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
Iclicker Question
A circular loop of wire carries a constant current. If the loop is
placed in a region of uniform magnetic field, the net magnetic
force on the loop is
A. perpendicular to the plane of the loop, in a direction given
by a right-hand rule.
B. perpendicular to the plane of the loop, in a direction given
by a left-hand rule.
C. in the same plane as the loop.
D. zero.
E. The answer depends on the magnitude and direction of
the current and on the magnitude and direction of the
magnetic field.
2
Iclicker Question
A circular loop of wire carries a constant current. If the loop is
placed in a region of uniform magnetic field, the net magnetic
force on the loop is
A. perpendicular to the plane of the loop, in a direction given
by a right-hand rule.
B. perpendicular to the plane of the loop, in a direction given
by a left-hand rule.
C. in the same plane as the loop.
D. zero.
E. The answer depends on the magnitude and direction of
the current and on the magnitude and direction of the
magnetic field.
3
Iclicker Question
𝜇 = 𝑁𝐴𝐼
3 turns
𝜏 =𝜇×𝐵
A. 4.510-2 Nm into board
B. 4.510-2 Nm out of board
C. 1.510-2 Nm up
D. 1.510-2 Nm down
E. Zero
4
Iclicker Question
𝜇 = 𝑁𝐴𝐼
3 turns
𝜏 =𝜇×𝐵
B. 4.510-2 Nm out of board
𝜏 = 3 ∙ 𝜋 0.05 2 ∙ 5 ∙ 1.2
= 4.5 ∙ 10−2 𝑁𝑚
C. 1.510-2 Nm up
Direction: upright = into board
A. 4.510-2 Nm into board
D. 1.510-2 Nm down
E. Zero
5
Iclicker Question
6
Iclicker Question
7
Magnetic Field of a Moving Charge
Positive charge moving with a constant velocity 𝒗 ≪ 𝒄:
𝑐 – the speed of light
𝜇0 𝑣 × 𝑟 𝜇0 𝑣 × 𝑟
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
the vacuum permeability
𝐵
𝑐2 =
1
𝜖0 𝜇 0
- a glimpse of a deep
sub-structure connecting E and B
8
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 𝑣 × 𝑟
𝐵 𝑟 =
𝑞 2
4𝜋
𝑟
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 𝑣 × 𝑟
𝐵 𝑟 =
𝑞 2
4𝜋
𝑟
10
Iclicker Question
+q
v
+q
v
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?
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 𝑣 × 𝑟
𝐵 𝑟 =
𝑞 3
4𝜋
𝑟
𝐹 = 𝑞𝑣 × 𝐵
11
Iclicker Question
+q
v
+q
v
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?
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 𝑣 × 𝑟
𝐵 𝑟 =
𝑞 3
4𝜋
𝑟
𝐹 = 𝑞𝑣 × 𝐵
12
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?
𝐵𝑚𝑎𝑥
𝜇0 𝑞𝑣
30
−7
−8
−12
=
=
10
∙
10
=
3
∙
10
𝑇
4𝜋 𝑟 2
0.12
𝐸𝑚𝑎𝑥
1 𝑞
10−8
10
4 𝑁/𝐶
=
≅
10
=
10
4𝜋𝜖0 𝑟 2
0.12
𝐵𝑚𝑎𝑥
𝑣
= 𝜇0 𝜖0 𝑣 = 2
𝐸𝑚𝑎𝑥
𝑐
Magnetic force could be regarded as a relativistic correction to
the electrical force.
13
Magnetic Field of Currents
𝐼
𝑂
Superposition:
𝐵 𝑟 =
𝑑𝑙
𝜇0
𝑣𝑑 × 𝑟
𝑞𝑠𝑒𝑔
4𝜋
𝑟2
the current segment is at the origin
𝐵 𝑟
1820 – first observation of the
magnetic field due to currents
The charge of carriers in a wire segment 𝐴𝑑𝑙:
𝑞𝑠𝑒𝑔 =
𝐼 = 𝑛𝑒𝑣𝑑 𝐴
𝑖 𝑞𝑖
= 𝑛𝑒𝐴𝑑𝑙
𝑞𝑠𝑒𝑔 𝑣𝑑 = 𝐼𝑑𝑙
𝜇0 𝑑𝑙 × 𝑟
𝐵 𝑟 =
𝐼
4𝜋
𝑟2
- proportional to 1/𝑟 2 , as the electric
field of a point charge
The magnetic field of a wire loop
with current:
𝜇0
𝐵 𝑟 =
𝐼
4𝜋
𝑑𝑙 × 𝑟 − 𝑟′
𝑟 − 𝑟′
3
14
Iclicker Question
𝜇0
𝐵 𝑟 =
𝐼
4𝜋
𝑑𝑙 × 𝑟 − 𝑟′
𝑟 − 𝑟′
3
15
Iclicker Question
𝜇0
𝐵 𝑟 =
𝐼
4𝜋
𝑑𝑙 × 𝑟 − 𝑟′
𝑟 − 𝑟′
3
16
Magnetic Field of a Circular Loop with Current
Find the magnetic field at the center of a circular
loop with current.
𝜇0
𝐵 𝑟 =
𝐼
4𝜋
𝑑𝑙
𝑅
O
𝛼
𝑑𝑙 × 𝑟 − 𝑟′
𝑟 − 𝑟′
3
Let’s place the origin at the center of the loop 𝑟 = 0 .
𝑟 − 𝑟′ = 𝑅
𝜇0
𝐵 0 =
𝐼
4𝜋
𝑑𝑙 = 𝑅 𝑑𝛼, 0 ≤ 𝛼 ≤ 2𝜋
𝑑𝑙
𝜇0 𝐼
=
𝑅2 4𝜋 𝑅
The field at the center of the loop:
2𝜋
0
𝜇0 𝐼
𝑑𝛼 =
2𝜋
4𝜋 𝑅
𝜇0 𝐼
𝐵 0 =
2𝑅
For N turns with current, the field is N times stronger.
17
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?
𝜇0
𝐵 𝑟 =
𝐼
4𝜋
𝑑𝑙 × 𝑟 − 𝑟′
𝑟 − 𝑟′
3
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
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?
𝜇0
𝐵 𝑟 =
𝐼
4𝜋
𝑑𝑙 × 𝑟 − 𝑟′
𝑟 − 𝑟′
3
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
Magnetic Field of a Straight Wire Segment
𝑜 𝐵 𝑟=0
Find 𝐵 at a distance 𝑟 from a straight wire segment carrying 𝐼.
𝛼
𝑟
Let’s choose the origin at the point where we measure 𝐵 (𝑟 = 0):
𝑟′
90 + 𝛼
𝑑𝑙
Express 𝑑𝑙 and 𝑟′ in terms of 𝑟 and 𝛼:
𝑑𝑙 × −𝑟′
𝜇0
𝐵 𝑟 =
𝐼
4𝜋
𝛼2
𝛼1
𝜇0
𝐵 𝑟=0 =
𝐼
4𝜋
𝑑𝑙 × −𝑟′
3
−𝑟′
1
𝑑𝑙 = 𝑟
𝑑𝛼
cos2 𝛼
𝑙 = 𝑟 tan𝛼
𝑟
𝑟′ =
cos𝛼
2 𝑑𝛼
𝑟𝑑𝛼
𝑟
= 𝑑𝑙 ∙ 𝑟 ′ ∙ sin 90 + 𝛼 = 𝑑𝑙 ∙ 𝑟 ′ ∙ cos 𝛼 =
∙𝑟 =
cos 2 𝛼
cos 2 𝛼
𝑟2
cos 3 𝛼
cos 2 𝛼
𝑟3
𝜇0 𝐼
𝑑𝛼 =
4𝜋𝑟
𝛼2
𝛼1
𝜇0 𝐼
cos𝛼 𝑑𝛼 =
sin𝛼2 − sin𝛼1
4𝜋𝑟
Example: Magnetic field of a square loop with current at the center
𝜋
𝜋
of the loop (𝛼2 = , 𝛼1 = − ):
4
𝑎
𝑜
𝐵 𝑟 =
𝐼
𝑖
4
𝜇0 𝐼
𝜋
4𝜇0 𝐼
𝐵𝑖 = 4
2sin =
4𝜋𝑎/2
4
2𝜋𝑎
20
Magnetic field of an infinitely long straight wire
𝑜 𝐵 𝑟=0
𝛼
𝑟
𝑟′
90 + 𝛼
𝑑𝑙
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)
21
𝜋
2
Problem
𝐼
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
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 𝜇 0 𝐼
=
2 2𝑅
4𝑅
𝛼2 =0
𝛼1 =−𝜋/2
𝜇0 𝐼
4𝜋𝑅
𝛼2 =𝜋/2
𝛼1 =0
(1/2 of the field of a circular loop)
𝜇0 𝐼
cos𝛼 𝑑𝛼 =
sin0 − sin −𝜋/2
4𝜋𝑅
𝜇0 𝐼
=
4𝜋𝑅
(1/2 of the field of an infinite wire)
𝜇0 𝐼
cos𝛼 𝑑𝛼 =
4𝜋𝑅
𝐵 𝑟 =
𝜇0 𝐼 𝜇 0 𝐼
𝜇0 𝐼
2
+
=
1+
4𝑅 2𝜋𝑅 4𝑅
𝜋
22
Interaction between Two Wires with Current
𝐼2
𝐼2𝐼
𝐼1
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?
𝐹𝑡𝑜𝑡
A.
C.
𝐼1 = 10𝐴
0.02 𝑚
𝐼2 = 5𝐴
0.1 𝑚
1𝑚
𝜇0 𝐼1 𝐼2 1 1
=𝐿
−
2𝜋 𝑟1 𝑟2
𝜇0
5000
2𝜋
𝜇0
2000
2𝜋
N to the right
N upward (toward AB)
B.
D.
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?
𝐹𝑡𝑜𝑡
A.
C.
𝐼1 = 10𝐴
0.02 𝑚
𝐼2 = 5𝐴
0.1 𝑚
1𝑚
𝜇0 𝐼1 𝐼2 1 1
=𝐿
−
2𝜋 𝑟1 𝑟2
𝜇0
5000
2𝜋
𝜇0
2000
2𝜋
N to the right
N upward (toward AB)
B.
D.
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 E field: point charge
(zero-dimensional object,
scalar)
Gauss’ Law:
𝐸 𝑟 ∙ 𝑑𝐴 =
𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝑞𝑒𝑛𝑐𝑙
𝜀0
Elementary source of the static B
field: current segment (onedimensional object, vector)
Absence of
magnetic
monopoles:
𝐵 𝑟 =
𝑙𝑜𝑜𝑝
- valid in electrostatics
only(!), will be modified in
electrodynamics.
𝑠𝑢𝑟𝑓𝑎𝑐𝑒
- valid in electrodynamics!
- valid in electrodynamics!
𝐸 𝑟 ∙ 𝑑𝑙 = 0
𝐵 𝑟 ∙ 𝑑𝐴 = 0
𝜇0 𝐼
2𝜋𝑟
𝐵 𝑟 ∙ 𝑑𝑙 = 𝜇0 𝐼 ≠ 𝟎
𝑙𝑜𝑜𝑝
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
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):
𝐹⊥ ′ = 𝐹⊥𝐸 ′ + 𝐹⊥𝐵 ’
Charges in motion (the lab. ref. frame, primed), both FE and FB:
According to the special theory of relativity:
𝐹⊥′
1
= 𝐹⊥
𝛾
𝐹⊥′
- force in the lab reference frame
𝛾=
𝐹⊥ - force in the proton reference frame
𝐸⊥′ = 𝛾𝐸⊥
𝐵⊥′ = 𝛾𝐵⊥
′
′
𝐹⊥𝐵
= 𝐹⊥′ − 𝐹⊥𝐸
𝛾
1
𝑣
1− 𝑐
2
1
𝑐 𝑣
In the lab ref. frame:
𝐹⊥′
1 1 𝑞2
=
𝛾 4𝜋𝜖0 𝑟 2
1
1 𝑞2
𝑣
= 𝛾−
=𝛾
𝛾 4𝜋𝜖0 𝑟 2
𝑐
2
1 𝑞2
𝐹⊥𝐸 ′ = 𝛾
4𝜋𝜖0 𝑟 2
1 𝑞2
𝜇0 𝑞𝑣
= 𝑞𝑣𝛾
= 𝑞𝑣𝐵⊥′
2
2
4𝜋𝜖0 𝑟
4𝜋 𝑟
Magnetic force could be regarded as a relativistic correction to the electrical force.
28
Magnetic Field of Currents
𝐼
𝑑𝑙
𝑟′
𝑜
𝑟
The charge of carriers in a wire segment 𝐴𝑑𝑙:
𝑞𝑠𝑒𝑔 = 𝑖 𝑞𝑖 = 𝑛𝑒𝐴𝑑𝑙
The current:
𝐼 = 𝐽𝐴 = 𝑛𝑒𝑣𝑑 𝐴
𝑞𝑠𝑒𝑔 𝑣𝑑 = 𝐼𝑑𝑙
𝐽- current density, 𝐼𝑑𝑙 - current segment
𝐵 𝑟
1820 – first observation of the
magnetic field due to currents
𝑣𝑑 × 𝑟 − 𝑟′
𝜇0
𝜇0 𝑑𝑙 × 𝑟 − 𝑟′
Superposition: 𝐵 𝑟 = 𝑞𝑠𝑒𝑔
=
𝐼
3
3
4𝜋
4𝜋
𝑟 − 𝑟′
𝑟 − 𝑟′
Assuming the
current segment
is at the origin:
𝐵 𝑟 =
𝜇0 𝑑𝑙 × 𝑟 𝜇0 𝑑𝑙 × 𝑟
𝐼
=
𝐼
4𝜋
𝑟3
4𝜋
𝑟2
- proportional to 1/𝑟 2 , as an electric
field of a point charge
The magnetic field
of a wire loop with
current:
𝜇0
𝐵 𝑟 =
𝐼
4𝜋
𝑑𝑙 × 𝑟 − 𝑟′
𝑟 − 𝑟′
3
𝜇0 𝐼
2𝜋𝑟
- proportional to 1/𝑟, as E(r) of an infinite charged wire)
The magnetic field at a distance r from a
straight wire with current :
𝐵 𝑟 =
29
Magnetic Field of a Circular Loop with Current
𝑑𝑙
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.
𝑑𝑙 × 𝑟 − 𝑟′
𝜇0
𝐵 𝑟 =
𝐼
3
4𝜋
𝑟 − 𝑟′
𝑟′
𝛼
𝑟
𝑅
Let’s place the origin at the center of the
loop and introduce two angles: 𝛼 and 𝜃.
𝑟 =𝑥∙𝑥
𝑟 − 𝑟′ =
𝜇0
𝐵𝑥 𝑥 =
𝐼
4𝜋
𝑅2
+
𝑥2
𝑑𝑙 = 𝑅 𝑑𝛼, 0 ≤ 𝛼 ≤ 2𝜋
𝑑𝑙
𝜇0
𝑅
𝑐𝑜𝑠𝜃
=
𝐼
𝑅2 + 𝑥 2
4𝜋 𝑅2 + 𝑥 2
2𝜋
3/2
0
𝑐𝑜𝑠𝜃 =
𝑅
𝑅2 + 𝑥 2
𝜇0
𝑅2
𝑅 𝑑𝛼 = 𝐼 2
2 𝑅 + 𝑥2
3/2
𝐵𝑦 component is zero due to the symmetry.
The field at the center of the loop (x=0):
𝐵 0 =
𝜇0 𝐼
2𝑅
For N turns with current, the field is N times stronger.
30