5 magnetism revised
TRANSCRIPT
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MagnetismMagnetism
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The ScienceThe Science
Not until 1819 was a connection between electrical andmagnetic phenomena shown. Danish scientist Hans Christian
Oersted observed that a compass needle in the vicinity of awire carrying electrical current was deflected!
In 1831, Michael Faraday discovered that a momentarycurrent existed in a circuit when the current in a nearbycircuit was started or stopped
Shortly thereafter, he discovered that motion of a magnettoward or away from a circuit could produce the same effect.
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The Connection is MadeThe Connection is Made
SUMMARY: Oersted showed that magnetic effects
could be produced by moving electrical charges;Faraday and Henry showed that electric currentscould be produced by moving magnets
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All magnetic phenomena
result from forcesbetween electric chargesin motion.
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For Every North, Thereisa SouthFor Every North, Thereisa South
Every magnet has at least one north pole and one south pole. Byconvention, we say that the magnetic field lines leave the North endof a magnet and enter the South end of a magnet.
If you take a bar magnet and break it into two pieces, each piece willagain have a North pole and a South pole. If you take one of thosepieces and break it into two, each of the smaller pieces will have aNorth pole and a South pole. No matter how small the pieces of themagnet become, each piece will have a North pole and a South pole.
S N S N S N
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Like repels likeLike repels like
Opposites attract!Opposites attract!
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No Monopoles AllowedNo Monopoles Allowed
It has not been shown to be possible to end up with a singleNorth pole or a single South pole, which is a monopole ("mono"means one or single, thus one pole).
Note: Some theorists believe that magnetic monopoles mayhave been made in the early Universe. So far, none have beendetected.
S N
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Magnets Have Magnetic FieldsMagnets Have Magnetic Fields
We will say that a moving charge sets up in the spacearound it a magnetic field,
andit is the magnetic field which exerts a force on anyother charge moving through it.
Magnetic fields are vectorquantities.that is, they have a magnitudeand a direction!
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Defining Magnetic Field DirectionDefining Magnetic Field Direction
Magnetic Field vectors written as B
Magnitude of the B-vector is proportional tothe force acting on the moving charge,magnitude of the moving charge, the magnitudeof its velocity, and the angle between v and theB-field. Unit is the Tesla or the Gauss (1 T =10,000 G).
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Magnetic Field LinesMagnetic Field Lines
Magnetic field lines describe the structure of magnetic fieldsin three dimensions. They are defined as follows. If at anypoint on such a line we place an ideal compass needle, free to
turn in any direction (unlike the usual compass needle, whichstays horizontal) then the needle will always point along thefield line.
Field lines converge where the magnetic force is strong, andspread out where it is weak. For instance, in a compact bar
magnet or "dipole," field lines spread out from one pole andconverge towards the other, and of course, the magneticforce is strongest near the poles where they come together.
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Magnetic Field Sources
Magnetic fields are produced by electric currents, which can bemacroscopic currents in wires, or microscopic currents associated withelectrons in atomic orbits.
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Magnetic Field Concepts, Interactions and Applications
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Force on the Charge: Lorentz ForceForce on the Charge: Lorentz Force
Right Hand Rule!
Put your fingers in the direction of motion ofthe charge, curl them in the direction of themagnetic field. Your thumb now points in thedirection of the magnetic force acting on thecharge. This force will bend the path of themoving charge appropriately.
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Magnetic Force LawMagnetic Force Law
magnitude: F magnetic = q v B sin(vB)
direction: right hand rule:
thumb = hand v fingers
Point your right hand in the direction of v,
curl you fingers in the direction of B, and
the force will be in the direction of your
thumb; if the charge is negative, the force
direction is opposite that ofyourthumb.
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The Lorentz ForceThe Lorentz Force
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Lorentz ForceLorentz Force
If a point charge is moving in a regionwhere both electric and magnetic fieldsexist, then it experiences a total forcegiven by
The Lorentz force equation is usefulfor determining the equation of motionfor electrons in electromagneticdeflection systems such as CRTs.
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BvEqFFF me v!!
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MagneticMagnetic fluxflux
Magnetic flux is definedexactly as electric flux
(Component of B B surface) x (Area element)
*B ! By dA
zero flux Maximum flux
SI unit of magnetic flux is the Weber ( = 1 T-m2
)
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Magnetic fluxMagnetic flux
Whatisthatmagneticflux throughthis
surface?
A. PositiveB. NegativeC. Zero
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Gauss law inGauss law in magnetostaticsmagnetostatics
Net magnetic flux through any closedsurface is always zero:
*magnetic ! 0
Basic magnetic element is the dipole
*electricQenclosed
Io
Compare to Gausslaw for electricfield
*B ! By dA = 0
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At any point P the magnitude of the magnetic fieldintensity produced by a differential element is
proportional to the product of the current, themagnitude of the differential length, and the sine ofthe angle lying between the filament and a lineconnecting the filament to the point P at which thefield is desired; also, the magnitude of the field is
inversely proportional to the square of the distancefrom the filament to the point P. The constant ofproportionality is 1/4T
MATHEMATICALLY:
BiotBiot--Savart LawSavart Law
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BiotBiot--Savart LawSavart Law
So, the magnetic field circulates around the wire
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Magnetic Field of g Straight Wire
Rewrite in terms ofRewrite in terms of R,R,UU::
Calculate field at point P usingBiot-Savart Law:
x
RrU
P
Idx
@
Which way is B?
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Magnetic Field ofMagnetic Field of gg Straight WireStraight Wire
@
x
Rr
U
P
Idx
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Magnetic field lines produced by aMagnetic field lines produced by a
current in a long straight wirecurrent in a long straight wire
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` What is the magnitude of the magneticfield at the centerof a loop ofradius R,
carrying current I?
(a) B = 0 (b) B !Q-R (c) B !Q-TR
To calculate the magnetic field at the center, wemust use the Biot-Savart Law:
Id
r
Two nice things about calculating B at the center ofthe loop:
Idx is always perpendicular to r r is a constant (=R)
R
IPROBLEM
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Magnetic field from a circular current loop
I
dBzdBperp
B
r z
R
dl E
d
r
B ! Q04T
Id
r
l
vr
r2
dB !Q0
4T
Idl
r
2
dBz! dBcosE !
Q0
4T
IdlcosEr2
r ! R2 z2 ,cosE ! RR
2 z2
Only z component
is nonzero.
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I
dBzdBperp
B
r z
R
dl E
At the center of the loop
At distance z on axisfrom the loop, z>>R
Mag tic fi l fr a circ lar c rr t l
B ! dBz!
Q0
4T
IR
(R2 z2 )32 dl
B !Q
0
4T
IR
(R2 z2 )32
dl !
B !Q
0
4TIR
(R2 z
2)32
2TR !Q
0
2
IR2
(R2 z
2)32
B !Q
0
I
2R
B !Q
0IR
2
23
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The magnetic dipole moment of the loopis defined as m = IA =ITR2.
The direction is given by the right handrule: with fingers closed in
the direction of the current flow, thethumb points along m.
Magnetic field in terms of dipole moment
I
B
z
R
Far away on the axis,
Q
BQ0IR
2
2z3
3
0
2 z
mB
T
Q!
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Magnetic Field Due to a Current in a Circular Arc of Wire
2J T!
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The wire in Fig. a carries a current i and consists of a
circular arc of radius R and central angle rad, andtwo straight sections whose extensions intersect thecenter C of the arc. What magnetic field does thecurrent produce at C?
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Sum up component ofB around path Equals
current through surface.
Ampereslaw
By ds QoIclosedpath
surfacebounded by path
I
rB
ComponentofB
alongpath
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rT
Q
2
0!
Ucosdd y!y
I
B
r
ds
1cos0 !! UU dsBdsB !y
dsrT2
0!ydsB
Using Biot-Savart Law
Take a shortvector on acircle, ds
Thus the dot product of B& the short vector ds is:
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I
B
r
ds
dsr
I
T
Q
2
0!yds
yds
Sum this around the whole ring
! dsr
I
T
Q
2
0 ! dsr
I
T
Q
2
0
rds T2! rr
I0
02
2Q
Q!!y d
Circumference of circle
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Sum B.ds around any path
0
path
Q!y dsB
this does not depend on r0circ.
Q!yds
In fact it does not depend onpath
ypathds
Amperes Law: on any closedloop
0Q!
where I is the current flowing through the loop
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Amperes Law
0
path
Q!y ds
B
I
I
0
path
2Q!y ds
0
path
!y dsBB
I
I
Sign comes fromdirection of loop,current & right handrule
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` No Different Physics from Biot-Savart Law
` Useful in cases where there is a high degree ofsymmetry
` Can be compared with Coulombs Law and
Gausss Law in electrostatics in functionality.
0
path
Q!y dsB
y! dsB
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Cal lat i ldat distance r ire singAmpere's aw:
Ampere's aw simpli ies thecalculati n thanks tsymmetry thecurrent! ( axial/cylindrical )
dl
vRI
Choose loop to be circle of radius Rcentered on the wire in a plane B to
wire.
Evaluate line integral inAmperes aw: Current enclosed by path = I
Apply Amperes Law:
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Magnetic field inside a wire current carrying I0
Assuming that the current density isuniform then the current flowing
through the loop is
A
We Take our Ampere loop to be a circle of radius r
0I
A
aI !
r
02
2
IRT
T!
02
2
IR
!
Now same as before
I0
Circle
B2 QT !!y dsB 020 2I
RTQ!
r
I
T
Q
2
0!
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Field from a long wire
B
r
020 2I
rB
TQ!
r
IB
T
Q
2
00!
R
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` A current I fl in an infinite strai tire in t e +z directi n as shown. A
concentric infinite cylinder ofradiuscarries current 2I in the -z direction.
hat is the magnetic field (a) at
oint a, just outside the cylinder asshown?
(a) Bx(b) < 0 (b) Bx(b) = 0 (c) Bx(b) > 0
What is the magnetic field Bx(b) at point b, justinside the cylinder as shown?
(a) Bx(a) < 0 (b) Bx(a) = 0 (c) Bx(a) > 0
xx
xxxx
x
x
2II
a
b
x
yPROBLEM
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This situation has massive cylindrical symmetry!Applying Amperes Law, we see that the field at point amust just be the field from an infinite wire with current Iflowing in the -z direction!
xI
B
B
B
xx
xx
xx
x
x
2II
a
b
y
(a) Bx(a) < 0 (b) Bx(a) = 0 (c) Bx(a) > 0
ANSWERI
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Just inside the cylinder, the total current enclosed by theAmpere loop will be I in the +z direction!
Therefore, the magnetic field at b will just be minus themagnetic field at a!!
xx
xx
xx
x
x
2II
a
b
y
(a) Bx(a) < 0 (b) Bx(a) = 0 (c) Bx(a) > 0
(a) Bx(b) < 0 (b) Bx(b) = 0 (c) Bx(b) > 0
What is the magnetic field Bx(b) at point b, justinside the cylinder as shown?
ANSWERII
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Magnetic field due to a Toroidal Coil
r
I0
Ampere Loop,circle radius r
No current flowingthrough loop thus B = 0inside the Toroid
Toroid has N loops of wire,carrying a current I0
TOROID
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Toroidal Coil
r
I0
Ampere Loop,circle radius r
For each wire going inthere is another wirecomeing out Thus nonett current flowingthrough loop thus B = 0outside the Toroid
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Toroidal Coil
rI0
Ampere Loop,circle radius r
For each loop of the coil
an extra I0 of currentpasses through theAmpere Loop
Zoom
B2Circle
rT!y dsB I0Q!
00NIQ!
Toroid has N loops of wire
r
NI
T
Q
2
00!
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Magnetic Field of a Solenoid
For a long ideal solenoid
where n is the number of turns per unit length of the solenoid
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Magnetic Force on CurrentMagnetic Force on Current--Carrying WireCarrying Wire
Since moving charges experience a force in a magnetic field, a current-
carrying wire will experience such a force, since a current consists of
moving charges. This property is at the heart of a numberof devices.
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x Calculate force on length L of wire a due tofield of wire b:
The field at a due to b is given by:
Calculate force on length L of wire bdue to field of wire a:
The field at b due to a is given by:
v
F
v
F
L dIb
Ia
L dIb
Ia
Force on b =
Force on a =
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y This expression is used to define a standard 1 Amperecurrent:
The constant current which will produce an attractiveforce of 2 107 Newton per metre of length between
two straight, parallel conductors of infinite length andnegligible circular cross section placed one metre apartin a vacuum.
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A current I flows in the positive ydirection in an infinite wire; a current I
also flows in the loop as shown in thediagram.
What is Fx, net force on the loop in the
x-direction?
(a) Fx < 0 (b) Fx =0 (c) Fx > 0
I
I
x
y
I
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You may have remembered from a previous ACT that the net force on acurrent loop in a constant magnetic field is zero. However, the magnetic field produced by the infinite wire is not aconstant field!!
Fleft
FrighX
The direction of the magnetic field at the current loop is in the -z direction.
Ftop
Fbottom
The forces on the top and bottom segments of the loop DO indeed cancel!!
The forces on the left and right segments of the loop DO OT cancel!! The left segment of the loop is in a larger magnetic field. Therefore, F
left
> Fright
I
I
x
y
(a) Fx < 0 (b) Fx =0 (c) Fx > 0