Oct. 2720.3-20.4

Oct. 29boardwork

Oct. 3120.5-20.8

Quiz 6

Nov. 321.1-21.2

Nov. 5boardwork

Nov. 721.3-21.5

Quiz 7

Nov. 10boardwork

Nov. 1221.6-21.6

Nov. 14boardwork

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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 = I ℓ 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.

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?

Back to our regularly scheduled lecture!

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,

Ir

0μ OSE : B = ,

2

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:

-7

0

T mμ = 4 ×10 .

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?

I=25 A

d=0.1 mB

Let’s make north be to the left in this picture, and up be up.

N

up

According to the right hand rule, the magnetic field is to the west, coming out of the plane of the “paper.”To calculate the magnitude, B:

Ir

0μ OSE : B = ,

2

-7

-5

T m4π ×10

25 AAB = = 5 ×10 T .2π 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…

0 11

μ IB = .

2 L

I1

L

Of course, the magnetic field exists everywhere. I just chose to indicate it at one point to avoid cluttering the figure.

B1

I2

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

Substituting B1 gives in the last equation gives

2 1

F = I B .

0 1 2μ I IF = .

2 L

I1

B1L

To OSE or not to OSE??

0 11

μ IB =

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?

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.

Newton’s third law says each wire exerts an equal and opposite force on the other.

No need for a picture.

0 1 2μ I IF =

2 L

0 1 2μ I IF =

2 L

-7

-3

T m4π ×10

8 A 8 AAF = 2 m 2π 3×10 m

both wires carry the same current (in magnitude)

-3F = 8.5 ×10 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.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.

I2

I1

w=mg

Fmag

Remember the litany for force problems?

Sketch (done on previous slide).

Free body diagram, showing forces on object.

I2

w=mg

Fmag

Label vectors.

Choose axes.

y

Draw components of vectors not along axes (not needed here).

OSE and solve. y yF = ma

mag,y y yF + w = ma 0

mag+ F + - mg = 0

0 1 2μ I I - mg = 0

2 L

(repeating last equation) mag+ F + - mg = 0

0 1 2μ I I = mg

2 L

0 1 2μ I I mg = 2 L

1 20

2 m g L I I =

μ

20 1

2 m g L I =

μ I

(algebraic solution)

-3 2

2-7

2π 0.12×10 kg 9.8 m/s 0.20 m I =

T m4π ×10 80 A 1 m

A

2 I = 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.

I’ve chosen to treat the wire as being 1 m long.

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.

Physics is constantly being “tweaked” as new knowledge and experimental techniques become available.

A coulomb is then defined as 1 A · 1 s.

I won’t put any of this on the test.

20.8 Ampere’s Law

What is a solenoid?

A solenoid is a coil of wire with many loops.

Each loop produces a magnetic field that looks like this.

Skip all of this section except what I present in lecture. You may use your text for reference.

“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 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

I

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.The slice is made perpendicular to the wires and parallel to the solenoid axis.

Textbooks show that the magnetic field inside the solenoid is

0 B = μ n I .

0B = μ 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:

0

NOSE: B = μ 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.

0

NOSE: B = μ I

L

-7 400T m

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.

0μ I OSE : B =

2π r

0

NOSE : B = μ 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!