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Transcript of Biot Savart Laws
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EEE1001/PHY1002 1
Fields, Materials & Devices
1
Biot-Savart Law• Jean-Baptiste Biot and Felix Savart also studied
the forces on wires around the same time asAmpere developed his theories.• To summarise their observations, at a point a
distance r from the current qu ,
• Θ is the angle between the direction of thecurrent and the point at which the magnetic fieldis being obtained.
B = µ qu sin θ
4π r 2
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EEE1001/PHY1002 2
Fields, Materials & Devices
2
Biot-Savart Law• There is a simple analogy here between
electric and magnetic fluxes:
• We can compare the source of electricfield being the charge with the source ofmagnetic field being a moving charge .
B = µ qu sin θ
4π r 2
q
4πε 0ε r r 2r E =
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EEE1001/PHY1002 3
Fields, Materials & Devices
3
Biot-Savart Law• The above expression for B is not the
standard way of viewing the Biot-SavartLaw.
• Instead we base it upon a current element .
• The charge passing a point in unit time isthe current, – qu=idl
dl i q u
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EEE1001/PHY1002 4
Fields, Materials & Devices
4
Biot-Savart Law• So
becomes
with dB being a small contribution to the totalmagnetic flux density arising from the small partof the current in the wire element “idl” .
B = µ qu sin θ
4π r 2
dB = µ i sin θ dl 4π r 2
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EEE1001/PHY1002 5
Fields, Materials & Devices
5
Biot-Savart Law• We can eliminate B in favour of H :
• The direction of the magneticfield is into the plane of thescreen at r
– This can be obtained using the “righthand screw rule”.
dH =i sin θ dl
4π r 2
+
dl
θ
r
i
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Biot-Savart Law• It would be a mistake to interpret the Biot-Savart
law as stated as indicating that there is a smallamount of magnetic field coming from each
element.
• The law is only complete in the integral form:
Fields, Materials & Devices
7
dH
∫= H =
i sin θ dl
4π r 2∫
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Biot-Savart Law
Fields, Materials & Devices
8
H =
∫4π r 3
i dl x r
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EEE1001/PHY1002 9
Application of the Biot-Savart Law
• The Biot-Savart law in this form can be used to
obtain the magnetic field due to some importantsystems.
• We shall now look at the example of the fieldabout a long straight wire…
Fields, Materials & Devices
9
dH
∫= H =
i sin θ dl
4π r 2
∫
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EEE1001/PHY1002 10
Application of the Biot-Savart Law
• The application
requires us toevaluate the integralfor a point relative tothe wire, effectivelysumming over all
current elementsalong an infinite wire.
Fields, Materials & Devices
10
Current, i
B , H
+
B , H
i
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EEE1001/PHY1002 11
Application of the Biot-Savart Law
• The sum is over
elements of thistype, so that
Fields, Materials & Devices
11
+
i
dl l
α
θ
r a H = i sin θ dl 4π r 2∫
∞
-∞
H = i cos α dl
4π r 2∫∞
-∞
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EEE1001/PHY1002 12
Application of the Biot-Savart Law
• To evaluate the
integral, we needeliminate theinterdependence ofterms such as rand α.
• We can express l and r in terms of α.
Fields, Materials & Devices
12
+
i
dl l
α
θ
r a
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Application of the Biot-Savart Law
1. a tan α = l
2. r cos θ = a
• Differentiating 1
gives – dl = a sec2α d α
Fields, Materials & Devices
13
+
i
dl l
α
θ
r a
H = i cos α a d α 4π r 2cos2α ∫
π /2
- π /2
Note, when l=- ∞, α =- π /2,
and when l= ∞, α = π /2
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Application of the Biot-Savart Law
1. a tan α = l
2. r cos θ = a
• Substituting 2
gives
Fields, Materials & Devices
14
+
i
dl l
α
θ
r a
H = i cos α a d α 4π a 2∫
π /2
-π /2
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EEE1001/PHY1002 15
Application of the Biot-Savart Law
can be integratedreadily, yielding:
Fields, Materials & Devices
15
+
i
dl l
α
θ
r a
H = i cos α a d α
4π a 2∫π /2
-π /2
H =i /2π a
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EEE1001/PHY1002 16
The infinite, straight, uniform wire
Fields, Materials & Devices
16
B , H
i a
H = i
2π a
B = µ i
2π a
Geometrically, the magnetic
field strength, H , is equal tothe current divided by theloop length, 2π a .
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EEE1001/PHY1002 17
The field for a finite wire…• What happens at the ends of a wire?
– Hint: consider a semi-infinite wire.
Fields, Materials & Devices
17
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EEE1001/PHY1002 18
Ampere’s Law• We can consider the generality of the
observed geometrical interpretation of theresult for the infinite wire.
• We integrate the magnetic field around a
closed loop enclosing the current…
Fields, Materials & Devices
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H = i 2π a
∮H.dl
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EEE1001/PHY1002 19
Ampere’s Law• Since H and dl are always parallel, the
scalar product is simply the product of themagnitudes:
Fields, Materials & Devices
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∮H.dl = i dl
2π a ∫0
2π a
= i
B , H
i a
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EEE1001/PHY1002 20
Ampere’s Law• Ampere showed that this is generally true,
so that
• This is Ampere Ampere ’ ’ s Law s Law .
• In words, the law states that “The lineintegral of H around any closed path isequal to the current enclosed”.
Fields, Materials & Devices
20
∮H.dl = i
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Gauss’ Law• The circulation of magnetic
field lines leads to an importantresult. – This is in stark contrast to
electrostatics where electricalcharge is a source of divergingor converging field lines
• The fact that magnetic fluxcirculates means that for anyclosed surface, as much fluxmust enter as leave
– This is a conservation orcontinuity law
• There are no isolated magneticflux sources or sinks
Fields, Materials & Devices
21
Electric dipole
Flux continuity
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EEE1001/PHY1002 22
Gauss’ Law
• This is Gauss’ Law of magnetic fields.
– Do not confuse this with Gauss’ law forelectrostatics
• There is no divergence of B
– This refers to a derivative form of the law.
Fields, Materials & Devices
22
∮B.ds = 0s
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EEE1001/PHY1002 23
Faraday’s Law – the beginnings of
electromagnetic machines• In 1831, Michael Faraday
established the principles whichlay the foundations of mostmodern electrical machines. – Up to this time, all electricity came
from chemical cells (batteries).• He inverted the idea
– If an electrical current can produce amagnetic field, then the oppositeshould also be true: a magnetic fieldshould be able to generate a currentin a wire.
Fields, Materials & Devices
23
F ar a d a y
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EEE1001/PHY1002 24
Faraday’s Law – the beginnings of
electromagnetic machines• He spend a long time trying to
show this idea in practice, placingmagnets in the vicinity of wiresand looking for in induced current
or voltage – there was none.• However, eventually it was found
that a moving magnet next to a
wire did induce an e.m.f.
Fields, Materials & Devices
24
F ar a d a y
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EEE1001/PHY1002 25
Faraday’s Law – the beginnings of
electromagnetic machines• This is a simple arrangement
that demonstrates the effect.
• Some of the magnet flux, ψ,from the permanent magnet linksthe coil.
• As the magnet moves theamount of flux linking the coilchanges.
• The voltage induced in the coil is
Fields, Materials & Devices
25
V
V = d ψ dt
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Question
• What wouldhappen if we
added a secondturn to the coil?
Fields, Materials & Devices
26
V
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Application of Faraday’s Law
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Motional EMF• Another view of the
same effect can beobtained by viewing
the impact upon a
moving metal rodpassing through a
magnetic field of flux
density, B .
x y
z E
u
B
F
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Motional EMF• For movement in the
x -direction, and B inthe y-direction, the
right-hand-rule for
the cross-productdirects the force in the
positive z -direction.
• The force acts on theelectrons with charge
-e .
x y
z E
u
B
F z
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Motional EMF• The force on the
electrons move the
electrons in the negativez -direction.
• This creates a spacecharge, and hence anelectric field, E , and aforce given by – F=qE=-eE
– E directed in the negativez- direction, so F is in thepositive z- direction
• C.f. Lenz’s Law
x y
z E
u
B
F z
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Motional EMF• In equilibrium , the
forces balance. – E=uB
• In terms of the vector
quantities, – E=B x u
x y
z E
u
B
e E
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32
Motional EMF• If the rod is of length L,
and moves in a directionperpendicular to themagnetic field, then theinduced voltage is given
by the integral of theelectric field along thelength of the rod.
• This is the same effect asthe Faraday Law we sawearlier.
x y
z E
u
B
= E dl = B L u ∫0
L
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33
Comparison of induced voltages• Faraday’s Law • Lorentz force
V = d ψ dt = E dl = B L u ∫0L
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Application of Ampere’s Law
Field in a solenoid• We imagine an ideal
toroidal solenoid to beuniform turn density,where all turns areperpendicular to the
axis of the ring andthe radius of the ringlarge enough for
variations in the fieldwith radius to benegligible.
Fields, Materials & Devices
34
Iron core, N turns.
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Application of Ampere’s Law
Field in a solenoid• The current enters
and leaves the circuitat (approximately) the
same place.
• We draw a circuit forAmpere’s law through
the axis of the ring.
Fields, Materials & Devices
35
i
r
B
dl
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Application of Ampere’s Law
Field in a solenoid• We have to integrage
B.ds over the circle ofradius r .
• By symmetry H andB are constant in
magnitude on thiscircle.
• B and H are
tangential to thecircle.
Fields, Materials & Devices
i
r
B
dl
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Application of Ampere’s Law
Field in a solenoid• We now apply
Ampere’s law:
• Since H and dl are
parallel, the scalar
product H.dl is justHdl.
Fields, Materials & Devices
i
r
B
dl
∮H.dl = i
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Application of Ampere’s Law
Field in a solenoid• The path length is
2π r .• Since the path
threads N loops
carrying current i , theright hand side is Ni :
Fields, Materials & Devices
i
r
B
dl
∮dl = H2 π r = Ni
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Application of Ampere’s Law
Field in a solenoid• We can note the MMF
is equal to theintegral, so is Ni .
• The magnetc flux
density can beobtained from B= µ H .
Fields, Materials & Devices
i
r
B
dl
∮dl = H2 π r = Ni
EEE1001/PHY1002
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Application of Ampere’s Law
Field in a solenoid• What is the field outside the coil?
• Is the answer to this subject to anyassumptions and/or approximations in thederivation?
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Example1. If we now apply an a.c. current of 50Hz
and peak amplitude 2.0A, calculate thevoltage at the coil terminals
2. Can you work out the coil inductance?