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Announcements
 Exam 2 is one week from tomorrow.
Contact me by the end of the Wednesday’s lecture if you
have special circumstances different than for exam 1.
 Exam 2 will cover chapters 24.3-27.
 Physics 2135 Test Rooms, Spring 2016:
Instructor
Dr. Hale
Dr. Kurter
Dr. Madison
Mr. Noble
Dr. Parris
Mr. Upshaw
Sections
G, J
D, F
H, N
B, E
K, M
A, C, L
Special Accommodations
Exam is from
5:00-6:00 pm!
Room
125 BCH
104 Physics
199 Toomey
B-10 Bert.
G-31 EECH
G-3 Schrenk
Testing Center
Know the exam time!
Find your room ahead of time!
If at 5:00 on test day you are lost, go to 104 Physics and check the exam
room schedule, then go to the appropriate room and take the exam there.
We’ve been working with the effects of magnetic fields
without considering where they come from. Today we
learn about sources of magnetic fields.
Today’s agenda:
Magnetic Fields Due To A Moving Charged Particle.
You must be able to calculate the magnetic field due to a moving charged particle.
Biot-Savart Law: Magnetic Field due to a Current Element.
You must be able to use the Biot-Savart Law to calculate the magnetic field of a currentcarrying conductor (for example: a long straight wire).
Force Between Current-Carrying Conductors.
You must be able to begin with starting equations and calculate forces between currentcarrying conductors.
Magnetic Field of a Moving Charged Particle
Let’s start with the magnetic field of a moving charged particle.
r
It is experimentally
observed that a moving
point charge q gives rise
to a magnetic field.
B
r̂
+
v
μ0 qv  rˆ
B=
.
2
4π r
0 is a constant, and its value is
0=4x10-7 T·m/A

Remember: the direction of r is always from the source point (the thing that
causes the field) to the field point (the location where the field is being measured.
cross products of unit vectors
We are going to be doing lots of cross products of unit vectors.
Here are some handy ways to do the cross product.
The “always works if you can do math” method
C = A B
Example:

 ˆi


C = det  A x

B

 x
ˆj
Ay
By

 ˆi ˆj
ˆk  ˆj = det  0 0


 0 -1


 
kˆ  

Av z  
B z  

kˆ  
 ˆ
1   = i  0 -  -1   = ˆi
0  

cross products of unit vectors
We are going to be doing lots of cross products of unit vectors.
Here are some handy ways to do the cross product.
The right hand rule method
link to image
http://en.wikipedia.org/wiki/Cross_product
cross products of unit vectors
We are going to be doing lots of cross products of unit vectors.
Here are some handy ways to do the cross product.
The “I learned this decades ago and forgot the name for it”
method
i j k i j k
ˆi  ˆj =kˆ
cross products of unit vectors
i j k i j k
ˆj  ˆi = -kˆ
i j k i j k
ˆi  kˆ = -jˆ
i j k i j k
 
ˆj  kˆ = - -iˆ = ˆi
Example: proton 1 has a speed v0 (v0<<c) and is moving along
the x-axis in the +x direction. Proton 2 has the same speed and
is moving parallel to the x-axis in the –x direction, at a distance
r directly above the x-axis. Determine the electric and magnetic
forces on proton 2 at the instant the protons pass closest to
each other.
y
This is example 28.1 in your text.
FE
Skip to next slide.
The electric force is
v0
1 q1q2 ˆ
FE =
r
2
4  r
E
r
1
2
1 e ˆ
FE =
j
2
4  r
Homework Hint: this and the next 3 slides!
2
z
r̂
v0
x
Alternative approach to calculating electric force. This is
“better” because we use the concept of field to calculate both
of electric and (later) magnetic forces.
At the position of proton 2 there is an electric field due to
proton 1.
y
1 q1 ˆ
1 eˆ
E1 =
r=
j
2
2
4   r
4   r
FE E
v0
This electric field exerts a force
on proton 2.
1 eˆ
1 e2 ˆ
FE = qE1 = e
j=
j
2
2
4  r
4  r
2
r
1
z
r̂
v0
x
To calculate the magnetic force: at the position of proton 2
there is a magnetic field due to proton 1.
 q1 v1  rˆ
B1 =
4 r 2
 ev 0 ˆi  ˆj
B1 =
4 r2
y
FE
v0
  ev 0 ˆ
B1 =
k
2
4 r
2
B1
r
1
z
r̂
v0
x
Proton 2 “feels” a magnetic force due to the magnetic field of
proton 1.
FB = q2 v2  B1
FB = ev 0
 
   ev 0 ˆ 
ˆ
i  
k
2
 4 r

y
FE
v0
 e2 v 02 ˆ
FB =
j
2
4 r
2
FB
B1
What would proton 1 “feel?”
r
1
Caution! Relativity overrules Newtonian mechanics!
However, in this case, the force is “equal & opposite.”
z
r̂
v0
x
Both forces are in the +y direction. The ratio of their
magnitudes is
   e 2 v 02 

2 
4

r
FB 

=
FE  1 e 2 

2 
4

r



y
FE
FB
=     v 20
FE
v0
2
FB
B1
Later we will find that
r
1
1
 = 2
c
z
r̂
v0
x
FB v 20
Thus
= 2
FE c
If v0=106 m/s, then
10 

FB
-5
=

1.11

10
FE  3 108 2
6 2
y
FE
Don’t you feel sorry for the poor,
weak magnetic force?
v0
What if you are a nanohuman, lounging on proton 1. You
rightfully claim you are at rest. There is no magnetic field
from your proton, and no magnetic force on 2.
B1
If you don’t like being confused,
Another nanohuman riding on proton 2 would say “I am
close your eyes
at rest, so there is no magnetic force on my proton,
and iscover
your
even though there
a magnetic
fieldears.
from proton 1.”
This calculation says there is a magnetic field and force.
Who
is here,
right?
Take
Physics
2305/107
totolearn
Or see
here,
and here
for a hint
about how
resolvethe
the answer.
paradox.
2
FB
r
1
z
r̂
v0
x
Today’s agenda:
Magnetic Fields Due To A Moving Charged Particle.
You must be able to calculate the magnetic field due to a moving charged particle.
Biot-Savart Law: Magnetic Field due to a Current
Element.
You must be able to use the Biot-Savart Law to calculate the magnetic field of a currentcarrying conductor (for example: a long straight wire).
Force Between Current-Carrying Conductors.
You must be able to begin with starting equations and calculate forces between currentcarrying conductors.
Biot-Savart Law: magnetic field of a current element
From the equation for the magnetic field of a moving charged
particle, it is “easy” to show that a current I in a little length dl
of wire gives rise to a little bit of magnetic field.
r
dB
r̂
dl
I
μ 0 I d  rˆ
dB =
4π r 2
The Biot-Savart Law
If you like to be more precise in your language, substitute
“an infinitesimal” for “a little length” and “a little bit of.”
I often use ds instead of dl because the script l does not
display very well.
You may see the equation written using r = r rˆ .
Applying the Biot-Savart Law
P
dB 
r

μ0 I ds  rˆ
r
ˆ
dB =
where r =
2
4π r
r
ds  rˆ = ds rˆ sin 
= ds sin because rˆ =1
r̂
ds
I
dB =
μ 0 I ds sin θ
4π
r2
B =  dB
Homework Hint: if you have a tiny piece of a wire, just calculate dB; no need to integrate.
Example: calculate the magnetic field at point P due to a thin
straight wire of length L carrying a current I. (P is on the
perpendicular bisector of the wire at distance a.)
y

dB
P
r
r̂ 
ds
μ 0 I ds  rˆ
dB =
4π r 2
a

z
x
x
I
ds  rˆ = ds sinθ kˆ
dB =
L
μ 0 I ds sinθ
4π
r2
ds is an infinitesimal quantity in the direction of dx, so
μ 0 I dx sinθ
dB =
4π
r2
sinθ =
a
r
2
r
r̂ 
ds
x
L
μ 0 I dx sinθ
4π
r2
μ 0 I dx a μ 0 I dx a
dB =
=
3
4π r
4π  x 2 + a2 3/2
a

z
dB =
r = x +a
y
dB
P
2
x
I
μ 0 I dx a
-L/2 4π
2
2 3/2
x +a 
B=
L/2
μ 0I a L/2
dx
B=
4π -L/2  x 2 + a2 3/2
μ 0I a L/2
dx
B=
4π -L/2  x 2 + a2 3/2
y
dB
P
r
r̂ 
ds
x
look integral up in tables, use the
web,or use trig substitutions
a
x

z

I
L
dx
x
2

2 3/2
+a
=
x
a x +a
2
L/2
μ 0I a
x
B=
4π a2  x 2 + a2 1/2
-L/2

μ 0I a 
L/2
=
4π  a2 L/2 2 +a2



1/2

-L/2


1/2 
2
2
2
a  -L/2  +a



2

2 1/2
y
dB
P
r
r̂ 
ds
x


μ 0I a 
2L/2

B=
1/2
4π  a2 L2 /4 +a2  


a

z
L
x
I
B=
μ 0I L
1
4πa L2 /4 + a2 1/2
μ 0I L
1
B=
2πa L2 + 4a2
μ 0I
B=
2πa
1
4a2
1+ 2
L
y
dB
P
r
r̂ 
ds
x
a

z
x
μ 0I
B=
2πa
1
4a2
1+ 2
L
I
L
When L, B =
μ 0I
.
2 πa
μ 0I
or B =
2 πr
The r in this equation has a different
meaning than the r in the diagram!
y
dB
P
r
r̂ 
ds
x
a

z
x
μ 0I
B=
2πa
1
4a2
1+ 2
L
I
L
When L, B =
μ 0I
.
2 πa
μ 0I
or B =
2 πr
The r in this equation has a different
meaning than the r in the diagram!
Magnetic Field of a Long Straight Wire
We’ve just derived the equation for the magnetic
field around a long, straight* wire…
μ0 I
B=
2 πr
I
B
r
…with a direction given by a “new” righthand rule.
link to image source
*Don’t use this equation unless you have a long, straight wire!
Looking “down” along the wire:
The magnetic field is not
constant (and not uniform).
I
B
At a fixed distance r from the wire, the magnitude of the
magnetic field is constant (but the vector magnetic field is
not uniform).
The magnetic field direction is always tangent to the
imaginary circles drawn around the wire, and perpendicular
to the radius “connecting” the wire and the point where the
field is being calculated.
Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two radial straight segments and a circular arc of
radius R that subtends angle .
A´
I see three “parts” to the wire.
A’ to A
A
A to C

O
C
I
C´
R
Thanks to Dr. Waddill for the use of the diagram.
C to C’
As usual, break the problem up
into simpler parts.
http://autosystempro.com/tag/generators/
Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two radial straight segments and a circular arc of
radius R that subtends angle .
μ 0 I ds  rˆ
dB =
4π r 2
A´
I
ds
r̂
A
For segment A’ to A:
C´

O
ds  rˆ = ds rˆ sin 0 = 0
C
R
dB A'A = 0
B A'A = 0
Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two radial straight segments and a circular arc of
radius R that subtends angle .
μ 0 I ds  rˆ
dB =
4π r 2
A´
A
r̂
ds

O
C
R
I
C´
For segment C to C’:
ds  rˆ = ds rˆ sin 180 = 0
dB CC' = 0
B CC' = 0
Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two straight segments and a circular arc of radius R
that subtends angle .
A´
r̂
A
Important technique, handy for
homework and exams:
ds
r̂

O
C
R
ds
C´
The magnetic field due to wire
segments A’A and CC’ is zero
because ds is either parallel or
antiparallel to r̂ along those
paths.
Example: calculate the magnetic field at point O due to the wire
segment shown. The wire carries uniform current I, and
consists of two radial straight segments and a circular arc of
radius R that subtends angle .
μ 0 I ds  rˆ
dB =
4π r 2
A´
A
For segment A to C:
ds
r̂

O
C
I
C´
ds  rˆ = ds rˆ sin 90 = 0
= ds 1 1
R
= ds
dB AC
μ 0 I ds
= dB AC =
4π R 2
A´
A
μ 0 I ds
dB AC =
4π R 2
r̂

O
μ 0I ds
B AC =  dB AC = 
2
4π
R
arc
arc
ds
C
I
C´
μ 0I
B AC =
4πR 2
R
μ 0I
B AC =
4πR 2
 ds
arc
The integral of ds is just
the arc length; just use
that if you already know it.
 R dθ
arc
μ 0I
B AC =
dθ

4πR arc
μ 0I
B AC =
θ
4πR
We still need to provide the
direction of the magnetic field.
A´
A
Cross ds into r̂ . The direction is
“into” the page, or .
ds
r̂

O
C
I
C´
If we use the standard xyz axes,
ˆ
the direction is -k.
y
R
z
B = B A'A + B AC + B CC' = -
μ 0 Iθ ˆ
k
4πR
x
Important technique, handy for
exams:
A´
A
Along path AC, ds is
perpendicular to r̂ .
ds
r̂

O
C
R
I
C´
ds  rˆ = ds rˆ sin 90
ds  rˆ = ds
Today’s agenda:
Magnetic Fields Due To A Moving Charged Particle.
You must be able to calculate the magnetic field due to a moving charged particle.
Biot-Savart Law: Magnetic Field due to a Current Element.
You must be able to use the Biot-Savart Law to calculate the magnetic field of a currentcarrying conductor (for example: a long straight wire).
Force Between Current-Carrying Conductors.
You must be able to begin with starting equations and calculate forces between currentcarrying conductors.
Magnetic Field of a Current-Carrying Wire
It is experimentally observed that parallel wires exert forces on
each other when current flows.
I1
I2
F12

F21

I1
I2
F12

F21

“Knowledge advances when Theory does battle with Real Data.”—Ian Redmount
Example: use the expression for B due to a current-carrying
wire to calculate the force between two current-carrying wires.
d
F12 = I1L1  B2
I1
L
ˆB = μ 0 I2 kˆ
2
2 πd
The force per unit length of wire is
Homework Hint: use this technique for any
homework problems with long parallel wires!
F12
B2
μ I
F12 = I1Ljˆ  0 2 kˆ
2 πd
μ 0 I1I2L ˆ
F12 =
i
2 πd
I2
L1
y

z
x
F12 μ 0 I1I2 ˆ
=
i.
L
2πd
L2

B
d
F21 =I2L2  B1
I1
L
μ 0 I1 ˆ
B1 = k
2 πd
The force per unit length of wire is
F12
F21
B1
 μ 0 I1 ˆ 
ˆ
F21 = I2Lj   
k
 2πd 
μ 0 I1I2L ˆ
F21 = i
2 πd
I2
L1
y

z
x
μ II
F21
= - 0 1 2 ˆi.
L
2 πd
L2

Be able to show that if the currents in the wires are in the
opposite direction, the force is repulsive.
F12 = F21 =
μ 0 I1I2L
2πd
d
L
I1
I2
F12
F21
4 π 10-7 I1I2L
L
-7
F12 =F21 =
= 2 10 I1I2
2πd
d
The official definition of the Ampere:
1 A is the current that produces a
force of 2x10-7 N per meter of length
between two long parallel wires placed
1 meter apart in empty space.
L1

L2
