Lesson 2 Sources of Magnetic Fields

7/27/2019 Lesson 2 Sources of Magnetic Fields

1/44

SOURCES OF

MAGNETIC FIELD


2/44

28.1

Magnetic Field of a Moving Charge

P

Consider

a

charge

q

moving

with

a

constant

velocity

v

.

The

location

of

the

charge

at

any

given

instant

is

called

the

source

point

.

The

point

P

where

the

magnetic

field

due

to

the

charge

is

being

evaluated

is

called

the

field

point

.

r

r

is

the

displacement

vector

from

source

point

to

field

point

.

is

the

angle

between

r

and

v

.

x

y

q

v

+


3/44

28.1


The

direction

of

the

magnetic

field

at

point

P

is

perpendicular

to

both

r

and

v

and

can

be

determined

using

the

right

hand

rule

.

P

r

B

x

y

q

v

+

0,// FthenBvIf


4/44

28.1


qB

2

1

rB

vB

sinB0,// FthenBvIf

Experimental

results

about

the

magnitude

of

B

.


5/44

The magnetic field

B

due to a moving charge

q

is

2

sin

4 r

vqB o

w

here:

*104 .7

AmT

o

ATmo 7

1014

28.1


0,// FthenBvIf

*permeability of free space is themeasure of the amount of resistanceencountered when forming a

magnetic field in a vacuum.


6/44

The magnetic field

B

due to a moving charge

q

is

2

sin

4 r

vqB o

28.1


2

4 r

rvqB o

w

here:

r

r

r

Vector Magnetic Field

0,// FthenBvIf


7/44

0,// FthenBvIf

28.1 Magnetic Field of a Moving Charge

NOTES:

B = 0 at all points alongthe line through the chargeparallel to the direction ofvelocity. sin = 0

at any distance r from

the charge, B is maximumat all points lying in theplane perpendicular to thedirection of velocity.sin = 1

if the charge is negative,the direction of B isopposite that for a positivecharge.


8/44

2

r

qqkFE

2

2199

)10.0(

)106.1()109( 2

2

m

CF

C

NmE

upwardNFE ,1030.226

Example 1. Two protons move parallel to the x-axis in oppositedirections at the same speed of1.00 x 105m/s. At the instantshown in the figure, where r = 10 cm, find the electric and

magnetic forces (magnitude and direction) on the upper proton.


9/44

2

sin

4 r

qvB o

2

5197

)10.0()/101)(106.1()101(

msmCB

ATm

TB 19106.1

qvBFM

)106.1)(/101)(106.1( 19519 TsmCFM

upwardNFM ,1056.233

Example 1. Two protons move parallel to the x-axis in oppositedirections at the same speed of1.00 x 105m/s. At the instantshown in the figure, where r = 10 cm, find the electric and

magnetic forces (magnitude and direction) on the upper proton.


10/44

Consider a segment dlof a current carrying conductor.

2

sin

4 r

vdQdB o

2

sin

4 r

dt

dldQ

dB o

2

sin

4 r

dldt

dQ

dB o

28.2 Magnetic Field of a Current Element


11/44

28.2 Magnetic Field of a Current Element

The magnetic field due to a current element is

2

sin

4 r

dlIdB o

24 r

rldIBd

o


12/44

Law ofBiot-Savart

The magnetic field ofa current-carrying conductor is

2sin

4 r

dlIB o

24 r

rldI

Bo


13/44

Example 2. Acopper wire carries a steady current of125A. Findthe magnetic field caused by a 1.00-cm segment of this wire at apoint 1.2 m from it ifthe point is (a) point P1straight to the side ofthe segment, and (b) point P

2on a line at 30oto the segment

2

sin

4)(

r

dlIBa o

TB 8107.8

2

7

)2.1(90sin)01.0)(125()101(

mmAB

ATm

TB 8103.4

2

7

)2.1(

30sin)01.0)(125(

)101()( m

mA

Bb ATm


14/44

Consider a conductor of length 2a carrying a current I.

Find the magnetic field Bat point Pat a distancexfrom theconductor on its perpendicular bisector.

22 yxr

22)sin(

yx

x

dydlLet

sin4

2r

dlIB o

28.3 Magnetic Field of a Straight Current-Carrying Conductor


15/44

sin4

2r

dlIB o

a

ao

yx

x

yx

dyIB 22224

a

a

o

yx

xdyIB

23

224


22

2

4 axx

aIB o


16/44

If the length 2a>>x, theconductor can be considered tobe infinitely long.

Ifa>>x, xa

aI

Bo 2

4


22

24 axx

aIB o

rIB o

2

x

IB

o

2


17/44

Example 3. The figure shows an end-on view oftwo longparallel straight wires perpendicular to the xy-plane, eachcarrying a current of10 A but in opposite directions. Findthe magnitude and direction ofthe magnetic field at pointsP1, P2, and P3.

x

y

0

10cm

20cm

-10cm

-30cm

I1

=1

0A

P1 P2 P3I2

=1

0A

B2

B1

r

IB

o

2For P1:

mAxB

ATm

20.010)102( 71

jTxB 101 51

m

AxB

ATm

40.0

10)102(

7

2

jTxB 105 62

jTxjTxBnet 10510165

jTxBnet 1056


18/44


r

IB

o

2For P2:

mAxB

ATm

10.010)102( 71

jTxB 102 51

m

AxB

ATm

10.0

10)102(

7

2

jTxB 102 52

jTxjTxBnet 10210255

jTxBnet 1045

x

y

0

10cm

20cm

-10cm

-30cm

I1

=1

0A

P1 P2 P3I2

=1

0A

B1

B2


19/44


r

IB

o

2For P3:

mAxB

ATm

30.010)102( 71

jTxB 1067.6 61

m

AxB

ATm

10.0

10)102(

7

2

jTxB 102 52

jTxjTxBnet 1021067.656

jTxBnet 1033.15

x

y

0

10cm

20cm

-10cm

-30cm

I1

=1

0A

P1 P2 P3I2

=1

0A

B2

B1


20/44

Consider two long and straight parallel current-carryingwires. Determine the magnitude and direction ofthe magnetic

force they exert on each other.

28.4 Magnetic Force Between Parallel Conductors


21/44

I1 I2

B1

B2

F12

F21


Force per unit length exerted

by wire 1 on wire 2 (F12)

r

1212 LBIF

r

ILIF o

2

1212

r

IIx

L

FATm 21712 )102(

Force per unit length exertedby wire 2 on wire 1 (F

21

)

2121 LBIF

r

ILIF o

2

2121

r

IIx

L

FATm 21721 )102(


22/44

I1 I2

B1B2

F12F

21


If the currents in the conductors are in opposite directions:

rx

Two parallel wires with currents in the same direction attracteach other.

Two parallel wires with currents in opposite directions repel

each other.


23/44

Example 4. Find the magnitude and direction of the Fnetper unitlength on wire 3 due to wires 1 and 2.

x

y

8 cm

I3 = 20 A

6 cm

I2 = 5 AI1 = 10 A

F13

F23

Fnet

mN

ATm

m

AA

L

F

4

713

104

10.0

)20)(10()102(

mN

ATm

m

AA

L

F

4

723

1033.3

06.0

)20)(5()102(


24/44

x

y

8 cm

I3 = 20 A

6 cm

I2 = 5 AI1 = 10 A

F13

F23

Fnet

mN

mNx

L

F

4

4

102.3

0)8.0)(104(

mN

mN

mNy

L

F

5

44

103.9

1033.3)6.0)(104(

mN

mN

mNnet

L

F

4

2524

1033.3

)103.9()102.3(

xaboveo

2.16

00032.0

000093.0tan

1

Example 4. Find the magnitude and direction of the Fnetper unitlength on wire 3 due to wires 1 and 2.


25/44

28.5 Magnetic Field of a Circular Current Loop

Consider a circular current loop ofradius a. Use the LawofBiot and Savart to find the magnetic field at point Pon the

axis ofthe loop.

24 r

dlIdB o

2222

2

)(4

cos4

ax

a

ax

dlI

r

dlI

dB

o

o

x

2222

2

)(4

sin4

ax

x

ax

dlI

r

dlIdB

o

oy


26/44

MAGNETIC FIELD OF A CIRCULAR CURRENT LOOP

0yB

a

ox dl

ax

IaB

2

022

2

3

)(4

xBB

)2()(4 2

322

aax

IaB ox

23

)(2 22

2

ax

INaB o

Where N is the number ofturns


27/44

MAGNETIC FIELD AT THE CENTER OF A CIRCULARCURRENT LOOP

23

)(222

2

axIaB o

Ifx=0, or at the center ofthe loop

a

IB

o

2


28/44

Example 5.

(a)


29/44

Example 5.

(b)


30/44

MAGNETIC FIELD OF OTHER CONDUCTORS


31/44

2

sin

4 r

dlIdB o

(a)

27 )05.0(90sin)0011.0)(10(101

mmAxdB

ATm

TxBd7

1040.4


32/44

(b)

27 )149.0(7.19sin)0011.0)(10(101

mmAxdB

ATm

TxBd 81067.1

mmmr 149.014.005.0 22 o

7.19tan1451

(c) 00 becausedB


33/44

For the 12-A wire:

Txm

mAxBd cm

cm

ATm 8

2

85.2

7 1079.8)08.0(

)0015.0)(12(101

For the 24-A wire:

Txm

mAxBd cm

cm

ATm 7

2

85.2

7 1076.1)08.0(

)0015.0)(24(101

x

TxBd net81079.8

x


34/44

eastTxm

Ax

r

IB

ATmo ,1091.2

50.5

800)102(

2

57


35/44

I1=25A

B1

B2

(a)

40cm -x

I2=75A

x

21 BB

2

2

1

1

22 r

I

r

I oo

xcm

A

x

A

40

7525

3

40

x

xcm

cmx 10

Bnet = 0 at a point 10cm from I1and 30 cm from I2.


36/44

I1=25A

B1

B2

(b)

40cm

I2=75A

x

21 BB

2

2

1

1

22 r

I

r

I oo

xcm

A

x

A

40

7525

3

40

x

xcm

cmx 20

Bnet = 0 at a point 20cm from I1and 60 cm from I2.

x


37/44

Txm

Ax

r

IB

ATmo 57 1011.1

045.0

50.2)102(

2

I

-

v

)1011.1)(/106)(106.1( 5419 TxsmxCxqvBF

NxF19

1007.1

same direction as the current


38/44

21 BB

2

2

1

1

22 r

I

r

I oo

2

1

1

2 rr

I

I

(a)

Amm

AI 2)50.0(

5.1

62


39/44

upwardTx

m

Ax

r

IB

ATmo

Q

67

1

11 104.2

5.0

6)102(

2

(b)

downwardTxm

Ax

r

IB

ATmo

Q

67

2

22 1067.2

5.1

2)102(

2

upwardTxBnetQ6

1013.2


40/44

(c)

TxTxTxBmm

mm

x

6

16.07

18.06 103.1)105()102(

B2 B1

Txm

AxB

ATm

S

67

1 1026.0

6)102(

Txm

AxB

ATm

S

77

2 1058.0

2)102(

TxTxTxBmm

mm

y

6

18.07

16.06 106.1)105()102(

TxBBB yxnetS622

1006.2


41/44

FBD:

x

mg

F

6o

T

F

6o

mg

T

I Ir

mxmr 31036.8)6sin04.0(2


42/44

FBD:

x

mg

F

6o

T

F

6o

mg

T

I I

FT 6sin mgT 6cos

mgIl

mgIlB

mgF r

Io

26tan

l

mgrI

o

2

)6(tan

AI 2.23


43/44

Due to segment a:

a

IB oa

4

Due to segment b:

b

IB ob

4

x

b

I

a

IB oonet

44

ba

IB onet

11

4


44/44

Due to vertical segment: 00sin4 2

1 rdlI

B o

Due to horizontal segment: 0180sin4 2

2 rdlI

B o

RIB o

83

Due to quarter-circle:

R

IB onet

8

Documents

Lesson 2 Sources of Magnetic Fields