L P X dL r Biot-Savard Law L P X dL r Biot-Savard Law.

ELEC 3105 BASIC EM AND POWER ENGINEERING

MAGNETIC FIELD OF A LONG STRAIGHT WIRE (FINITE

LENGTH/ INFINITE LENGTH)

AMPERE’S LAW

MAGNETIC FIELD OF A LONG SOLENOID

B FIELD FOR A LONG STRAIGHT WIRE (FINITE LENGTH)

LP

X

dL 𝑑𝜃

𝜃2

𝜃1

��𝐻��

r ��𝐵

��1

��2

��

Biot-Savard Law

��= ∫𝐿 𝑖𝑛𝑒

❑

��𝐵 ∫𝐿𝑖𝑛𝑒

❑ 𝜇𝑜

4𝜋��𝑑𝐿× ��𝑅2


❑

��𝐻 ∫𝐿𝑖𝑛𝑒

❑1

4𝜋��𝑑𝐿× ��𝑅2

LP

X

dL 𝑑𝜃

𝜃2

𝜃1

��𝐻��

r ��𝐵

��1

��2

��

Biot-Savard Law


❑

��𝐵 ∫𝐿𝑖𝑛𝑒

❑ 𝜇𝑜

4𝜋��𝑑𝐿× ��𝑅2

��𝑑𝐿=𝑑𝑧 ��

��𝑑𝐿× ��=Isin (𝜃 )dz ��

��=��𝐼 𝜇𝑜

4 𝜋 ∫𝐿𝑖𝑛𝑒

❑ 𝑠𝑖𝑛 (𝜃 )𝑅2 𝑑𝑧


LP

X

dL 𝑑𝜃

𝜃2

𝜃1

��𝐻��

r ��𝐵

��1

��2

��

��=��𝐼 𝜇𝑜

4 𝜋 ∫𝐿𝑖𝑛𝑒

❑ 𝑠𝑖𝑛 (𝜃 )𝑅2 𝑑𝑧

R=r csc (𝜃 )

𝑧=−𝑟 𝑐𝑜𝑡 (𝜃 )

𝑑𝑧=𝑟 𝑐𝑠𝑐2 (𝜃 )𝑑𝜃

��=��𝐼 𝜇𝑜

4 𝜋 ∫𝜃 1

𝜃 2 𝑠𝑖𝑛 (𝜃 )𝑟 𝑐𝑠𝑐2 (𝜃 )𝑟 2𝑐𝑠𝑐2 (𝜃 )

𝑑𝜃


LP

X

dL 𝑑𝜃

𝜃2

𝜃1

��𝐻��

r ��𝐵

��1

��2

��

R=r csc (𝜃 )

𝑧=−𝑟 𝑐𝑜𝑡 (𝜃 )

𝑑𝑧=𝑟 𝑐𝑠𝑐2 (𝜃 )𝑑𝜃

��=��𝐼 𝜇𝑜

4 𝜋 ∫𝜃 1

𝜃 2 𝑠𝑖𝑛 (𝜃 )𝑟 𝑐𝑠𝑐2 (𝜃 )𝑟 2𝑐𝑠𝑐2 (𝜃 )

𝑑𝜃

��=��𝐼 𝜇𝑜

4𝜋𝑟∫𝜃1

𝜃2

𝑠𝑖𝑛 (𝜃 )𝑑𝜃

��=��𝐼 𝜇𝑜

4𝜋𝑟 [𝑐𝑜𝑠 (𝜃1 )−𝑐𝑜𝑠 (𝜃2 ) ]


LP

X

dL 𝑑𝜃

𝜃2

𝜃1

��𝐻��

r ��𝐵

��1

��2

��

��=��𝐼 𝜇𝑜

4𝜋𝑟 [𝑐𝑜𝑠 (𝜃1 )−𝑐𝑜𝑠 (𝜃2 ) ]

B FIELD FOR A LONG STRAIGHT WIRE (INFINITE LENGTH)

𝜃1→0

𝜃2→𝜋

��=��𝐼 𝜇𝑜

2𝜋𝑟infinite length

AMPERE’S LAW

r

�� 𝐼��=��𝐼 𝜇𝑜

2𝜋𝑟

infinite length

𝑑𝑙

∮ �� ∙𝑑𝑙=𝐼 𝜇𝑜RHR gives ��

AMPERE AND THE LONG STRAIGHT WIRE

This is how it is done

RHR for field direction

��=��𝐼 𝜇𝑜

2𝜋𝑟

𝐵=𝐼 𝜇𝑜

2𝜋𝑟

9

DIFFERENTIAL FORM OF AMPERE’S LAW

Lecture 16 slide 3

Area enclosed A

J

Current density flowing through loop.

enclosedoIdB ∫

10

Ampere’s Law:

enclosedoIdB ∫

Area enclosed A

J


∫ ∫ S

oadJdB

∫ ∫ S

odanJdanB ˆˆ

JBo

∫ ∫ S

adFdF Using Stoke’s theorem

Differential equation for B

Can we solve this equation?

DIFFERENTIAL FORM OF AMPERE’S LAW

MAGNETIC FIELD OF A INFINITELY LONG SOLENOID

Current out of page

Current into page

Infinite coil of wire carrying a current I

Axis of solenoid

P

Evaluate B field here

1

In the vicinity of the point P

P

2 3 4 5

3

12

45Axis of solenoid

resultant

Expect B to lie along axis of the solenoid

Current out of page

0 B

Implies that B field has no radial component. I.e. no component pointing towards or away from the solenoid axis.

B


Current out of page

Current into page

P

Claim: The magnetic field outside of the solenoid is zero.

1 Closed pathaB

bB


Current out of page

Current into page

P

Must have since there is no net enclosed current.1 Closed pathaB

bB

01

∫

dB

Conceivably there might be some non-zero field components outside of the solenoid, and as shown.aB

bB


Current out of page

Current into page

P

We would need to have to give .

1 Closed pathaB

bB

01

∫

dB

Conceivably there might be some non-zero field components outside of the solenoid, and as shown.aB

bB

ba BB


MAGNETIC FIELD OF A LONG SOLENOID

Current out of page

Current into page

P

Now we distort the path by moving one side away from the solenoid.

2Closed path

aB

bB

02

∫

dBStill valid


Current out of page

Current into page

P

To have ,B would have to have the same magnitude no matter how far we moved away from the solenoid. This is possible only if B = 0 outside of the solenoid. 2

Closed pathaB

bB

02

∫

dBStill valid

02

∫

dB

0 ba BB


Current out of page

Current into page

Using the closed path shown we can now obtain an expression for the magnetic field inside the long solenoid.

3 Closed path

0bB

P

L


Current out of page

Current into page

3 Closed path

0bB

P

L

NIBLdB o∫3

enclosedo IdB ∫

Ampere’s Law

L

NIB o

N : number of turns enclosed by length L


Current out of page

Current into page

0bB

P

L

NIB o

N : number of turns enclosed by length L

• B is independent of distance from the axis of the long solenoid as we are inside the solenoid!• B is uniform inside the long solenoid.


21


MAGNETIC VECTOR POTENTIAL

22


To find for general problems, we need more sophisticated techniques than Ampere’s law and a few postulates.

B

7

MagnetostaticsPOSTULATE POSTULATE 11FOR THE MAGNETIC FIELDFOR THE MAGNETIC FIELD

A current element immersed in a magnetic field will experience a force given by:

dBIFd

dI

B

Fd

Units of Newtons {N}

26

MagnetostaticsPOSTULATE POSTULATE 22FOR THE MAGNETIC FIELDFOR THE MAGNETIC FIELD

A current element produces a magnetic field which at a distance Ris given by:

d

R

RIBd o

2

ˆ

4

dI

B

Bd

Units of {T,G,Wb/m2}

3

A m p e r e ’s L a w

T h e s t a r t i n g p o in t i s a m o d i f i c a t i o n o f p o s tu la t e 2 .

W e n e e d a m e a n s o f c o m p u t in g t h e m a g n e t i c f i e ld f o r a k n o w n c u r r e n t d i s t r ib u t i o n .

enclosedo IdB ∫

L in e in te g r a l a r o u n d c lo s e d p a th

C u r r e n t e n c lo s e d b y p a th

∫ S

adJI

Suppose we can find a function such that: AB

zyxA ,,

zyxA ,, Is called the Magnetic Vector Potential.

It is a function of the coordinates.It has direction.

It is possible to show that for any zyxA ,,

: 0 A

0 B


24

Recall: JBo

Differential form of Ampere’s Law

Ampere’s Law:enclosedoIdB ∫

Area enclosed A

J


∫ ∫ S

oadJdB

∫ ∫ S

odanJdanB ˆˆ

JBo

∫ ∫ S

adFdF

Using Stoke’s theorem

Differential equation for B

Can we solve this equation?And: AB

Then:

JAo


Add in mathematical manipulations

25

But for any vector function :

JAo

A

AAA

2

And if 0 A

AA

2

use here


26

JAo

2

zozJA 2

GIVES

In component form:

yoyJA 2

xoxJA 2

We have managed to generate independent equations governing the x, y and z components of

zyxA ,,


27

zyxA ,,

Is called the Magnetic Vector Potential.It is a function of the coordinates.It has direction.Is not unique.Three scalar equations involving derivatives


zozJA 2

yoyJA 2

xoxJA 2

Natural progression of the student’s perspective of the magnetic vector potential

Today Next week Before exam After exam

I’m doing the deferred exam

28

Similar to Poisson’s equation:

zozJA 2

yoyJA 2

xoxJA 2

zyxAAA ,,

3 4

P o i s s o n ’s / L a p l a c e ’s E q u a t i o n

I n m a n y r e g i o n s o f s p a c e = 0 , n o n e t c h a r g e d e n s i t y . I n t h i s c a s e :

o

V

2

L a p l a c e ’ s E q u a t i o nL a p l a c e ’ s E q u a t i o n02 V

P o i s s o n ’ s E q u a t i o nP o i s s o n ’ s E q u a t i o n

∫∫∫

volo zzyyxx

dzdydxzyxzyxV

2

21

2

21

2

21

222222111

,,

4

1,,

Solution of this form in (x, y, z)

Integration over volume containing charge .

Same form of equation, same form of solutionFigure next page

FINDING THE MAGNETIC VECTOR POTENTIAL

Need to obtain all three components

General solution for scalar potential

29

RECAL FOR POTENTIAL V

3 4

P o i s s o n ’s / L a p l a c e ’s E q u a t i o n

I n m a n y r e g i o n s o f s p a c e = 0 , n o n e t c h a r g e d e n s i t y . I n t h i s c a s e :

o

V

2

L a p l a c e ’ s E q u a t i o nL a p l a c e ’ s E q u a t i o n02 V

P o i s s o n ’ s E q u a t i o nP o i s s o n ’ s E q u a t i o n

∫∫∫

volo zzyyxx

dzdydxzyxzyxV

2

21

2

21

2

21

222222111

,,

4

1,,

x y

z

222

,, zyx

),,(111zyxV

2r

1r

Charge distribution

∫∫∫

volo rr

dzdydxrrV

21

22221 4

1

Could write solution in short hand form as:

21rr

30zyx AAA ,,

In the short hand vector notation

FINDING THE MAGNETIC VECTOR POTENTIAL

∫∫∫

vol

xox

zzyyxx

dzdydxzyxJzyxA

2

21

2

21

2

21

222222111

,,

4,,

∫∫∫

vol

yoy

zzyyxx

dzdydxzyxJzyxA

2

21

2

21

2

21

222222

111

,,

4,,

∫∫∫

vol

zoz

zzyyxx

dzdydxzyxJzyxA

2

21

2

21

2

21

222222111

,,

4,,

∫∫∫

vol

o

rr

dzdydxrJrA

21

22221 4

31

EXAMPLE B USING AFind and for a long wire aligned along the z-axis.

A

B

y

z

I

wire

111

,, zyxP

r

Evaluate hereA

By geometry:

0yx

JJ

0yx

AASimplifies solving for A

∫∫∫

vol

xox

zzyyxx

dzdydxzyxJzyxA

2

21

2

21

2

21

222222111

,,

4,,

∫∫∫

vol

yoy

zzyyxx

dzdydxzyxJzyxA

2

21

2

21

2

21

222222

111

,,

4,,

∫∫∫

vol

zoz

zzyyxx

dzdydxzyxJzyxA

2

21

2

21

2

21

222222111

,,

4,,

32

y

z

I

wire

111

,, zyxP

r

Becomes

A

Where

Current in the wire

∫∫∫

vol

zoz

zzyyxx

dzdydxzyxJzyxA

2

21

2

21

2

21

222222111

,,

4,,

EXAMPLE B USING A

∫

wire

oz

zzyx

dzIzyxA

2

21

2

1

2

1

2111 4

,,

IdydxJ z ∫ sectioncross

wire22

33

y

z

I

wire

111

,, zyxP

r

Az should have the same value for any z1.(Once again by a symmetry argument)

A

Chose to evaluate for z1 = 0 for simplicity.

Wire extends for - to +

EXAMPLE B USING A

∫

wire

oz

zzyx

dzIzyxA

2

21

2

1

2

1

2111 4

,,

∫

wire

oz

zyx

dzIzyxA

2

2

2

1

2

1

2111 4

,,

2

1

2

1111 ln2

,, yxI

zyxA oz

34

y

z

I

wire

111

,, zyxP

rA

Since is the distance from the point P to the wire

EXAMPLE B USING A

2

1

2

1111 ln2

,, yxI

zyxA oz

rIzyxA o

z ln2

,, 111

2

1

2

1 yxr

35

y

z

I

wire

111

,, zyxP

rA

Now for B

We have

and

AB

Then we can evaluate the curl of A to get B.Note that A has only a z component.

EXAMPLE B USING A

rIzyxA o

z ln2

,, 111

22 r

IyB o

x

22 r

IxB o

y

0z

B

38

y

z

I

wire

111

,, zyxP

rA

Now for B

We have

and

AB

Then we can evaluate the curl of A to get B.Note that A has only a z component.

EXAMPLE B USING A

rIzyxA o

z ln2

,, 111

22 r

IyB o

x

22 r

IxB o

y

0z

B

39

y

z

I

wire

111

,, zyxP

rB

22 r

IyB o

x

22 r

IxB o

y

Gives

22

yxBBB

r

IB o

2

As would be obtained from Ampere’s Law

circles wire

EXAMPLE B USING A

THE FORCE AND TORQUE ON A DIPOLE IS TAKEN UP AT THE START OF THE NEXT LECTURE


FORCE AND TORQUE ON MAGNETIC DIPOLE

Magnetic dipole = product of current in

loop with surface area of loop

FORCE ON A MAGNETIC DIPOLE B

Consider a circular ring of current I placed at the end of

a solenoid as shown in the figure. The current in the

solenoid produces a magnetic field in which the current loop is placed into.

By postulates 1 and 2 of magnetic fields, the current ring will be subjected to a

magnetic force.out of page into page

I

z


I

z

outF

outF

downF

downF

Circular ring

Cancel in pairs around the ringout

F

outF

downF

downF

Will add in same direction on ring giving a net force.


I

z

outF

outF

downF

downF

Circular ring

downF

downF

Will add in same direction on ring giving a net force.

Using postulate 1: dBIFd

r2

B

zB

rB

downF

We need for find Br

Gives:

rdown rIBF 2

FORCE ON A MAGNETIC DIPOLE

z

z

Gaussian cylinder

We will relate Br to z

Bz

Total magnetic flux through Gaussian cylindrical surface must be zero. As many magnetic field lines that enter the surface, leave the surface. No magnetic charges or monopoles.0 B

11

Another important property of B

0 B

Recall everywhere

No net magnetic flux through any closed surface.

Closed surface S

∫ S

adB 0

∫ vol

voldvB 0

Using divergence theorem

0 3-D view


z

z

Flux through side:

3-D view

siderzBr 2

Flux through top:

topz

zzBr 2

Flux through bottom:

bottomz

zBr 2

0bottomtopside


z

z

3-D view

02 22 zBrzzBrzBrzzr

0bottomtopside

z

zBzzBrB zz

r

2

z

BrB z

r

2

We can now use this in our force on current ring expression


I

z

outF

outF

downF

downF

Circular ring

rdownrIBF 2

r2

B

zB

rB

downF

We have found Br

z

BrB z

r

2

z

BrrIF z

down

2

2


I

z

outF

outF

downF

downF

Circular ring

r2

B

zB

rB

downF

z

BrrIF z

down

2

2

z

BIrF z

down

2

z

BIAF z

down

z

BmF z

down

z

BmF z

z

Force pulls dipole into region of stronger magnetic field


3-D view

z

In general

xxBmF

yyBmF

zzBmF

BmF

TORQUE ON A MAGNETIC DIPOLE

We will consider a dipole in a uniform magnetic field. We can use any shape we want for the dipole. Here we will select a square loop of wire.

I out of page

I into page

m

B

a

a

I

Side view

Topview

Wire loop

a

a

I

Topview

Wire loop

a

2

a


TORQUE ON A MAGNETIC DIPOLEm

B

a

a

I

Side view

Topview

Wire loop

F

F

Torque attempts to align dipole

moment with .m

B

Pivot point

Pivot line


m

B

Side view

F

F

Torque attempts to align dipole

moment with .m

B

Pivot point

sin2

2a

F

Fr

2

a

Total torque

F => Magnetic force on wire of length a


m

B

Side view

F

F

Pivot point

sin2

2a

F

2

a

F => Magnetic force on wire of length a

IBaF Through postulate 1 for magnetic fields

sin2IBaThen


TORQUE ON A MAGNETIC DIPOLEm

B

Side view

F

F

Pivot point

2

a

sin2IBa

a

a

I

Wire loopIam 2

sinBm

Bm


L P X dL r Biot-Savard Law L P X dL r Biot-Savard Law.

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Transcript of L P X dL r Biot-Savard Law L P X dL r Biot-Savard Law.