Announcements Contact me by the end of the Wednesday’s lecture if you have special circumstances different than for exam 1. Exam 2 is one week from

Announcements

Contact me by the end of the Wednesday’s lecture if you have special circumstances different than for exam 1.

Exam 2 is one week from tomorrow.

Today’s agenda:

Magnetic Fields.You must understand the similarities and differences between electric fields and field lines, and magnetic fields and field lines.

Magnetic Force on Moving Charged Particles.You must be able to calculate the magnetic force on moving charged particles.

Magnetic Flux and Gauss’ Law for Magnetism.You must be able to calculate magnetic flux and recognize the consequences of Gauss’ Law for Magnetism.

Motion of a Charged Particle in a Uniform Magnetic Field.You must be able to calculate the trajectory and energy of a charged particle moving in a uniform magnetic field.

Magnetism

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

Similarly, there are two kinds of magnetic poles (north and south), and like poles repel, opposites attract.

S N

S N SN

S N

Repel

Attract

S N

SNS N

Repel Attract

S N

There is an important 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 matter how many times you “chop it up.”

S NS N S N

*But see here, here, here, and here.

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 Fields

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?

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

For those of you who aren’t going to pay attention until you have been told the secret behind the naming of the earth’s magnetic poles…

The north pole of a compass needle is defined as the end that points towards the geographic “Santa Claus” north pole, which experts in the field of geomagnetism call “the geomagnetic north” or “the Earth’s North Magnetic Pole.”

If you think about it, the people to whom compasses meant the most—sailors—defined magnetic north as the direction the north poles of their compass needles pointed. Their lives depended on knowing where they were, so I guess it is appropriate that we acknowledge their precedence. Thus, by convention, the thing we call Earth’s North Magnetic Pole is actually the south magnetic pole of Earth’s magnetic field.

For those of you who are normal humans, just ignore the above.

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.

We use the symbol B for magnetic field.

Remember: magnetic field lines point away from north poles, and towards south poles.

S N

The SI unit* for magnetic field is the Tesla.

1 kg

1 T =C s

In a bit, we’ll see how the units are related to other quantities we know about, and later in the course we’ll see an “official” definition of the units for the magnetic field. *Old unit, still sometimes used: 1 Gauss = 10-4 Tesla.

These units come from the magnetic force equation, which appears three slides from now.

The earth’s magnetic field has a magnitude of roughly 0.5 G, or 0.00005 T. A powerful perm-anent magnet, like the kind you might find in headphones, might produce a magnetic field of 1000 G, or 0.1 T.

The electromagnet in the basement of Physics that my students (used to) use in experiments can produce a field of 26000 G = 26 kG = 2.6 T. Superconducting magnets can produce a field

of over 10 T. Never get near an operating super-conducting magnet while wearing a watch or belt buckle with iron in it!

http://liftoff.msfc.nasa.gov/academy/space/mag_field.html

Today’s agenda:





A charged particle moving in a magnetic field experiences a force.

Magnetic Fields and Moving Charges

The following equation (part of the Lorentz Force Law) predicts the effect of a magnetic field on a moving charged particle:

F=qv B

force on particle

magnetic field vector

velocity of charged particle

What is the magnetic force if the charged particle is at rest?

Oh nooo! The little voices are back.

What is the magnetic force if v is (anti-)parallel to B?

Vector notation conventions:

is a vector pointing out of the paper/board/screen (looks like an arrow coming straight for your eye).

is a vector pointing into the paper/board/screen (looks like the feathers of an arrow going away from eye).

Use right hand rule:

fingers out in direction of v, thumb perpendicular to them

(or bend fingers through smallest angle from v to B)thumb points in direction of F on + charge

Direction of magnetic force---

Your text presents two alternative variations (curl your fingers, imagine turning a right-handed screw). There is one other variation on the right hand rule. I’ll demonstrate all variations in lecture sooner or later.

still more variations: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html

rotate your hand until your palm points in the direction of B

Complication: your Physics 1135 text uses the right hand rule shown in the figure, so we’ll use the same rule in Physics 2135.

There are a number of variations of this rule.

Unfortunately, most of the Youtube videos I find say to use your palm for , your thumb for , and your outstretched fingers for This includes the MIT Open Courseware site.

A

B

A B .

I’ll show you later that both ways are correct. I am going to use our textbook’s technique. You can use whatever works for you!

Direction of magnetic force:

θF= q vB sin

+

F

B

v

F

B

v-

v

B-F?

Fingers of right hand out in direction of velocity, thumb perpendicular to them. Bend your fingers until they point in the direction of magnetic field. Thumb points in direction of magnetic force on + charge.

Force is “down” because charge is -.

I often do this: fingers out in direction of velocity, thumb perpendicular to them. Rotate your hand until your palm points in the direction of magnetic field. Thumb points in direction of magnetic force on + charge.

“Foolproof” technique for calculating both magnitude and direction of magnetic force.

F=qv B

ˆˆ ˆ

x y z

x y z

i j k

F=q det v v v

B B B

All of the right-hand rules are just techniques for determining the direction of vectors in the cross product without having to do any actual math.

http://www.youtube.com/watch?v=21LWuY8i6Hw

Example: a proton is moving with a velocity v = v0j in a region of uniform magnetic field. The resulting force is F = F0i. What is the magnetic field (magnitude and direction)?

^

^

To be worked at the blackboard.

Blue is +, red is -. Image from http://www.mathsisfun.com/algebra/matrix-determinant.htm


^

^

ˆ0v=v j ˆ

0F=F i

F=qv B

ˆ ˆˆ ˆ ˆ ˆ

ˆ

0 x y z 0

x y z x y z

i j k i j k

F=F i=q det v v v = e det 0 v 0

B B B B B B

ˆˆ ˆ ˆ 0 0 z y z x y x 0F i= e i v B B 0 j 0 B B 0 k 0 B B v


^

^

ˆ ˆˆ ˆ ˆ ˆ ˆ 0 0 z x 0 0 z x 0F i= ev B i 0j eB v k = ev B i+ 0 j+ eB v k

0 0 z x 0F = ev B and 0= eB v

0z x

0

FB = and B =0

ev

What is By???? It could be anything. Not enough information is provided to find By.What is By???? It could be anything. Not enough information is provided to find By. We can’t find the magnitude and direction of the magnetic field using only the information given!

Example: a proton is moving with a velocity v = v0j in a region of uniform magnetic field. The resulting force is F = F0i. What is the minimum magnitude magnetic field that can be present?

^

^

yB = 0 for a minimum magnitude magnetic field, so...

0min

0

FB =

ev

0z x

0

FB = and B =0

ev

A charged particle moving along or opposite to the direction of a magnetic field will experience no magnetic force.

Homework Hint

θF= q vB sin

Conversely, the fact that there is no magnetic force along some direction does not mean there is no magnetic field along or opposite to that direction.

It’s OK to use if you know that v and B are perpendicular, or you are calculating a minimum field to produce a given force (understand why).

F= q vB

Alternative* view of magnetic field units.

θF= q vB sin

θ

FB=

q v sin

N N

B =T = =C m/s A m

*“Official” definition of units coming later.

Remember, units of field are force per “something.”

Today’s agenda:





Magnetic Flux and Gauss’ Law for Magnetism

Magnetic Flux

We have used magnetic field lines to visualize magnetic fields and indicate their strength.

We are now going to count the number of magnetic field lines passing through a surface, and use this count to determine the magnetic field.

B

The magnetic flux passing through a surface is the number of magnetic field lines that pass through it.

Because magnetic field lines are drawn arbitrarily, we quantify magnetic flux like this: M=BA.

If the surface is tilted, fewer lines cut the surface.

BA

If these slides look familiar, refer back to lecture 4!

B

B

A

The “amount of surface” perpendicular to the magnetic field is A cos .

A = A cos so M = BA = BA cos .

We define A to be a vector having a magnitude equal to the area of the surface, in a direction normal to the surface.

Because A is perpendicular to the surface, the amount of A parallel to the electric field is A cos .

Remember the dot product from Physics 1135?

M B A

If the magnetic field is not uniform, or the surface is not flat…

divide the surface into infinitesimal surface elements and add the flux through each…

dAB

iM i iA 0

i

lim B A

M B dA

your starting equation sheet hasB B dA

M B A BA cos if possible, use

If the surface is closed (completely encloses a volume)…

B

…we count lines going out as positive and lines going in as negative…

M B dA

dA

a surface integral, therefore a double integral

But there are no magnetic monopoles in nature (jury is still out on 2009 experiments, but lack of recent developments suggests nothing to see). If there were more flux lines going out of than into the volume, there would be a magnetic monopole inside.

B

Therefore

M B dA 0

dA Gauss’ Law for Magnetism!

Gauss’ Law for magnetism is not very useful in this course. The concept of magnetic flux is extremely useful, and will be used later!

This law may require modification if the existence of magnetic monopoles is confirmed.

You have now learned Gauss’s Law for both electricity and magnetism.

enclosed

o

qE dA

These equations can also be written in differential form:

0

E

B dA 0

B 0

Congratulations! You are ½ of the way to being qualified to wear…

This will not be tested on the exam.

The Missouri S&T Society of Physics Student T-Shirt!

Today’s agenda:





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

Motion of a charged particlein a uniform magnetic field

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

The above paragraph is a description of uniform circular motion.

The electron will move in a circular path with a constant speed and acceleration = v2/r, where r is the radius of the circle.

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).

-

B

vF

v

F

-

-

Motion of a proton in a uniform magnetic field

+

Bout

v

FB

+ +

v

v

FBFB

r

The force is always in the radial direction and has a magnitude qvB. For circular motion, a = v2/r so

The rotational frequency f is called the cyclotron frequency

The period T is

2mvF= q vB=

r

qrB mv

v= r=m qB

π π2 r 2 mT = =

v qBπ

qB1f = =

T 2 m

Thanks to Dr. Waddill for the use of the picture and following examples.

Remember: you can do the directions “by hand” and calculate using magnitudes only.

Helical motion in a uniform magnetic field

If v and B are perpendicular, a charged particle travels in a circular path. v remains constant but the direction of v constantly changes.

If v has a component parallel to B, then v remains constant, and the charged particle moves in a helical path.

v

v

B

+

There won’t be any test problems on helical motion.

*or antiparallel

Electrons confined to move in a plane

Thanks to Dr. Yew San Hor for this and the next slide.

Apply B-field perpendicular to plane

Apply B-field perpendicular to plane

Lorentz Force Law

If both electric and magnetic fields are present, . F=q E+v B

Applications

See your textbook for numerical calculations related the next two slides.

If I have time, I will show the mass spectrometer today.

The energy calculation in the mass spectrometer example is often useful in homework.

Velocity Selector

E

- - - - - - - - - - - - - - - -

v

Bout

+

The magnetic field is uniform inside the dashed-line region. What speed will allow the proton to pass undeflected through region of uniform electric field?

Velocity Selector

E

- - - - - - - - - - - - - - - -

v

Bout

+

When the electric and magnetic forces balance then the charge will pass straight through.

+

EF =qE

BF =qv B

v

E BF = F

E

qE= q vB or v=B

Velocity Selector

E

- - - - - - - - - - - - - - - -

v

Bout

+

This only works if the electric and magnetic fields are perpendicular and oriented so that is antiparallel to

+

EF =qE

BF =qv B

v

qv BE.

Also, we simplified the calculation by making perpendicular to

v

B.

B

S

V

+q

xR

Mass Spectrometers

Mass spectrometers separate charges of different mass.

When ions of fixed energy enter a region of constant magnetic field, they follow a circular path.

The radius of the path depends on the mass/charge ratio and speed of the ion, and the magnitude of the magnetic field.

Thanks to Dr. Waddill for the use of the picture.

mvx=2r and r=

qB

Worked example with numbers to follow.

B

S

V

+q

xR

Example: ions from source S enter a region of uniform magnetic field B that is perpendicular to the ion path. The ions follow a semicircle and strike the detector plate at x = 1.7558 m from the point where they entered the field. If the ions have a charge of 1.6022 x 10-19 C, the magnetic field has a magnitude of B = 80.0 mT, and the accelerating potential is V = 1000.0 V, what is the mass of the ion?

B

S

V

+q

xR

Step 1: find the final speed of the ions if they have a charge of 1.6022 x 10-19 C and the accelerating potential is V = 1000.0 V?

Let’s focus on the part of the diagram where the ions from the source are being accelerated.If the ions are “boiling” off a source filament, their initial velocity directions will be almost random, so their average speed will be almost zero. Let’s assume the ions start at rest (corresponding to zero average speed).

S

V

+q +qvi = 0

v = ?

f i other i fE E W

f f i i other i fK U K U W

0 0

f f i f iK U +U U U U

21mv U q V

2

2q Vv

m

Notes: we don’t know m, so let’s keep this symbolic for now. Besides, we want to solve for m later. Also, V must be negative, which makes sense (a + charge would be accelerated away from a higher potential).

Don’t get your v’s and V’s mixed up! Use a script v for speed!

Step 1: find the final speed of the ions if they have a charge of 1.6022 x 10-19 C and the accelerating potential is V = 1000.0 V?

B

S

V

+q

xR

Step 2: find the radius of the circular path followed by the ions in the region of uniform magnetic field of magnitude B.

B

+q

xR


Let’s focus on the magnetic field region of the diagram. v

B

+q

xR


vFB

For motion in a circular path...

BF qv B qvBsin 90

2

B

vF ma m

R

2vq vB m

R

mvR

q B

On an exam, we’ll probably let you get away without using the absolute value signs, even though it makes our brains hurt when we do that (let you get away with something).

B

S

V

+q

xR

Step 3: solve for m, using q=1.6022x10-19 C, V=1000.0 V, B = 80.0 mT, and R=x/2=1.7558/2=0.8779 m.

Now we can put it all together.

2q Vv

m

mvR

q B

We know q, R, B, and V. We have two unknowns, v and m, and two equations, so we can solve for m.

2

2 22

2 2 2 22 2 2

2q Vm

2q V m 2 V mm v mR

q Bq B q B q B

B

S

V

+q

xR

22

2 V mR

q B

2 2q B R

m2 V

Step 3: solve for m, using q=1.6022x10-19 C, V=-1000.0 V, B = 80.0 mT, and R=x/2=1.7558/2=0.8779 m.

2 219 31.6022 10 80 10 0.8779m

2 1000

25m 3.9515 10 kg 1 atomic mass unit (u) equals 1.66x10-27 kg, so m=238.04 u.

uranium-238!

Documents

Announcements Contact me by the end of the Wednesday’s lecture if you have special circumstances different than for exam 1. Exam 2 is one week from