IV. ЕЛЕКТРОМАГНЕТИЗАМ

Analogijata pome|u elektri~no i magnetno kolo :

1. Na torus od `elezo so sreden obem 140srl cm= i popre~en presek 212S cm= , namotani se 500N = navivki bakarna `ica niz koja te~e struja 2I A= .

Ovaa struja vo torusot sozdava magneten fluks 1,2mWbφ = . Kolkav e relativ-niot magneten permeabilitet na `elezoto.

3

4

4

4

70

1

1,2 10 1, 4500 2 12 10

14 10

14 104 10

1114,6

m

m

r

r

U N IlRS

N I Sl

lN I S

Hm

φ

µ

φ µ

φµ

µ

µµ

µ πµ

−

−

−

−

−

⋅= =

⋅

⋅= ⋅ ⋅

⋅ ⋅ ⋅= =

⋅ ⋅ ⋅ ⋅ ⋅ = ⋅

⋅= =

⋅=

EIR

=

Omov zakon

mm

m

UR

φ =

Hopkinsov zakon

[ ]

[ ]

[ ]

1m

m

mm

m

m

U I N

l ARs VS

U WbR

B TS

µ

φ

φ

= ⋅ →

= ⋅ →

= →

= →

magnetomotorna sila ampernavivki

magnetomotoren otpor

magneten fluks

magnetna indukcija

2. Jadro od dinamolim so sredna dol`ina 84l cm= , so presek 210S cm=

treba magnetno da se pobudi taka da magnetniot fluks bide 31,6 10 Wb−Φ = ⋅ . Krivata na magnetizirawe e dadena na slikata.

• Kolkava treba da bide strujata vo namotkata okolu jadroto, ako brojot na navivkite e 100N = .

• Kolkava e magnetnata otpornost na jadroto Rm . • Kolku treba da bide elektri~nata otpornost na namotkata ako taa e

priklu~ena na napon od 220V .

( ) ( )

5

100 29, 40,0016

18.4 10

220 7, 4829, 4

Кирхофов закон за маг. кола

mm

m m

m

k k kC C

U N I N IRR R

A ARWb Vs

UR RI

H l K I II

⋅ ⋅ ⋅Φ = = ⇒ = =

Φ

= ⋅

= = = Ω

± ⋅ = ± ⋅ ⋅ − ∑ ∑

b)

v)

2

84100.0016100

l cmS cm

WbN

=

=Φ =

=

4

0.0016) 1,610 10

a B TS −

Φ= = =

⋅Од крива на магнетиз. за 1,6 3500 AB T H

m= ⇒ =

3500 0,84 29, 4100

H lNI H l I AN

⋅ ⋅= ⋅ ⇒ = = =

3. Na jadro napraveno od dva feromagnetni materijali se nao|aat dve na-motki so 1 250N = namotki i 2 500N = namotki. Niz namotkite te~at strui

1 20, 4 , 0,7I A I A= = . Relativnite magnetni permeabilnosti se 1 2

950, 600r rµ µ= = .

Dimenziite na magnetnite kola se dadeni na slikata vo mm. Da se presmeta mag-netniot fluks niz magnetnoto kolo.

1

2

35

7 4

35

7 4

35

0 7 4

35

0 7 4

1 1 2 2

76.5 10 2,137 10950 4 10 3 10

71.5 10 4,74 10600 4 10 2 10

1 10 39,5 104 10 2 10

1.10 26,53 104 10 3 10

100 35073 10

m

m

m

m

m

ARV sAR

V sAR

V sAR

V sN I N I

R

π

π

π

π

−

− −

−

− −

−

− −

−

− −

⋅= = ⋅

⋅⋅ ⋅ ⋅ ⋅ ⋅⋅

= = ⋅⋅⋅ ⋅ ⋅ ⋅ ⋅

⋅′ = = ⋅⋅⋅ ⋅ ⋅ ⋅

′′ = = ⋅⋅⋅ ⋅ ⋅ ⋅

⋅ − ⋅ −Φ = =

⋅∑5

5 3, 415 10 Wb−= − ⋅

0 Кирхофов закон за магнетни колаmN I R II⋅ ⋅ − ⋅ Φ = −∑ ∑

02 4 2

22 4 2

1

1

10 20 200 2 10

15 20 300 3 10

m

r

lRS

S mm mS mm m

µµ µ µ

−

−

= ⋅

= ⋅

= ⋅ = = ⋅

= ⋅ = = ⋅

( ) ( )

( ) ( )

1

1

2

2

39 7.5 50 12.5 7.576.534 5 50 12.5 5

71.5

ll mmll mm

= − + − +

=

= − + − +

=

4. Vo magnetno jadro dadeno na slikata potrebno e, pri struja 2I A= , vo namotkite da se ostvari magneten fluks 410 Wbφ −= . Krivata na magnetizirawe e dadena na sl. Da se odredi vkupniot broj na navivki vo jadroto. Dimenziite na magnetnoto kolo se vo mm.

5. Za magnetnoto kolo na slikata da se odredi magnetomotornata sila koja vo srednata granka sozdava magnetna indukcija 3 0,8B T= . Krivata na magnet-izirawe e dadena na slikata. Dimenziite se vo mm.

4

4

2

1 2

1 2

10 0, 254 10

20 20 40, 25 900 /

24

108

sr sr

B TS

S cmB T H A m

H l N I N I l cmH lN N

I

φ −

−= = =⋅

= ⋅ == ⇒ =

⋅ = + =⋅

+ = =

od dijagramot za

,

navivki

( )

2 21 2 3 1 2 3

1 1 3 3

2 2 3 3

1 2 3

3 3

32 3

2

140 , 40 , 40 20 800 , 1600

0

0 0 I

0,8 100 /

2

l l mm l mm S S mm S mmH l H l N IH l H l

B T H A ml

H Hl

φ φ φ φ

= = = = = ⋅ = =⋅ + ⋅ = ⋅⋅ + ⋅ =

− + + = =

= ⇒ =

= =

∑ - Kirhofov zakon za mag. kola

Od krivata na magnetizirawe, za

2 2

42 2 2

43 3 3

41 2 3

11

1

1 1

1 1 3 3

8,6 /

28,6 0,2

1,6 10

12,8 10

14, 4 10

1,8

1,8 20000

28

A m

AH B TmB S WbB S Wb

Wb

B TS

AB T H mN I H l H l

φ

φ

φ φ φφ

−

−

−

= ⇒ =

= ⋅ = ⋅

= ⋅ = ⋅

= + = ⋅

= =

= ⇒ =

⋅ = ⋅ + ⋅ =

Od krivata na magnetizirawe za

Od krivata na magnetizirawe za

04 навивкиA ⋅

6. ^etvoroagolno zatvoreno kolo se nao|a nad beskrajno dolg provodnik niz koj te~e struja 100 .I A= . Kolkava struja treba da te~e niz zatvorenoto kolo za da koloto lebdi nad provodnikot ako negovata te`ina iznesuva 0,3G N= . Koloto i provodnikot se nao|aat vo ista ramnina vo vozduh (µ0 = 4π⋅10-7 H/m).

9401

a cmb cmc cm

===

( )

( )

( )

3 4

3 4

3 1 3 1 0

4 1 4 1 0

3 4 1 0

1

0

маг. поле околу 2

долг проводник

,2 2

2

2

1 12

1 12

rez

rez

F I l B Био Саваров закон

F GF F F

IHr

I IH Hc a c

IF I b B I bcIF I b B I ba c

IF F I b Gc a c

GIIb

c a

π

π π

µπ

µπ

µπ

µπ

= × − −

== −

= −⋅

= =⋅ ⋅ +

= ⋅ ⋅ = ⋅ ⋅ ⋅⋅

= ⋅ ⋅ = ⋅ ⋅+

− = ⋅ ⋅ ⋅ ⋅ − = +

=⋅ ⋅ ⋅ −

417A

c

= −

7. Mnogu dolg pravoliniski provodnik niz koj te~e postojana struja 1I i

kruto pravoagolno kolo niz koe te~e postojana struja 2I se nao|aat vo vozduh kako na slikata. Da se odredi rezultantnata magnetna sila so koja provodnikot dejstvuva na konturata.

( )

( )

1 3

1 2 1 3 2 3

0 1 0 11 2 3 2

0 1 2

,

,2 2

2

rez

rez

F F FF I b B F I b B

I IF I b F I bc a c

I I b aFc a c

µ µπ π

µπ

= −= ⋅ ⋅ = ⋅ ⋅

⋅ ⋅= ⋅ ⋅ = ⋅ ⋅

⋅ +

⋅ ⋅ ⋅= ⋅

⋅ +

8. Mnogu dolg, tenok, pravoliniski provodnik so postojana ednonaso~na struja 1I i kruto pravoagolno kolo so postojana ednonaso~na struja 2I se nao|aat vo vozduh kako na slikata. Pravoagolnoto kolo e povrzano za nepodvi`na oska i mo`e da se vrti okolu nea. Da se odredi momentot na magnetnite sili na koloto sprema oskata 1 20 0 , vo polo`ba prika`ana na slikata.

Kosinusna teorema: 2 2 2 cosr a c a c α= + − ⋅ ⋅ ⋅

Sinusna teorema: sin sincr

ϕ α= ⋅

( )1 2

0 2 2

sin2 2 cos

I I a b cMa c ac

αµ

π α⋅ ⋅ ⋅ ⋅ ⋅

=+ − ⋅

Za koloto na slikata: 2π

α = ⇒ ( )1 2

0 2 22I I a b cM

a cµ

π⋅ ⋅ ⋅ ⋅

=+

9. Da se odredi magnetniot fluks niz pravoagolna kontura koja se nao|a

vo magnetno pole to go sozdava mnogu dolg provodnik niz koj te~e struja I .

0 0

cos0

ln2 2

S S Sd a

d

Bd S BdS B b dx

I b I b d adxx d

φ

µ µφ

π π

+

= = ⋅ = ⋅ ⋅

⋅ ⋅ ⋅ ⋅ += =

∫ ∫ ∫

∫

10. Mnogu dolg pravoliniski provodnik niz koj te~e konstantna struja I

i pravoagolno kolo se nao|aat vo ista ramnina vo vozduhot. Pravoagolnoto kolo se oddale~uva od provodnikot so postojana brzina ϑ ostanuvaji vo ista ramnina so nego. Vo momentot 0t = , provodnikot i pravoagolnoto kolo imaat polo`ba kako na slikata. Da se odredi izrazot za induciranata elektromotorna sila vo pravoagolnoto kolo.

2 2 1

0 12 2

2

2

2

sin

F I b B

IF I br

M a FM a F

µπ

ϕ

= ⋅ ⋅ ⋅

= ⋅ ⋅⋅

= ⋅= ⋅ ⋅

( )

( ) ( )

( )( )

0

00

00

0

0

2

ln2

ln2

12

Sa c

S c

B dS

I bB b dx dx

xI b a c

cI b a xx

xd x d x dxe

dt dx dtdxx c tdt

d x I a be

dx x a x

φ

µπ

µπ

µπ

ϑ ϑ

µ ϑϑ

π

+

= ⋅

⋅ ⋅Φ = ⋅ ⋅ =

⋅

⋅ ⋅ +Φ = ⋅ ⋅

⋅ ⋅ +Φ = ⋅ ⋅

Φ Φ= − = − ⋅

= + ⋅ ⇒ =

Φ ⋅ ⋅ ⋅ ⋅= − ⋅ = ⋅

⋅ +

∫

∫ ∫

11. Pokraj dolg pravoliniski provodnik niz koj te~e struja cos 60 cosmi I t tω ω= ⋅ = ⋅ , 50f Hz= se nao|a kalem so 100N = navivki so pravoago-

len oblik vo polo`ba kako na slikata. Da se odredi maksimalnata vrednost na induciranata EMS vo kalemot.

145

a cmb cmc cm

===

0 01

1 0

7 2

7

0

4

ln2 2

1 ln cos2

1 44 10 100 4 10 ln 60 cos1002 1665 10 cos100

1 ln sin2

209 10 si

Sb b

a a

VK m

VK

VK

VKm

B d S

i i b bd b dxx a

bN N b I ta

t

t Wbd be N b I t

dt ae

φ

µ µφ φ

π π

φ φ µ ωπ

φ π ππ

φ π

φµ ω ω

π

− −

−

−

= ⋅

⋅ ⋅ ⋅= = ⋅ ⋅ =

= ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

= − = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅

∫

∫ ∫

4max

n100209 10

tE V

π−

⋅

= ⋅

12. Na tenok kartonski torus so sredna dol`ina b i popre~en presek S ramnomerno se namotani N navivki `ica so zanemarliva elektri~na otpornost. Niz namotkata te~e struja ( ) cosmi t I tω= . Da se odredi izrazot za napon pome|u kraevite na namotkata.

00

2

0

2

0

2

0

2

0

2

0

( )cos

cos

sin

sin

cos2

cos2

m

m

m

mab

mab

ab m

N i t N IB t

b bdedt

N IN B S S t

bN I S

e tb

N I SU e t

bN I S

U tb

U LI t

N SLb

µµ ω

φ

φ µ ω

ωµ ω

ωµ ω

ω πµ ω

πω ω

µ

⋅ ⋅ ⋅= = ⋅

= −

= ⋅ ⋅ = ⋅ ⋅ ⋅

⋅ ⋅= ⋅

⋅ ⋅= = ⋅ ⋅

⋅ ⋅ = ⋅ ⋅ +

= ⋅ +

⋅= ⋅

13. Na drven torus daden na slikata ramnomerno se namotani N navivki na bakarna `ica niz koja te~e struja I . Da se odredi induktivnosta na namot-kata.

7 40

1

22

0

7

500 0,8 2000 /0, 2

20004 10 8 10

4

125 10

VK

VK

H l N IN I N IH A m

l D

B H T

B SN L I

dNN B SLI I D

L H

π π π

µ ππ

φφ φ

πµφ

ππ

− −

−

⋅ = ⋅⋅ ⋅ ⋅

= = = =⋅ ⋅

= ⋅ = ⋅ ⋅ = ⋅

= ⋅= ⋅ = ⋅

⋅ ⋅⋅ ⋅= = =

⋅= ⋅ ⋅

20 ; 1 ; 0,8 ; 500D cm d cm I A N= = = =

14. Da se odredi vektorot na magnetnata indukcija to go sozdava kru`na strujna kontura so radius a vo svojot centar.

0 02

0 02 2

0 0 0 02 2 2

4

sin2

4 4

224 4 4C C

I dl rd B

r

I dlI dl

dBa a

Idl I I IB dl a

aa a a

µπ

πµ µπ πµ µ µ µ

ππ π π

⋅ ×= ⋅

⋅ ⋅ ⋅ = ⋅ =

= = = =∫ ∫

15. Namotka bez jadro so 1000 navivki ima dol`ina 25 cm i dijametar

4 cm. Na nea e namotana druga namotka so 3000 navivki. Kolkava EMS e se in-ducira vo vtorata namotka ako strujata vo prvata namotka ja zgolemime ramno-merno od 1 A na 6 A za vreme od 0,5 s.

16. Vo namotka so 100N = navivki magnetniot fluks se menuva vo tekot na vremeto kako to e prika`ano na slikata. Da se presmeta i grafi~ki da se prika`e promenata na induciranata elektromotorna sila vo namotkata.

( )

[ ]

[ ]

10 2 1

7

6

2

6

1000 0,001254 10 50,25

31,4 10

31,4 103000 0,1880,5

N S I Il

Wb

E Nt

E V

φ µ

φ π

φ

φ

−

−

−

⋅∆ = ⋅ −

⋅∆ = ⋅ ⋅ ⋅

∆ = ⋅

∆= ⋅

∆⋅

= ⋅ =

0

0

0

0 1

2 1

2 1 1 10 0

B H

HS

H SI N S

l

I N S I N Sl l

µ

φµ

φ µµ

φ

φ φ φ

φ µ µ

=

=

= ⋅ ⋅⋅ ⋅ ⋅

=

∆ = −⋅ ⋅ ⋅ ⋅

∆ = ⋅ − ⋅

2 11

2 1

2

2 0100 504

0 2100 2001

e N Vt t

e V

φ φ− −= − = − ⋅ = −

−−

= − ⋅ =

de Ndtφ

= −

17. Da se odredi potrebniot broj na navivki i strujata niz namotkata za da elektromagnetot prika`an na slikata mo`e da nosi te`ina 3000G N= . Elek-tromagnetot e napraven od dinamolim ~ija kriva na magnetizirawe e dadena na slikata, a dimenziite se : 2

1 2100 , 50 , 21S cm l cm l cm= = = .

23

00

6.08 10F S F WbS

φφ µ

µ−= ⇒ = ⋅ ⋅ = ⋅

⋅

0.608B TSφ

= =

Od krivata na magnetizirawe, za 0,608 140B T H A m= ⇒ =

140 0,7 98 навивкиN IH N I H l Al⋅

= ⇒ ⋅ = ⋅ = ⋅ = ⋅

18. Daden e elektromagnet so slednite podatoci: • presek na jadroto ( jarem i kotva) 23j kS S cm= = ,

• sredna dol`ina na magnetnoto kolo vo kotvata i jaremot 30j kl l cm+ = ,

• vkupnata dol`ina na dvata vozduni procepi 2 2mmδ = , • broj na navivki 10000N = so elektri~en otpor 500R = Ω .

Karakteristikata na magnetizirawe e dadena tabelarno. Da se odredi: a) Naponot na kraevite od namotkata koga elektromagnetot ja privlekuva

kotvata so sila 343F N= . b) Snagata koja se troi vo namotkata.

a) 2

0

70

4

4 10 343 1,23 10

jB SF

FB Т

S

µ

µ π −

−

⋅=

⋅ ⋅ ⋅= = =

⋅

od tabelata, za 1,2 1300 /B T H A m= ⇒ =

b) 2 2500 0,23 26,5P R I W= ⋅ = ⋅ =

( )AH m 680 930 1300 2200

( )B T 0,8 1,0 1,2 1,4

( )

0

0 0 0

400 7

0

0

0

1300

1,2 96 10 /4 10

2

20, 23

115

j k

k j

k j

AH H mB B constB H

BH A m

H l N IH l H l H N I

H l l HI A

NU R I V

µ

µ π

δ

δ

−

= =

= == ⋅

= = = ⋅⋅

⋅ = ⋅

⋅ + ⋅ + ⋅ = ⋅ ⇒

+ + ⋅= =

= ⋅ =

∑ ∑