Magnetism

Magnets and Magnetic Fields

We will study ferromagnetism in this lecture.

It is so named because many ferromagnetic materials (including the first-discovered ones) are “ferrous materials,” i.e., they contain iron.

We studied electric forces and electric fields yesterday…

Now we study magnetism and magnetic fields.

Rather than defining magnetism, we begin by discussing properties of magnetic materials.

Recall how there are two kinds of charge (+ and -), and likes repel, opposites attract.

Similarly, there are two kinds of magnetic poles (North and South), and like poles repel, opposites attract.*

*Recall also that I have a mental defect which often causes me to say “likes attract and unlikes repel” when I mean the opposite. I am not to be penalized for a mental defect!

S N

S N SN

S N

Repel

AttractThanks to Dr. Waddill for the nice pictures!

S N

SNS N

Repel Attract

S NS N

There is one difference between magnetism and electricity: it is possible to have isolated + or – electric charges, but isolated N and S poles have never been observed.

+- S N

I.E., every magnet has BOTH a N and a S pole, no how many times you “chop it up.”

S N S N S N+=

The earth has associated with it a magnetic field, with poles near the geographic poles.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magearth.html

The pole of a magnet attracted to the earth’s north geographic pole is the magnet’s North pole.

The pole of a magnet attracted to the earth’s south geographic pole is the magnet’s South pole.

N

S

Just as we used the electric field to help us “explain” and visualize electric forces in space, we use the magnetic field to help us “explain” and visualize magnetic forces in space.

Magnetic field lines point in the same direction that the north pole of a compass would point.

Magnetic field lines are tangent to the magnetic field.

The more magnetic field lines in a region in space, the stronger the magnetic field.

Outside the magnet, magnetic field lines point away from N poles (*why?).

*The N pole of a compass would “want to get to” the S pole of the magnet.

Huh?

Nooooooo….

Is the earth’s north pole a magnetic N or a magnetic S?

It has to be a S, otherwise, the compass N would not point to it.

Unless the N of a compass needle is really S. Dang! This is too much for me!

Yup, it’s confusing.

Here’s a “picture” of the magnetic field of a bar magnet, using iron filings to map out the field.

The magnetic field ought to “remind” you of the earth’s field.

Later I’ll give a better definition for magnetic field direction.

Here’s what the magnetic field looks like when you put unlike or like poles next to each other.

The magnetic field B is a vector which points in the direction of magnetic field lines. We will quantify the magnitude of B later.

Understand the declination, or angle of dip, if you plan to get lost in the woods and use a compass to find your way out.

Electric Current Produces Magnetism

An electric current produces a magnetic field.

The direction of the current is given by the right-hand rule. Grasp the current-carrying wire in your right hand, with your thumb pointing in the direction of the current. Curl your fingers around the wire. Your fingers indicate the direction of the magnetic field.

Picture on previous page is from http://physics.mtsu.edu/~phys232/Lectures/L12-L16/L17/Current_Loops/current_loops.html

This picture also illustrates the magnetic field due to a current-carrying loop of wire.

Field comes out of page here. “Turns around” and

goes into page here.

These symbols mean “out of page” and “into page.” See next section.

Here’s a simpler case: the magnetic field due to a straight wire.

If the wire is grasped in the right hand with the thumb pointing in the direction of the current, the fingers will curl in the direction of B.

Thanks again to Dr. Waddill for the nice pictures!

Field comes out of page here.

Field turns around and goes into page here.

Force on an Electric Current in a Magnetic Field; Definition of B

As seen above, an electric current gives rise to a magnetic field, which must exert a force on a magnet (e.g. compass needle).

Does a magnet exert a force on a current-carrying wire? (Newton’s 3rd Law says it should.)

Yes—a current-carrying wire in a magnetic field “feels” a force. The direction is given by the right-hand rule:

Point your outstretched fingers in the direction of the current. Bend your fingers 90º and orient your hand to point the bent fingers in the direction of the magnetic field. Your thumb points in the direction of the force.

You may need to re-orient your hand as you go through this procedure. During exams, I see all sorts of gyrations as students try to figure out directions.

I’ll demonstrate another right-hand rule in class.

Here is a web “physics toy” to help you visualize the force on a current-carrying conductor.

Below is another picture to help you visualize. It came from http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html.

The web page even has a built-in calculator that gives you numerical answers to your force problems!

You would expect the magnitude of the force to depend on the magnitudes of the magnetic field and the current. In fact, it does.

The force also depends on how much of the wire is in the magnetic field.

The force is perpendicular to both the current and the magnetic field.

If the direction of the current is perpendicular to the magnetic field, then F = I ℓ B. I’ll write the lowercase l in italics (ℓ) to help you distinguish it from the number 1.

If the current and magnetic field are not perpendicular, the force is given by

OSE: F = I ℓ B sin ,

where is the angle between the current vector and the magnetic field vector. (Smallest angle from the current vector to the magnetic field vector.)

B

I

ℓ

The right-hand rule and the equation above actually serve as the definition of the magnetic field B.

The SI unit for magnetic field is the tesla: 1 T = 1 N / (1 A · 1 m).

An older unit for magnetic field (which you might see occasionally) is related to the weber (weber is 1 Wb = 1 N / 1 A). 1 T = 1 Wb / m2.

Magnetic fields are also given in units of gauss: 1 G = 10-4 T. Argh! Confusing. Let’s try to stick with SI units, OK?

Here is a little movie I found on a web site (forgot where) illustrating force on a wire due to magnetic field.

I have to go to a lot of effort to explain magnetic field and force direction when I teach the non-calculus course.

It’s so much easier with calculus and vectors. The force on a charge q moving with a velocity v in a magnetic field B is found to obey

F = qv B.

The magnitude of the cross product is qvB sin . But it’s so much easier learning the right-hand rule for the vector cross product, and applying it to torques, charged particles, etc., instead of learning a seemingly new right hand rule for each new topic. The elegance of math!

If you take a number of charged particles in a volume of wire that has a length ℓ in a magnetic field, it is easy to derive the vector form of our OSE:

IF = B.

Example In the figure two slides back, B=0.9 T, I=30 A, ℓ=12 cm, and =60°. What is the force on the wire?

F = I ℓ B sin = (30 A) (0.12 m) (0.9 T) (sin 60°)

F = 2.8 N

B

I

ℓ

Here’s a repeat of the figure, for handy reference.

We need to have a way to draw 3-d vectors on 2-d paper. We will use the symbol for a vector pointing directly out of the page, towards us (that is supposed to look like the sharp point of an arrowhead coming right towards your eye).

We will use the symbol for a vector pointing directly into the page, away from us (that is supposed to look like the feathered end of an arrow going away from your eye).

Hold it! Force is a vector quantity! What is the direction.

The force is perpendicular to both current direction and magnetic field direction, so it is either into paper or out of the paper. Apply either right-hand rule you and find it is into the paper.

I’ll also show you the right-hand screw rule, which is the way I best visualize the direction.

F

B

I I10 cm

Example A rectangular loop of wire hangs vertically in a magnetic field B as shown. B is uniform along the 10 cm horizontal length of wire, and the top portion of the wire is outside the field. The loop hangs from a balance which measures a downward force F=3.48x10-2 N in excess of the wire weight when the current is 0.245 A. What is the magnitude of B?

Huh?

Sorry, that’s not an acceptable answer. You do know how to work this problem!

F

B

I I10 cm

The forces on these vertical segments of the wire are equal and opposite in direction. We need not worry about them further.

The force on the lower horizontal segment is downward, as shown in the drawing. Could you verify that?

I can figure out the directions of those two forces. Can you?

The angle between current and magnetic field is 90°. sin(90°)=1.

OSE: F = I ℓ B sin solve for B!

B = F ( I ℓ sin(90) )

B = 1.42 T

Hey, that wasn’t so bad! Only the direction bit is hard work at this point.

B = (3.48x10-2 N) (0.245 A) (0.1 m)

Force on an Electric Charge Moving in a Magnetic Field

We’ve kind of done this already—what’s the difference between a moving charge and a current in a wire?

The current was confined to a wire, but we don’t expect that to alter the forces involved.

If the charges are moving with a speed v, then the distance ℓ they travel in the time t is ℓ = vt.

Let’s think of a moving charge as a “current” not necessarily confined to a wire (actually, a perfectly reasonable thing to do—how about lightning?).

If N charges of charge q pass a point in space during a time t, the current is I = Q / t = Nq/t.

Plugging I = Nq/t and ℓ = vt back into F = I ℓ B sin gives

Our current OSE--remember?

net

NqF = vt B sin θ = N q v B sin θ .

t

The above equation gives the total force on the N charges. The force on a single charge is Fnet/N, or

OSE : F = q v B sin θ .

If the charged particle is moving perpendicular to B, = 90° and the force is greatest: F = q v B.

The above OSE gives the magnitude of the force. The right hand rule gives the direction for positive charges.

For negative charges, just reverse the direction (determine the direction as if it were for a positive charge and the force on the negative charge is in the opposite direction).

-+

Bout

vv

FBFB

Thanks again to Dr. Waddill for the nice picture. Don’t you wish you were taking Physics 24 too?

Example A proton having a speed of 5x106 m/s in a magnetic field feels a force of 8x10-14 N toward the west when it moves vertically upward. When moving horizontally in a northern direction, it feels zero force. What is the magnitude and direction of the magnetic field in this region? The charge on a proton is +1.6x10-19 C.

OSE : F = q v B sin θ .

The statement “when moving horizontally in a northern direction, it feels zero force” tells you =0° or =180° so the magnetic field points either north or south.

Applying the right-hand rule to the figure to the right shows that B must point north.

N

W

up

B

F v

Now we can calculate the magnitude of B.

F = q v B

When the proton is moving west, the angle between v and

B (which points north) is 90°. sin(90°)=1 so

FB =

q v

-14

-19 6

8 10 NB =

1.6 10 C 5 10 m/s

B = 0.1 T

Example An electron travels at 2x107 m/s in a plane perpendicular to a 0.01 T magnetic field. Describe its path.

The force on the electron (remember, its charge is -) is always perpendicular to the velocity. If v and B are constant, then F remains constant (in magnitude).

The above paragraph is a description of uniform circular motion.

The electron will move in a circular path with an acceleration equal to v2/r, where r is the radius of the circle.

B

v

F

v

F

-

-

B and v are perpendicular, so F = q v B.2v

F = ma = m = q v B r

2mvr =

q v B

mvr =

q B

-31 7

-19

9.11 10 kg 2 10 m/sr =

1.6 10 C 0.01 T

-2r = 1.1 10 m

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,

0μ I OSE : B = ,

2π 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:

-70

T mμ = 4π 10 .

A

This “funny” definition of 0 allows us to more elegantly define current (later).

Example 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:

0μ I OSE : B = ,

2π r

-7

-5

T m4π 10

25 AAB = = 5 10 T .

2π 0.1 m

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 = .

2π 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 = .

2π L

I1

B1L

OSE or not OSE??

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 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 =

2π L

0 1 2μ I IF =

2π 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.

Example 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

2π L

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

0 1 2μ I I = mg

2π L

0 1 2μ I I mg = 2π L

1 20

2π m g L I I =

μ

20 1

2π 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.

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.

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.

“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 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