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20.3-20.4
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20.5-20.8
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“I can't remember my telephone number, but I know it is in the high
numbers,”—mathematician and economist John Keynes
A note on the right hand rule.

I personally find the three-fingered axis system
to generally (but not always) be the most useful.


“In F= I ℓ B sin  and F = q v B sin  does it matter which finger I
use for what?”
F= Iℓ B sin 
F = q v B sin 
 

No, as long as
you keep the
right order. All
three of these
will work:












This works:


This doesn’t:

Switching only two is wrong!

“This is unfair! Physics is discriminating against left-handers!”
No, we could get the same results with left-hand axes and left
hand rules. See this web page.
Back to our regularly scheduled lecture!
20.5 Magnetic Field due to a Straight Wire
We already saw how the magnetic
field due to a current “curls
around” a wire. This tells us the
direction of the magnetic field.
What about the magnitude?
Experimentally it is found (and verified by theory) that the
larger the current, the larger the magnetic field, and the further
away from the wire, the weaker the magnetic field.
Mathematically,
OSE :
μ0 I
B=
,
2 r
where I is the current in the wire, r is the distance away from
the wire at which B is being measured, and 0 is a constant:
μ0 = 4 ×10-7
Tm
.
A
This “funny” definition of 0 allows us to more elegantly define
current (later).
Example 20-7. A vertical electric wire in the wall of a building
carries a current of 25 A upward. What is the magnetic field at
a point 10 cm due north of this wire?
Let’s make north be to the left in this
picture, and up be up.
According to the right hand rule, the
magnetic field is to the west, coming out
of the plane of the “paper.”
up
I=25 A
N
To calculate the magnitude, B:
OSE :
B=
μ0 I
,
2 r

-7 T  m 
4π
×10

  25 A 
A

B= 
= 5 ×10-5 T .
2π
 0.1 m
B
d=0.1 m
20.6 Force Between Two Parallel Wires
Current in a wire produces a magnetic field.
A wire carrying a current in a magnetic field feels a force.
I wonder what would happen
if you put two current-carrying
wires next to each other.
Maybe the magnetic field from
one creates a force on the other
and vice-versa?
A current I1 in wire 1 gives rise a distance L away to a magnetic
field…
μ0 I1
B1 =
.
2 L
I1
B1
L
Of course, the magnetic field exists everywhere. I just chose to indicate it at
one point to avoid cluttering the figure.
A second, parallel conductor of length ℓ a distance L away
carrying a current I2 “feels” a force F = I2 ℓ B1. The force per
unit length on the second wire is
F
= I 2 B1 .
μ0 I1
B1 =
2 L
I2
B1
I1
L
Substituting B1 gives in the last equation gives
To OSE or not to OSE??
F
μ0 I1 I2
=
.
2 L
What about the direction of the force?
If the current in the two wires is in the same direction, the force
is attractive. Otherwise the force is repulsive. This is
somewhat counterintuitive, isn’t it?
Newton’s third law says each wire exerts an equal and opposite
force on the other.
Example 20.8. The two wires of a 2 m appliance cord are 3
mm apart and carry a current of 8 A. (Assume dc current.)
Calculate the force between these two wires.
No need for a picture.
F
=
μ0 I1 I2
2 L
μ I I 
F=  0 1 2 
 2 L 
both wires carry
the same current
(in magnitude)

-7 T  m 
 4π ×10

A  8 A 8 A 

F =  2 m
-3
2π
3×10
m

F = 8.5 ×10-3 N
The force will be repulsive because the wires carry current in
opposite directions.
Example20.9. A horizontal wire carries a current I1=80 A dc.
A second parallel wire 20 cm below it must carry how much
current I2 so that it doesn’t fall due to gravity? The lower wire
has a mass of 0.12 g per meter of length.
The currents need to be in the
same direction to produce an
attractive force (doesn’t matter
which direction).
This is just an equilibrium problem
from our mechanics semester.
Fmag
I1
I2
w=mg
The magnetic attraction provides the force that balances the
weight of the levitated wire.
We could do our calculations per unit length, or just pick a
meter length of wire. You would probably do the latter.
y
Remember the litany for force
problems?
 Sketch (done on previous slide).
 Free body diagram, showing
forces on object.
 Label vectors.
Fmag
I2
w=mg
 Choose axes.
 Draw components of vectors not along axes (not needed
here).
 OSE and solve.
F
y
Fmag,y
= may
0
+ w y = may
+ Fmag +  - mg = 0
+ Fmag +  - mg = 0
(repeating last equation)
 μ0 I1 I2 

 - mg = 0
 2 L 
 μ0 I1 I2 

 = mg
 2 L 
μ0 I1 I2
mg
=
2 L
2 m g L
I1 I2 =
μ0
2 m g L
I2 =
μ0 I1
(algebraic solution)
I2 =
 2π  0.12×10-3 kg  9.8 m/s 2   0.20 m

-7 T  m 
 4π ×10
  80 A 1 m 
A 

I’ve chosen to treat the wire
as being 1 m long.
I2 = 15 A .
“This sure looks like a lot more work than the textbook went
through to get the answer.”
Wrong! It just skipped a lot of steps. Don’t skip steps. It only
leads to mistakes.
Caution: in the equation, L is the distance between wires and ℓ
is the wire length. Possible source of confusion and error.
20.7 Definition of the Ampere and the Coulomb
We defined the ampere of current in chapter 16 as being 1 C of
charge flowing past a point in 1 s: 1 A = 1 C / 1 s.
That’s the way I learned it many years ago.
Now we find the ampere is actually defined as the current
flowing in two parallel wires 1 m apart which produces a force
per unit length of 2x10-7 N/m.
A coulomb is then defined as 1 A · 1 s.
Physics is constantly being “tweaked” as new knowledge and
experimental techniques become available.
I won’t put any of this on the test.
20.8 Ampere’s Law
Skip all of this section except what I present in lecture. You
may use your text for reference.
What is a solenoid?
A solenoid is a coil of wire with many
loops.
Each loop produces a magnetic field that
looks like this.
“When the coils of the solenoid are closely spaced, each turn
can be regarded as a circular loop, and the net magnetic field is
the vector sum of the magnetic field for each loop. This
produces a magnetic field that is approximately constant inside
the solenoid, and nearly zero outside the solenoid.”
Thanks again to Dr. Waddill for the pictures and text.
The slice is made perpendicular to the
wires and parallel to the solenoid axis.

I

“The ideal solenoid is approached when the coils are very
close and the length of the solenoid is much greater than its
radius. Then we can approximate the magnetic field as
constant inside and zero outside the solenoid.”
B
The vectors in and out of the page
represent the current (and therefore
the wires), so imagine this picture as a
slice through the center of the solenoid,
perpendicular to the wires.
Textbooks show that the magnetic field inside the solenoid is
B = μ0 n I .
B = μ0 n I
B is the magnitude of the magnetic field inside the solenoid (the
direction is given by the right-hand rule), n is the number of
loops per unit length (loops per meter), and I is the current in
the wire.
I’ll write the “official” version like this:
OSE :
B = μ0
N
I
L
N is the total number of loops (sometimes called “turns”) and L
is the total length of the solenoid.
More about solenoids on-line here.
The magnetic field of a solenoid looks like the magnetic field of
a bar magnet.
(http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/elemag.html#c1)
Example 20.10. A thin 10-cm long solenoid has a total of 400
turns of wire and carries a current of 2 A. Calculate the
magnetic field inside near the center.
OSE :
N
B = μ0
I
L

-7 T  m   400 
B =  4π ×10
2 A 

A   0.1 m 

B = 0.01 T
That’s all we’ll do that’s testable in Chapter 20! You can read
the rest yourself if you want.
fs2002 lecture 13 ends here
What follows on the next five slides is for your cultural
gratification only.
You will be better humans for having seen it. But it won’t help
your grade any.
The BIG IDEAS
There are two BIG IDEA equations buried in this chapter. It is
not obvious where they are, because we are so focused on
details when we learn this material for the first time.
One of the big ideas arises from the observation that magnetic
poles always come in pairs, unlike + and – charged particles.
In the next lecture, I’ll introduce the idea of magnetic flux,
which is “like” the idea of electric flux.
You can calculate the magnetic flux through a given area:
If you integrate the magnetic flux over a closed area (e.g., a
sphere, or a cylinder closed at both ends), the result is zero:
The integral is zero because wherever you find a N pole, you
also find a S pole, and the net flux going out of the surface
must equal the net flux going into the surface (kind of like the
N “cancels” the S).
The equation is called Gauss’ law for magnetism, and is one of
Maxwell’s four equations.
It also says there is no such thing as a magnetic monopole.
Some quantum theories suggest that magnetic monopoles
might exist. We have not found them. If we do, then the
right hand side of the equation above will need modified.
You also saw Ampere’s law, which appeared in the context of
a solenoid. The law is far more general than that. It also
appeared in the equation for a magnetic field due to a current
in a wire, except then we didn’t call it Ampere’s law.
OSE :
μ0 I
B=
2π r
OSE :
B = μ0
N
I
L
If you integrate this expression over a closed path, you get a
result proportional to the current I and the total path length.
In the next chapter we will find that electric fields which
change with time also give rise to magnetic fields, so the full
version of the Ampere’s law Maxwell equation is
You’ve seen three out of Maxwell’s four equations. One more
chapter on E&M, one more equation!