Transcript Slide 1

Announcements
 Physics 2135 spreadsheets for all sections, with Exam 2 scores,
will be available today on the Physics 2135 web site. You need
your PIN to find your grade.
 Preliminary exam average is about 75.0% (12 sections out of 12
reporting). Reasonable! Scores ranged from a low of 37 to a high
of 200 (3 students). I will fill in the ??’s during the “live” lecture and in its “.ppt” file.
Physics 2135 Exam 2 will be returned in recitation Thursday.
When you get the exam back, please check that points were added
correctly. Review the course handbook and be sure to follow
proper procedures before requesting a regrade. Get your regrade
requests in on time! (They are due by next Thursday’s recitation.)
On a separate sheet of paper, briefly explain the reason for your regrade request. This should be based on the work actually
shown on paper, not what was in your head. Attach to the exam and turn it in by the end of your next Thursday’s recitation.
Announcements
 Midterm grades include your first 3 labs. If you are missing one
of them, this does impact your grade. Remember, the one lowest
lab score is dropped, and there are no makeups.
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
rˆ
+
v
μ 0 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.
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 8.1 in your text.
FE
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: slides 6-9!
2
z
rˆ
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
rˆ
v0
x
Also at the position of proton 2 there is a magnetic field due to
proton 1.
 q1 v1  rˆ
B1 =
4 r2
 ev 0 ˆi  ˆj
B1 =
4 r 2
y
FE
v0
  ev 0 ˆ
B1 =
k
2
4 r
2
B1
r
1
z
rˆ
v0
x
Proton 2 “feels” a magnetic force due to the magnetic field of
proton 1.
FB = q2 v2 B1
 
  ev 0 ˆ 
ˆ
FB = ev 0  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
rˆ
v0
x
Both forces are in the +y direction. The ratio of their
magnitudes is
   e2 v 02 

2 
4

r
FB 

=
FE  1 e2 

2 
4

r



y
FE
FB
=  v 20
FE
v0
2
FB
B1
Later we will find that
r
1
1
 = 2
c
z
rˆ
v0
x
FB v 20
Thus
= 2
FE c
If v0=106 m/s, then
10 

FB
-5
=

1.11

10
FE  3  10 8 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)
answer.
Or see
here,
and here
for a hint
about
howtotolearn
resolvethe
the paradox.
2
FB
r
1
z
rˆ
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
rˆ
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
dB 
r

rˆ
ds
I
μ 0 I ds  rˆ
r
ˆ
dB =
where
r
=
4π r 2
r
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
rˆ 
ds
x
μ 0 I ds  rˆ
dB =
4π r 2
a

z
ds  rˆ = ds sinθ kˆ
x
I
dB =
μ 0 I ds sinθ
4π
r2
L
ds is an infinitesimal quantity in the direction of dx, so
μ 0 I dx sinθ
dB =
4π
r2
sinθ =
a
r
2
r
rˆ 
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
μ0
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
rˆ 
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
rˆ 
ds
x


μ 0I a 
2L/2

B=
1/2
4π  a2 L2 /4 + a2  


a

z
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
rˆ 
ds
a

z
x
When L, B =
x
μ 0I
B=
2πa
1
4a2
1+ 2
L
I
μ 0I
.
2πa
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
with a direction given by a “new” right-hand
rule.
I
B
r
*Don’t use this equation unless you have a long, straight wire!
Looking “down” along the wire:
I
B
The magnetic field is not
constant.
At a fixed distance r from the wire, the magnitude of the
magnetic field is constant.
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 straight segments and a circular arc of radius R
that subtends angle .
A´
A
rˆ

R
Important technique, handy for
exams:
ds
C
I
C´
Thanks to Dr. Waddill for the use of the diagram.
The magnetic field due to wire
segments A’A and CC’ is zero
because ds is parallel to rˆ along
these paths.
A´
A
rˆ

R
Important technique, handy for
exams:
ds
C
I
Along path AC, ds is
perpendicular to rˆ.
C´
Thanks to Dr. Waddill for the use of the diagram.
ds  rˆ = -ds kˆ
ds  rˆ = ds
μ 0 I ds  rˆ
dB =
4π r 2
dB =
μ 0I ds
4π R 2
If we use the “usual” xyz axes.
A´
A
rˆ

R
ds
C
I
dB =
μ 0I ds
4π R 2
B=
μ 0I ds
4π R 2
C´
Thanks to Dr. Waddill for the use of the diagram.
B=
μ 0I
ds
2 
4πR
μ 0I
B=
R dθ
2 
4πR
B=
μ 0I
dθ

4πR
μ 0I
B=
θ
4πR
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

I1
F21
I2
F12


F21

Skip ahead to slide 29.
“Knowledge advances when Theory does battle with Real Data.”—Ian Redmount
We showed that a long straight wire carrying a current
I
I gives rise to a magnetic field B at a distance r from
the wire given by
μ0 I
B=
2 πr
B
The magnetic field of one wire exerts a force on a
nearby current-carrying wire.
d
I1
L
I2
F12

F21

r
The magnitude of the force depends on
the two currents, the length of the wires,
and the distance between them.
μ I I L This is NOT a
F= 0 1 2
starting equation
2πd
The wires are electrically neutral,
so this is not a Coulomb force.
Remember, the direction of the field is given another (different) right-hand rule: grasp the wire and point the thumb of your
right hand in the direction of I; your fingers curl around the wire and show the direction of the magnetic field.
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
μ I
Bˆ 2 = 0 2 kˆ
2 πd
μ I
F12 = I1Ljˆ  0 2 kˆ
2 πd
μ 0 I1I2L ˆ
F12 =
i
2 πd
The force per unit length of wire is
Homework Hint: use this technique for any
homework problems with long parallel wires!
I2
F12
B2
y
L1


x
F12 μ 0 I1I2 ˆ
=
i.
L
2 πd
L2
d
F21 =I2L2 B1
I1
L
μ 0 I1 ˆ
B1 = k
2πd
 μ0 I1 ˆ 
ˆ
F21 = I2Lj   
k
 2πd 
F21 = -
μ 0 I1I2L ˆ
i
2 πd
The force per unit length of wire is
I2
F12
F21
B1
y
L1

L2

x
μ 0 I1I2 ˆ
F21
=i.
L
2πd
If the currents in the wires are in the opposite direction, the
force is repulsive.
d
L
I1
I2
F12
F21
y
L1

L2

F12 = F21 =
μ 0 I1I2L
2 πd
d
I1
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 force per meter of length between two
long parallel wires placed 1 meter apart in
empty space.
I2
F12
F21
L1

L2
