Biot-Savard Law Applications of Biot- Savard Law 1.

1

ELEC 3105 Basic EM and Power Engineering

Biot-Savard LawApplications of Biot-Savard Law

2

Biot-Savard Law

Usually we deal with current localized to wires rather than spread over space.

I

I

d

1r

2r

P

Origin

When the current is contained by thin wires, “we can use the magnetic vector potential to derive an integral from which the magnetic field can be found directly

from the wire location and currents”.

3

Biot-Savard Law KISS Rule

I

I

d

1r

2r

P

Origin

We will find the magnetic field at some point P located at vector from the origin due to a short segment of wire located at from the origin, carrying

current I, where the current is in the direction of the vector , as shown below.

B

1111

,, zyxr

d 2222

,, zyxr

d

small segment of wire at ),,(

222zyx

find hereB

4

Biot-Savard Law

Consider a small segment of wire of overall length

I

d

P

21r

This expression isThe Biot-Savard Law

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}

d

Bd

Same result as postulate 2 for the magnetic field

2

21

211

ˆ

4 r

rdIrBd o

5

Biot-Savard Law

The magnetic field produced by all the short current elements along the line is the Biot-Savard law.


I

d

P

R


d

Bd

Line

o

Line R

RdIrBdrB 21

ˆ

4

An integral is another way of saying “principle of superposition” applies to magnetic fields.

6

Biot-Savard Law & Current Density J (A/m2)

I

I

d

1r

2r

P

Origin

find hereB

1rBd

dIdxdydzJ The current in the differential volume element dxdydz is:

volume

o

volume R

RJdvolrBdrB 21

ˆ

4

𝑅

7

Biot-Savard Law & Surface Current Density K (A/m)

I

I

d

1r

2r

P

Origin

find hereB

1rBd

dIKdA The current in the differential surface element dA is:

surface

o

surface R

RKdArBdrB 21

ˆ

4

𝐾

𝑅dA

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ELEC 3105 Basic EM and Power EngineeringSTART Applications of Biot-Savard Law

BIOT from

RAMA

9

Magnetic field produced by a circular current ring

y

10


We will first consider the magnetic field present at the

center of a current carrying ring.

I

x

y

a

We can use the Biot-Savard Law

11


All contributions from each line segment point in the same direction: in the direction of the axis of the ring.

I

x

y

ad

add

Top viewSide view

I out of page

I into page

a aBdBd

Bd

12


I

x

y

ad

add Top viewSide view

a aStarting from the Biot-Savard Law for a single segment.

We get: ar 21

ˆadd rr ˆ

21

r

12r

2

21

211

ˆ

4 r

rdIrBd o

24 a

adIdB o

13


I

x

y

ad

add Top view

a a

ar 21

ˆadd rr ˆ

21

r

12r

Direction:

zr ˆˆˆ

Side view

As shown by arrows in side view

24 a

adIdB o

2

21

211

ˆ

4 r

rdIrBd o

14


I

x

y

ad

add Top view

Summation over all segments around the ringd

Magnetic field at the center of the circular current ring

2

0 4 a

dIdBB o

a

IB o

2

za

IB o ˆ

2

15


We will now consider the magnetic field present on

the central axis of a current carrying ring.

Biot-Savard Law

The Biot-Savard law applied to the small segment gives an element of magnetif field at the point P.


I

d

P

21r


26

MagnetostaticsPOSTULATE POSTULATE 22 FOR THE MAGNETIC FIELDFOR THE MAGNETIC FIELD

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

d

R

RIBd o

2

ˆ

4

dI

B

Bd

Units of {T,G,Wb/m2}

d

Bd

Same result as postulate 2 for the magnetic field

Bd

2

21

211

ˆ

4 r

rdIrBd o

We can use the Biot-Savard Law

I

x

y

a

3-D viewz

z

16


d add

I out of page

I into page

a a

Bd

Bd

I

x

y

a

3-D viewz

z

All contributions from each line segment point in different directions: We must consider components of .

Bd

Bd

21r

17


I out of page

I into page

a a

Bd

Bd

z21

r Bd Bd

zdB

zdB

hdB

hdB

hdB

hdB

Horizontal components will cancel in pairsaround the segments of the current ring

zdB

zdB

Z components will add around the segments of the current ring

18


hdB

hdB

Horizontal components will cancel in pairsaround the segments of the current ring

zdB

zdB

Z components will add around the segments of the current ring

19


I out of page

I into page

a a

Bd

Bd

z21

r Bd z

dB

cosdBdBz

22

2121zarr

21

cosr

a

3-D view

Can use Biot-Savard Law

We need only concern ourselves with the magnitude of since we know the final direction of is along the z direction.

Bd

B

2

21

211

ˆ

4 r

rdIrBd o

20


I out of page

I into page

a a

Bd

Bd

z21

r Bd z

dB

cosdBdBz

22

2121zarr

21

cosr

a

3-D view

Expression for magnetic field produced by one current segment of the ring.

2

214 r

dIdB o

2

214 r

adIdB o

cos4 2

21r

adIdB o

z

321

2

4 r

daIdB o

z

21


d add

I out of page

I into page

a a

Bd

Bd

Ix

y

a

3-D viewz

z21

r

Summation around ring:

Magnetic field on the central axis of the circular current ring

2

03

21

2

4 r

daIdBB o

zz

2

03

22

2

4d

za

aIB o

z

322

2

2 za

aIB o

z

22

Magnetic field produced by a circular current ringWe only considered the central axis field

The full field is obtained when we will discuss the magnetic dipole

magnetotherapy

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Magnetic field of a long FINITE solenoid

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Magnetic field of a FINITE solenoid

Current out of page

Current into page

finite coil of wire carrying a current I

Axis of solenoid

P

Evaluate B field here

a

Radius of solenoid is a. Cross-section cut through solenoid axis

L

25


1 2 3 4 5

Axis of solenoid

Current out of page

zdB

dSegment of the solenoid coil

r

d

sin

rdd

arc length rd

a r

asin

26


L

The solenoid consists on N turns of wire ”rings of current” each carrying a current I.

The number of turns per unit length of solenoid can be expressed as:

L

N

27


LThe current in a segment (dI) can be expressed as the current in one ring (I) multiplied by the number of turns per unit length (N/L) and multiplied by the segment length .

dL

NIdI

d

d

dL

NIrdI

sin

sin

rdd

28

Return to this slide for a moment

d add

I

x

y

a

3-D viewz

Magnetic field on the central axis of the circular current ring

Each ring of the solenoid contributes a magnetic field. Can use this expression for our finite solenoid. Sum over the rings.

322

2

2 za

aIB o

z

29


1 2 3 4 5

Axis of solenoid

Current out of page

zdB

dSegment of the solenoid coil

r

d

sin

rdd

arc length rd

a r

asin 3

2

2 r

adIdB o

z

30


1 2 3 4 5

zdB

d

r

d

a

dL

NIrdI

sin

r

asin

sub in

3

2

2 r

adIdB o

z

d

L

NI

r

ad

La

NIr

r

adB oo

z 22

2

3

2

d

L

NIdB o

z sin2

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L

d

12

We can now sum (integrate) the expression for over the angular extent of the coil. I.e. sum over all the rings of the finite length solenoid.

zdB

z

2

1

sin2

d

L

NIB o

z

21 coscos2

L

NIB o

z

zL

NIB o ˆcoscos

2 21

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Magnetic field of a INFINITE solenoid

L

d

01180

2

z

INFINITE SOLENOID RESULT

infinite solenoid result

180cos0cos2

L

NIB o

z

zL

NIB o ˆL

NIB oz

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Magnetic field at one end of a FINITE solenoid

L

d

901180

2

z

Magnetic field is ½ that of center

180cos90cos2

L

NIB o

z

zL

NIB o ˆ

2

L

NIB oz 2

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Magnetic flux

Φ𝑚= ∬𝑆𝑢𝑟𝑓𝑎𝑐𝑒

❑𝐵 ∙𝑑𝐴

𝐵Goes by the name of magnetic flux density vector

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Magnetic flux: CLOSED SURFACE

Φ𝑚= ∯𝑆𝑢𝑟𝑓𝑎𝑐𝑒

❑𝐵 ∙𝑑𝐴=0

Magnetic field lines loop back on themselvesNo magnetic sourceNo magnetic sink

No magnetic monopoleNo magnetic charge

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Divergence of a Magnetic Field

Φ𝑚= ∯𝑆𝑢𝑟𝑓𝑎𝑐𝑒

❑𝐵 ∙𝑑𝐴=0

For a closed surface

No magnetic monopole

No magnetic charge∯

𝑆𝑢𝑟𝑓𝑎𝑐𝑒

❑𝐵 ∙𝑑𝐴= ∰

𝑉 𝑜𝑙𝑢𝑚𝑒

❑𝛻 ∙𝐵𝑑𝑣𝑜𝑙

Divergence theorem

𝛻 ∙𝐵=0𝛻 ∙𝐷=𝜌 𝑣

Electric charge exist𝛻 ∙𝐻=0

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Magnetic flux


❑𝐵 ∙𝑑𝐴

We have the expression for the magnetic field AND the limits for the integral Both specified in cylindrical coordinates.

Given that the radial magnetic field is , calculate the magnetic flux through the surface defined by and .

Solution

Remainder of solution presented in class

39

Magnetic flux


❑𝐵 ∙𝑑𝐴

We have the expression for the magnetic field AND the limits for the integral Both specified in cylindrical coordinates.

In cylindrical coordinates , calculate the magnetic flux through the surface defined by and .

Solution

Remainder of solution presented in class

Biot-Savard Law Applications of Biot- Savard Law 1.

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