biot savart laws

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Fields, Materials & Devices

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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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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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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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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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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:


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dH

∫= H =

i sin θ dl

4π r 2∫


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Biot-Savart Law


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H =

∫4π r 3

i dl x r


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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…


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dH

∫= H =

i sin θ dl

4π r 2

∫


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• The application

requires us toevaluate the integralfor a point relative tothe wire, effectivelysumming over all

current elementsalong an infinite wire.


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Current, i

B , H

+

B , H

i


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• The sum is over

elements of thistype, so that


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+

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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• To evaluate the

integral, we needeliminate theinterdependence ofterms such as rand α.

• We can express l and r in terms of α.


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+

i

dl l

α

θ

r a


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1. a tan α = l

2. r cos θ = a

• Differentiating 1

gives – dl = a sec2α d α


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+

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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1. a tan α = l

2. r cos θ = a

• Substituting 2

gives


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+

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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can be integratedreadily, yielding:


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+

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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The infinite, straight, uniform wire


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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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The field for a finite wire…• What happens at the ends of a wire?

– Hint: consider a semi-infinite wire.


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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…


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H = i 2π a

∮H.dl


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Ampere’s Law• Since H and dl are always parallel, the

scalar product is simply the product of themagnitudes:


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∮H.dl = i dl

2π a ∫0

2π a

= i

B , H

i a


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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”.


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∮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


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Electric dipole

Flux continuity


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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.


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∮B.ds = 0s


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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.


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F ar a d a y


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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.


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F ar a d a y

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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


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V

V = d ψ dt

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Question

• What wouldhappen if we

added a secondturn to the coil?


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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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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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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.


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Iron core, N turns.

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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.


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i

r

B

dl

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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.


i

r

B

dl

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Field in a solenoid• We now apply

Ampere’s law:

• Since H and dl are

parallel, the scalar

product H.dl is justHdl.


i

r

B

dl

∮H.dl = i

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Field in a solenoid• The path length is

2π r .• Since the path

threads N loops

carrying current i , theright hand side is Ni :


i

r

B

dl

∮dl = H2 π r = Ni

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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 .


i

r

B

dl

∮dl = H2 π r = Ni

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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?

biot savart laws

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