current segment
is at the origin
𝑞𝑠𝑠𝑠 = ∑𝑖 𝑞𝑖 = 𝑛𝑛𝑛𝑛𝑛
𝐼 = 𝑛𝑛𝑣𝑑 𝐴
𝑞𝑠𝑠𝑠 𝑣⃗𝑑 = 𝐼𝐼𝑙⃗
𝜇0 𝑑𝑙⃗ × 𝑟̂
𝐵 𝑟 =
𝐼
4𝜋
𝑟2
- proportional to 1/𝑟 2 , as the electric
field of a point charge
15
Magnetic Field of a Circular Loop with Current
𝑅𝑟̂
O
𝑑𝑙⃗
Find the magnetic field at the center of a circular loop
with current.
All contribution – in the
same direction. Total 𝐵:
Contribution of a small
portion of the loop:
𝜇0 𝑑𝑙⃗ × 𝑟̂
𝛿𝐵 𝑟 =
𝐼
4𝜋
𝑅2
𝜇0 ∑ 𝑑𝑑
� 𝛿𝐵𝑖 =
𝐼 2
4𝜋 𝑅
� 𝑑𝑑 = 2𝜋𝑅
The field at the center of the loop:
𝜇0 𝐼
𝐵 0 =
2𝑅
For N turns with current, the field is N times
stronger.
16
Iclicker Question
17
Iclicker Question
18
Iclicker Question
A wire consists of two straight
sections with a semicircular
section between them. If
current flows in the wire as
shown, what is the direction
of the magnetic field at P due
to the current?
A. to the right
B. to the left
C. out of the plane of the figure
D. into the plane of the figure
E. misleading question — the magnetic field at P is zero
19
Iclicker Question
A wire consists of two straight
sections with a semicircular
section between them. If
current flows in the wire as
shown, what is the direction
of the magnetic field at P due
to the current?
A. to the right
B. to the left
C. out of the plane of the figure
D. into the plane of the figure
E. misleading question — the magnetic field at P is zero
20
Magnetic field of an infinitely long straight wire
0 𝐵 0
𝑟
Find the magnetic field at a distance 𝑟 from an infinitely
long straight wire with current 𝐼.
Contribution of a small
portion of the wire:
𝐼𝑑𝑙⃗
All contribution – in the same direction. Total 𝐵:
𝜇0 𝐼
𝐵 𝑟 =
2𝜋𝑟
𝜇0 𝑑𝑙⃗ × 𝑟̂
𝛿𝐵 𝑟 =
𝐼
4𝜋
𝑟2
∑ 𝛿𝐵𝑖 (see Appendix)
(the r-dependence resembles that for E(r) of an infinite charged wire)
This equation will be obtained next time using Ampere’s Law.
21
Problem
Consider a wire bent in the hairpin shape. The
wire carries a current 𝐼 . What is the
approximate magnitude of the magnetic field
at point a?
𝑅
𝐼
𝑎
𝐼
Superposition: the field at point a is a superposition of three 𝐵 fields
produced by the current in the semi-circle and two straight wires.
Semi-circle:
Top wire:
Bottom wire:
𝐵 𝑟 =
1 𝜇0 𝐼 𝜇0 𝐼
=
4𝑅
2 2𝑅
𝜇0 𝐼
𝐵 𝑟 =
4𝜋𝑅
𝜇0 𝐼
𝐵 𝑟 =
4𝜋𝑅
(1/2 of the field of a circular loop)
(1/2 of the field of an infinite wire)
(1/2 of the field of an infinite wire)
𝜇0 𝐼 𝜇0 𝐼
𝜇0 𝐼
2
𝐵 𝑟 =
+
=
1+
4𝑅 2𝜋𝑅 4𝑅
𝜋
22
𝐼2𝐼
𝐼1
Interaction between Two Wires with Current
𝐼2
𝐼1
The magnetic field due
to 𝐼1 at the position of
the wire with 𝐼2
Force per unit length:
𝜇0 𝐼1
𝐵 𝑟 =
2𝜋𝑟
𝐹
𝜇0 𝐼1 𝐼2
= 𝐼2 𝐵 =
𝐿
2𝜋𝑟
Parallel (anti-parallel) currents attract (repel)
each other.
SI unit and definition for electric current: The ampere is that constant current
which, if maintained in two straight parallel conductors of infinite length, of
negligible circular cross section, and placed 1 meter apart in vacuum, would
produce between these conductors a force equal to 2×10-7 newton per meter
of length.
23
Iclicker Question
The long, straight wire AB carries a 10-A
current as shown. The rectangular loop
has long edges parallel to AB and carries
a clockwise 5-A current.
What is the magnitude and direction of
the net magnetic force that the straight
wire AB exerts on the loop?
𝐹𝑡𝑡𝑡
A.
C.
𝐼2 = 5𝐴
𝜇0
2000
2𝜋
N to the right
N upward (toward AB)
B.
D.
0.1 𝑚
1𝑚
𝜇0 𝐼1 𝐼2 1 1
=𝐿
−
2𝜋 𝑟1 𝑟2
𝜇0
5000
2𝜋
𝐼1 = 10𝐴
0.02 𝑚
𝜇0
2000
2𝜋
𝜇0
5000
2𝜋
N to the left
N downward (away from AB)
E. misleading question — the net magnetic force is zero
24
Iclicker Question
The long, straight wire AB carries a 14-A
current as shown. The rectangular loop
has long edges parallel to AB and carries
a clockwise 5-A current.
What is the magnitude and direction of
the net magnetic force that the straight
wire AB exerts on the loop?
𝐹𝑡𝑡𝑡
A.
C.
𝐼2 = 5𝐴
𝜇0
2000
2𝜋
N to the right
N upward (toward AB)
B.
D.
0.1 𝑚
1𝑚
𝜇0 𝐼1 𝐼2 1 1
=𝐿
−
2𝜋 𝑟1 𝑟2
𝜇0
5000
2𝜋
𝐼1 = 10𝐴
0.02 𝑚
𝜇0
2000
2𝜋
𝜇0
5000
2𝜋
N to the left
N downward (away from AB)
E. misleading question — the net magnetic force is zero
25
Electrostatics vs. Magnetostatics
Elementary source of the static B
field: current segment (onedimensional object, vector)
Elementary source of the
static E field: point charge
(zero-dimensional object,
scalar)
Gauss’ Law:
�
𝑠𝑢𝑢𝑢𝑢𝑢𝑢
𝐸 𝑟⃗ ∙ 𝑑𝐴⃗ =
𝑞𝑒𝑒𝑒𝑒
𝜀0
Absence of
magnetic
monopoles:
𝑙𝑙𝑙𝑙
- valid in electrostatics
only(!), will be modified in
electrodynamics.
𝑠𝑢𝑢𝑢𝑢𝑢𝑢
𝐵 𝑟⃗ ∙ 𝑑𝐴⃗ = 0
- valid in electrodynamics!
- valid in electrodynamics!
� 𝐸 𝑟⃗ ∙ 𝑑𝑙⃗ = 0
�
𝐵 𝑟 =
𝜇0 𝐼
2𝜋𝑟
� 𝐵 𝑟⃗ ∙ 𝑑𝑙⃗ = 𝜇0 𝐼 ≠ 𝟎
𝑙𝑙𝑙𝑙
circulation of the field B
along the loop
𝐵 𝑟 ∝
For a loop that doesn’t enclose
any current, the circulation is 0.
In magnetostatics, currents are the only
source of the non-zero circulation of B. This
will be modified in electrodynamics.
1
𝑟
Next time: Lecture 16: Ampere’s Law.
§§ 28.6- 28.8
27
Appendix I: Magnetic Force as a Relativistic Correction to the Electric Force
In general, electrical repulsion + magnetic attraction.
𝑟
1 𝑞2
𝐹𝐸 =
4𝜋𝜖0 𝑟 2
Charges at rest (the proton ref. frame),
only electric force (repulsion):
Charges in motion (the lab. ref. frame, primed), both FE and FB:
According to the special theory of relativity:
𝐹⊥′
1
= 𝐹⊥
𝛾
𝐸⊥′ = 𝛾𝐸⊥
𝐹⊥′
- force in the lab reference frame
𝐹⊥ - force in the proton reference frame
𝐵⊥′ = 𝛾𝐵⊥
′
′
𝐹⊥𝐵
= 𝐹⊥′ − 𝐹⊥𝐸
𝛾=
1
𝐹⃗⊥ ′ = 𝐹⃗⊥𝐸 ′ + 𝐹⃗⊥𝐵 ’
𝑣
1−
𝑐
2
1
𝛾
𝑐 𝑣
In the lab ref. frame:
𝐹⊥′
1 1 𝑞2
=
𝛾 4𝜋𝜖0 𝑟 2
1
1 𝑞2
𝑣
= 𝛾−
=
𝛾
𝑐
𝛾 4𝜋𝜖0 𝑟 2
2
1 𝑞2
𝐹⊥𝐸 ′ = 𝛾
4𝜋𝜖0 𝑟 2
1 𝑞2
𝜇0 𝑞𝑞
=
𝑞𝑞𝛾
= 𝑞𝑞𝐵⊥′
2
2
4𝜋 𝑟
4𝜋𝜖0 𝑟
Magnetic force could be regarded as a relativistic correction to the electrical force.
28
Appendix II: Integral Form of the Magnetic Field of Currents
𝑜
𝑟𝑟
𝑑𝑙⃗
𝑟⃗
𝐵 𝑟
𝐼
The charge of carriers in a wire segment
𝑞𝑠𝑠𝑠 = ∑𝑖 𝑞𝑖 = 𝑛𝑛𝑛𝑛𝑛
The current:
𝐼 = 𝐽𝐴 = 𝑛𝑛𝑣𝑑 𝐴
𝐴𝐴𝐴 :
𝑞𝑠𝑠𝑠 𝑣⃗𝑑 = 𝐼𝐼𝑙⃗
𝐽- current density, 𝐼𝑑𝑙⃗ - current segment
𝑣⃗𝑑 × 𝑟⃗ − 𝑟𝑟
𝜇0
𝜇0 𝑑𝑙⃗ × 𝑟⃗ − 𝑟𝑟
Superposition: 𝐵 𝑟 = 𝑞𝑠𝑠𝑠
=
𝐼
3
3
4𝜋
4𝜋
𝑟⃗ − 𝑟𝑟
𝑟⃗ − 𝑟𝑟
Assuming the
current segment
is at the origin:
The magnetic field
of a wire loop with
current:
𝜇0 𝑑𝑙⃗ × 𝑟⃗ 𝜇0 𝑑𝑙⃗ × 𝑟̂
𝐵 𝑟 =
𝐼
=
𝐼
4𝜋
𝑟3
4𝜋
𝑟2
- proportional to 1/𝑟 2 , as an electric
field of a point charge
𝑑𝑙⃗ × 𝑟⃗ − 𝑟𝑟
𝜇0
𝐵 𝑟 =
𝐼�
3
4𝜋
𝑟⃗ − 𝑟𝑟
29
Appendix III: Magnetic Field of a Circular Loop with Current
Find the magnetic field at the center of a circular
loop with current.
O
𝑅
𝛼
𝑑𝑙⃗
𝑑𝑙⃗ × 𝑟⃗ − 𝑟𝑟
𝜇0
𝐵 𝑟 =
𝐼�
3
4𝜋
𝑟⃗ − 𝑟𝑟
Let’s place the origin at the center of the loop 𝑟 = 0 .
𝑟⃗ − 𝑟𝑟 = 𝑅
𝑑𝑙 = 𝑅 𝑑𝛼, 0 ≤ 𝛼 ≤ 2𝜋
2𝜋
𝜇0
𝑑𝑙
𝜇0 𝐼
𝜇0 𝐼
𝐵 0 =
𝐼� 2 =
� 𝑑𝑑 =
2𝜋
4𝜋
4𝜋 𝑅
4𝜋 𝑅
𝑅
0
The field at the center of the loop:
𝜇0 𝐼
𝐵 0 =
2𝑅
For N turns with current, the field is N times stronger.
30
Appendix IV: Magnetic Field of a Circular Loop with Current
𝑑𝑙⃗
𝛼
𝑟⃗ = 𝑥 ∙ 𝑥�
𝑟𝑟
𝑟⃗
𝑅
𝑟⃗ − 𝑟𝑟 =
A wire loop with current has a shape of a
circle of radius a. Find the magnetic field at
a distance x from its center along the axis of
symmetry.
𝑑𝑙⃗ × 𝑟⃗ − 𝑟𝑟
𝜇0
𝐵 𝑟 =
𝐼�
3
4𝜋
𝑟⃗ − 𝑟𝑟
𝑅2
+
𝑥2
Let’s place the origin at the center of the
loop and introduce two angles: 𝛼 and 𝜃.
𝑑𝑙 = 𝑅 𝑑𝛼, 0 ≤ 𝛼 ≤ 2𝜋
𝜇0
𝑑𝑙
𝜇0
𝑅
𝐵𝑥 𝑥 =
𝐼� 2
𝑐𝑐𝑐𝑐
=
𝐼
4𝜋
4𝜋 𝑅2 + 𝑥 2
𝑅 + 𝑥2
𝐵𝑦 component is zero due to the symmetry.
The field at the center of the loop (x=0):
2𝜋
𝑐𝑐𝑐𝜃 =
𝑅
𝑅2 + 𝑥 2
𝜇0
𝑅2
� 𝑅 𝑑𝑑 = 𝐼 2
3/2
2 𝑅 + 𝑥2
0
𝜇0 𝐼
𝐵 0 =
2𝑅
For N turns with current, the field is N times stronger.
3/2
31
Appendix V: Magnetic Field of a Straight Wire Segment
𝑜 𝐵 𝑟⃗ = 0
𝑟
𝛼
𝑟𝑟
Find 𝐵 at a distance 𝑟 from a straight wire segment carrying 𝐼.
Let’s choose the origin at the point where we measure 𝐵 (𝑟⃗ = 0):
𝑑𝑙⃗ × −𝑟𝑟
𝜇0
𝐵 𝑟⃗ = 0 =
𝐼�
3
4𝜋
−𝑟𝑟
𝑑𝑙⃗
1
Express 𝑑𝑙⃗ and 𝑟𝑟 in terms of 𝑟 and 𝛼:
𝑙 = 𝑟 tan𝛼 𝑑𝑑 = 𝑟
𝑑𝛼
cos 2 𝛼
90 + 𝛼
𝑑𝑙⃗ × −𝑟𝑟
𝐵 𝑟 =
𝑎
𝑜
𝛼2
𝑟
𝑟𝑟 =
cos𝛼
2 𝑑𝑑
𝑟𝑑𝑑
𝑟
= 𝑑𝑙 ∙ 𝑟 ′ ∙ sin 90 + 𝛼 = 𝑑𝑑 ∙ 𝑟 ′ ∙ cos 𝛼 =
∙𝑟 =
cos 2 𝛼
cos 2 𝛼
2
3
𝛼2
cos 𝛼
𝜇0
𝑟
𝜇0 𝐼
𝜇0 𝐼
𝐼 �
𝑑𝛼
=
�
cos𝛼
𝑑𝑑
=
sin𝛼2 − sin𝛼1
2
3
cos 𝛼 𝑟
4𝜋𝑟
4𝜋𝑟
4𝜋
𝛼1
𝛼1
Example: Magnetic field of a square loop with current at the center
𝜋
𝜋
of the loop (𝛼2 = , 𝛼1 = − ):
4
𝐼
4
𝜇0 𝐼
𝜋
4𝜇0 𝐼
𝐵 𝑟 = � 𝐵𝑖 = 4
2sin =
4𝜋𝜋/2
4
2𝜋𝜋
𝑖
32
Magnetic field of an infinitely long straight wire
𝑜 𝐵 𝑟⃗ = 0
𝑟
𝛼
𝑟𝑟
90 + 𝛼
𝑑𝑙⃗
𝜇0 𝐼
𝐵 𝑟 =
sin𝛼2 − sin𝛼1
4𝜋𝑟
𝜋
2
Infinitely long straight wire: 𝛼2 = , 𝛼1 = −
𝜇0 𝐼
𝐵 𝑟 =
2𝜋𝑟
(the r-dependence resembles that for E(r) of an infinite charged wire)
33
𝜋
2
Problem
Consider a wire bent in the hairpin shape. The
wire carries a current 𝐼 . What is the
approximate magnitude of the magnetic field
at point a?
𝑅
𝐼
𝑎
𝐼
Superposition: the field at point a is a superposition of three 𝐵 fields
produced by the current in the semi-circle and two straight wires.
Semi-circle:
Top wire:
𝐵 𝑟 =
𝜇0 𝐼
𝐵 𝑟 =
4𝜋𝑅
Bottom wire: 𝐵 𝑟 =
1 𝜇0 𝐼 𝜇0 𝐼
=
4𝑅
2 2𝑅
𝛼2 =0
�
𝛼1 =−𝜋/2
𝛼2 =𝜋/2
𝜇0 𝐼
�
4𝜋𝑅
𝛼1 =0
(1/2 of the field of a circular loop)
𝜇0 𝐼
cos𝛼 𝑑𝑑 =
sin0 − sin −𝜋/2
4𝜋𝑅
𝜇0 𝐼
=
4𝜋𝑅
(1/2 of the field of an infinite wire)
𝜇0 𝐼
cos𝛼 𝑑𝑑 =
4𝜋𝑅
𝐵 𝑟 =
𝜇0 𝐼 𝜇0 𝐼
𝜇0 𝐼
2
+
=
1+
4𝑅 2𝜋𝑅 4𝑅
𝜋
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