Lecture14 Magnets; Magnetic Field; Magnetic Forces on Charged Particles; Magnetic Flux; Gauss' Law...

8/3/2019 Lecture14 Magnets; Magnetic Field; Magnetic Forces on Charged Particles; Magnetic Flux; Gauss' Law for Magnetism

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Announcements

Contact me by the end of the Wednesdays lecture if youhave special circumstances different than for exam 1.

y xam 2 is one week from tomorrow.

If you took the Exam 1 at 4:30, 5:30, or with bandmembers, but will take Exam 2 at the regular time (5:00),also let me know.


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Todays 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 MovingC

harged 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 Lawfor 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 auniform magnetic field.


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There is an important difference between magnetism andelectricity: 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 howmany times you chop it up.

S N

S N S N

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


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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 tothe earths north geographic pole isthe magnets north pole.

The pole of a magnet attracted tothe earths south geographic poleis the magnets south pole.

N

S

Just as we used the electric field to help us explain andvisualize electric forces in space, we use the magnetic field to

help us explain and visualize magnetic forces in space.

Magnetic Fields


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Magnetic field lines point in the same direction that the northpole 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 strongerthe 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

Later Ill give a better definition for magnetic field direction.


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For those of you who arent going to pay attention until youhave been told the secret behind the naming of the earthsmagnetic poles

The north pole of a compass needle is defined as the end that pointstowards the Santa Claus north pole, which experts in the field ofgeomagnetism call the geomagnetic north or the Earths NorthMagnetic Pole.

If you think about it, the people to whom compasses meant the mostsailorsdefined magnetic north as the direction the north poles of theircompass 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 Earths North Magnetic Pole isactually the south magnetic pole of Earths magnetic field.

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


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Heres a picture of the magnetic

field of a bar magnet, using ironfilings to map out the field.

The magnetic field ought to remindyou of the earths field.


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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 kg1 T =

Cs

In a bit, well see how the units are related to other quantitieswe know about, and later in the course well see an officialdefinition of the units for the magnetic field.

*Old unit, still sometimes used: 1 Gauss = 10-4 Tesla.

These units come from themagnetic force equation, whichappears two slides from now.


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The earths magnetic field has amagnitude of roughly 0.5 G, or0.00005 T. A powerful perm-

anent magnet, like the kind youmight find in headphones, mightproduce a magnetic field of 1000G, or 0.1 T.

The electromagnet in the basementof Physics that my students use inexperiments can produce a field of26000 G = 26 kG = 2.6 T.

Superconducting magnets can produce a field ofover 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


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Todays agenda:


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




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A charged particle moving in a magnetic field experiences aforce.

Magnetic Fields and Moving Charges

The magnetic force equation predicts the effect of a

magnetic field on a moving charged particle.

v

& &

&

F= qv B

force onparticle

magnetic fieldvector

velocity ofchargedparticle

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

Oh nooo! Thelittle voicesare back.


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Vector notation conventions:

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

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


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Magnitude of magnetic force (and the meaning of vB and BB):

B BF = q vB sin = q v B = q vB

+ UB

v+ U

B

v

vB BB

Cross product as presented in Physics 23.

BF = q v B BF = q vB


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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 othervariation on the right hand rule. Ill demonstrate all variations inlecture 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


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Direction of magnetic force:

B BF = q vB sin = q v B = q vB

+ U

F

B

v U

F

B

v-

Uv

B-F?

Fingers out in direction of velocity, thumb perpendicularto them. rotate your hand until your palm points in the

direction of magnetic field. Thumb points in direction ofmagnetic force on + charge.


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Foolproof technique for calculating both magnitude anddirection of magnetic force.

v& &

&

F = qv B

-

&

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.


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Example: a proton is moving with a velocity v = v0j in a region

of uniform magnetic field.The resulting force is F

=F0i. Whatis the magnetic field (magnitude and direction)?

^

^

To be worked at the blackboard.

Example: in the above example, what is the minimum

magnetic field that can be present (magnitude and direction)?


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

Homework Hint

F = q vB sin

Conversely, the fact that there is no magnetic force along somedirection does not mean there is no magnetic field along oropposite to that direction.

Its OK to use if you know that v and B areperpendicular, or you are calculating a minimum field toproduce a given force (understand why).

F = q vB


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Alternative* view of magnetic field units.

F = q vB sin

FB =

q v sin

? A N N

B = T = =C m/s A m

*Official definition of units coming later.

Remember, units of field areforce per something.


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Quiz time (maybe for points, maybe just for practice!)


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Todays agenda:


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




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Magnetic Flux and Gauss Law for Magnetism

Magnetic Flux

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

We are now going to count thenumber of magnetic field lines passing

through a surface, and use this countto determine the magnetic field.

B


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

Because magnetic field lines aredrawn arbitrarily, we quantify

magnetic flux like this: *M=BA.

If the surface is tilted, fewer lines cutthe surface.

BA

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

B

U


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B

U

U A

The amount of surface perpendicular

to the magnetic field is Acos U.

A`` = A cos U so *M = BA`` = BA cosU.

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

Because A is perpendicular to the surface, the amount of Aparallel to the electric field is AcosU.

Remember the dot product from Physics 23?M

B A* ! &&


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

divide the surface intoinfinitesimal surfaceelements and add theflux through each

dAB

i

M i iA 0

ilim B A( p* ! (

&&

MB dA* !

&&

your starting equationsheet has

B B dA* ! &&

M B A BA cos* ! ! U&&

if possible, use


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If the surface is closed (completely encloses a volume)

B

we count lines going outas positive and lines goingin as negative

M B dA* ! &&

dAa surface integral, therefore adouble integral

But there are no magnetic monopoles in nature (jury is still outon 2009 experiments, but lack of recent developments suggestsnothing to see). If there were more flux lines going out of thaninto the volume, there would be a magnetic monopole inside.


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B

Therefore

M B dA 0* ! !&&

dA Gauss Law for Magnetism!

Gauss Law for magnetism is not very useful in this course. Theconcept of magnetic flux is extremely useful, and will be usedlater!

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


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You have now learned Gausss Law for both electricity andmagnetism.

enclosed

o

qE dA !

I

&&

These equations can also be written in differential form:

0

EV

!I

& &

B dA 0 !&&

B 0 !& &

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


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This will not be tested on the exam.

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


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Todays agenda:


Magnetic Force on Moving Charged Particles.

You must be able to calculate the magnetic force on moving charged particles.


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


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

Motion of a charged particle

in a uniform magnetic field


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

The above paragraph is adescription of uniform

circular motion.

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

The force on the electron(remember, its charge is -) isalways perpendicular to thevelocity. If v and B are

constant, then F remainsconstant (in magnitude).

-

B

v

F

v

F

-

-


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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 amagnitude qvB. For circularmotion, a = v2/r so

The rotational frequency f iscalled the cyclotron frequency

The period T is

2mvF = q vB =

r

q rB mvv = r =

m q B

2 r 2 mT = =

v q B

q B1f= =

T 2 m

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

Remember: you can dothe directions by hand

and calculate usingmagnitudes only.


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Helical motion in a uniform magnetic field

If v and B are perpendicular, acharged particle travels in a circularpath. v remains constant but thedirection of v constantly changes.

If v has a component parallel to B,then v`` remains constant, and thecharged particle moves in a helicalpath.

v``

vB

B

+

There wont be any test problems onhelical motion.


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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


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Apply B-field perpendicular to plane


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Lorentz Force Law

If both electric and magnetic fields are present, .v& & &

&

F = q E+v B

Applications

The applications in the remaining slides are in your textbooksections for the next lecture.

If I have time, I will show them today.

The energy calculation in the mass spectrometer example isoften useful in homework.


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Velocity Selector

E

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

v

Bout

+ + q

When the electric and magnetic forces balance then the chargewill pass straight through. This occurs when FE = FB or

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

& &

EF = qE

v& &

&

BF = qv B

EqE = qvB or v =

B


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B

S

(V

+q

x

r

Mass Spectrometers

Mass spectrometers separate charges of different mass.

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

The radius of the pathdepends on themass/charge ratio andspeed of the ion, and

the magnitude of themagnetic field.

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


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B

S

(V

+q

x

r

Example: Ions from source S enter a region of constant magneticfield B that is perpendicular to the ions path. The ions follow asemicircle and strike the detector plate at x = 1.7558 m from the

point where they entered the field. If the ions have a charge of1.6022 x 10-19 C, the magnetic field has a magnitude of B = 80.0mT, and the accelerating potential is (V= 1000.0 V, what is themass of the ion?

Radius of ion path:

Unknowns are m and v.

mvx = 2r and r =

qB

(from slide 36,with positive q)


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B

S

(V

+q

x

r

Conservation of energy gives speed of ion. The ions leave thesource with approximately zero kinetic energy

i i f f K +U =K +U

f i i f K =K + U -U

(f f i f iK = - U -U = -q V -V = -q V

(21 mv = -q V

2

(-2q Vv =

m

(fK = -q V

0

A proton accelerates when itgoes from a more positive Vto a less positive V; i.e., when(V is negative. Thats whatthis minus sign means.

Caution! V is potential,v (lowercase) is speed.


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B

S

(V

+q

x

r

(-2q Vv =

m

2mvx = 2r =

qB

(2m -2q Vx = qB m

(2 -2m Vx =

B q

(

2 2B x qm = -

8 V


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B

S

(V

+q

x

r

(

2 2B x qm = -

8 V

v2 2 -190.08 T 1.7558 m 1.6022 10 C

m = -8 -1000 V

v -25m = 3.9515 10 kg

m = 238.04 u

1 atomic mass unit(amu) equals

1.66x10-27 kg, so

uranium-238!

Lecture14 Magnets; Magnetic Field; Magnetic Forces on Charged Particles; Magnetic Flux; Gauss' Law...

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Transcript of Lecture14 Magnets; Magnetic Field; Magnetic Forces on Charged Particles; Magnetic Flux; Gauss' Law...