CHAPTER 7 MAGNETIC CIRCUITS EXERCISE 33, Page 83 2
1. What is the flux density in a magnetic field of cross-sectional area 20 cm having a flux of 3 mWb?
Φ Flux dens!", # = A
=
3 ×1 0−3 Wb 20 × 10−4 m 2 = 1.$ T
%. etermine the total flux emerging from a magnetic !ole face having dimensions " cm by # cm$ if the flux density is 0%& '%
(=
Φ
10−4 = %.7 '() A from )hich$ &lux$ Φ = ( × A = 0%& × " × # ×10
3. 'he maximum )or*ing flux density of a lifting electromagnet is 1%& ' and the effective area of a !ole face is circular in cross-section% +f the total magnetic flux !roduced !roduced is #11 mWb determine determine the radius of the !ole face%
Φ
Φ (= = 2 A π r
2
r from )hich$
Φ = (π
and *adus, * =
Φ ( π÷ =
#11 #11× 10−3 1% & × π
= 0%32 m or 3% +'
)ound on a circular magnetic circuit of radius . A current of " A is !assed through a 1000-turn coil )ound 120 mm% ,alculate a. the magnetomotive force$ and b. the magnetic field strength% a. Magne!-'-!e &-*+e, '.'.&. = /+ = 1000.". = $/// A b. ength of magnetic field$ l = 2 πr = 2π0%120. m /+ -1000.-". = l 2π-0%120. = 31 A2' ence$ 'agne!+ &eld s!*eng!0, H = densit y of 0%4" '% +f the magnetic flux is 20 Wb $. An electromagnet of suare cross-section !roduces a flux density find the dimensions of the electromagnet cross-section% Φ 5lux density$ ( = A from )hich$ Φ cross-sectional area$ are a$ A = (
=
20 ×10−# 0 %4 "
= 1%# × 10 10 −3 m 2 = 1%# ×10 −3 ×104 cm 2 = 1# cm cm 2
2 +f each side of the suare cross-section is x$ then x = 1# and x = 1# = 4 ence$ !0e d'ens-ns -& !0e ele+!*-'agne! +*-ssse+!-n s +' )" +'
)hen a coil of 400 400 turns . 5ind the magnetic field strength a!!lied to a magnetic circuit of mean length "0 cm )hen is a!!lied to it carrying a current of 1%2 A
/+ Magne!+ &eld s!*eng!0, H = l
=
-400.-1%2. "0 ×10−2 = 4/ A2'
5 6-0n #*d Pu)ls0ed )" Ta"l-* and F*an+s
7. A solenoid 20 cm long is )ound )ith "00 turns of )ire% 5ind the current reuired to establish a magneti6ing force of 2"00 A7m inside the solenoid%
/+ 8agnetic field strength$ = l l from )hich$
current$ + = /
=
-2"00.-20 ×10−2 . "00
=1A
8. A magnetic field strength of "000 A7m is a!!lied to a circular magnetic circuit of mean diameter 2"0 mm% +f the coil has "00 turns find the current in the coil%
ength of magnetic field$ l = πd = π0%2". m / + × l "000 × π× 0%2" = = l / "00 from )hich$ +u**en!, I = = 7.8$ A
EXERCISE 3, Page 8$ 1. 5ind the magnetic field strength and the magnetomotive force needed to !roduce a flux density of 0%33 ' in an air-ga! of length 1" mm%
(
= µ 0µ r
( from )hich$ 'agne!+ &eld s!*eng!0, H =
µ0 µr
=
0%33 4π× 10
−
×1 = %%// A2'
−3 Magne!-'-!e &-*+e, '.'.&. = × l = 2#2#00 × 1" ×10 = 3434 A
%. An air-ga! bet)een t)o !ole !ieces is 20 mm in length and the area of the flux !ath across the 2 ga! is " cm % +f the flux reuired in the air-ga! is 0%" mWb find the m%m%f% necessary%
Φ (
(
= µ0
µ
=
A
µ
=
Φ A µ0
0 for air$ from )hich$ = 0 Φ 0%" × 10−3 × 20 ×10−3 ×l = A µ0 " ×10−4 × 4π×10− and '.'.&. = × l = = %387/ A
3. a. etermine the flux density !roduced in an air-cored solenoid due to a uniform magnetic field strength of 9000 A7m% b. +ron having a relative !ermeability of 1"0 at 9000 A7m is inserted into the solenoid of !art a.% 5ind the flux density no) in the solenoid%
( a. ( b.
= µ0
for air$ from )hich$ &lux dens!", # =
= µ0 µ r
from )hich$
µ0 = 4π×10− × 9000 = 1/./$ 'T
= 4π×10− ×1"0 × 9000 µ µ 0 r &lux dens!", # = = 1.$/8 T 5 6-0n #*d Pu)ls0ed )" Ta"l-* and F*an+s
. 5ind the relative !ermeability of a material if the absolute !ermeability is 4%094 x 10-4 7m%
Absolute !ermeability$ =
µ0 µ r
µ r =
from )hich$ *ela!e e*'ea)l!"$
µ µ0
=
4%094 ×10−4 4π×10 −
= 3%$
$. 5ind the relative !ermeability of a !iece of silicon iron if a flux density of 1%3 ' is !roduced by a magnetic field strength of 00 A7m%
(
= µ0 µ r
(
from )hich$ *ela!e e*'ea)l!",
*
=
µ0
=
1%3 4 π×10 − × 00
= 178
. A steel ring of mean diameter 120 mm is uniformly )ound )ith 1"00 turns of )ire% When a current of 0%30 A is !assed through the coil a flux density of 1%" ' is set u! in the steel% 5ind the relative !ermeability of the steel under these conditions% −3 ength of magnetic field$ l = π× d = π×120 mm = π×120 ×10 m = 0%12 π m
/+ 8agnetic field strength$ = l
(
= µ0 µ r
=
-1"00.-0%3. 0%12π
= 11&3%##
A7m (
from )hich$ *ela!e e*'ea)l!",
*
=
µ0
=
1%" 4π×10
−
×11&3%## = 1///
2 7. A uniform ring of cast steel has a cross-sectional area of " cm and a mean circumference of 1" cm% 5ind the current reuired in a coil of 1200 turns )ound on the ring to !roduce a flux of 0%9 mWb% :se the magnetisation curve for cast steel sho)n on !age 93 of textboo*.
−2 2 −4 −3 ength of magnetic field$ l = 1" ×1 0 m = 0%1" m $ c%s%a%$ A = " ×10 m $ flux$ Φ = 0%9 ×10 Wb
Φ
5lux density$ ( = A
=
0%9 × 10 −3 " ×10 −4 = 1%# '
5rom the gra!h of cast steel on !age 93$ )hen ( = 1%# '$ = 4900 A7m 8agnetomotive force$ m%m%f% = × l = / × +
×l from )hich$ +u**en!, I = /
=
4900 × 0%1" 1200
= /./ A
5 6-0n #*d Pu)ls0ed )" Ta"l-* and F*an+s
2
8. a. A uniform mild steel ring has a diameter of "0 mm and a cross-sectional area of 1 cm % etermine the m%m%f% necessary to !roduce a flux of "0 µWb in the ring% :se the (- curve for mild steel sho)n on !age 93.% b. +f a coil of 440 turns is )ound uniformly around the ring in !art a. )hat current )ould be reuired to !roduce the flux? 2 −3 −4 −# a. ength of magnetic field$ l = π× "0 × 10 m $ c%s%a%$ A = 1× 10 m $ flux$ Φ = "0 × 10 Wb
Φ
5lux density$ ( = A
=
"0 ×10−# 1× 10−4 = 0%" '
5rom the gra!h of mild steel on !age 93$ )hen ( = 0%" '$ = 00 A7m −3 8agnetomotive force$ '.'.&. = × l = 00 × π× "0 × 10 = 11/ A
m%m%f% b. m%m%f% = / + from )hich$ +u**en!, I =
/
=
110 440 = /.%$ A
4. 5rom the magnetisation curve for mild steel sho)n on !age 93$ derive the curve of relative !ermeability against magnetic field strength% 5rom your gra!h determine a. the value of r )hen the magnetic field strength is 1200 A7m$ and b. the value of the magnetic field strength )hen r is "00
( ( = 0 r hence r =
1
µ0 = µ0
×
(
10
= 4π
×
(
A number of co-ordinates are selected from the (- curve and r is calculated for each as sho)n in the follo)ing table% ( '. 0%10 0%3" 0%"" 0%2 1%03 1%24 1%3 A7m. 2"0 "00 "0 1000 1"00 2000 2"00 319 " "93 "3 "41 4&3 43# 10 (
µ r =
4π
×
r is !lotted against as sho)n belo)%
5 6-0n #*d Pu)ls0ed )" Ta"l-* and F*an+s
a. 5rom the gra!h$ the value of r )hen the magnetic field strength is 1200 A7m is a*-und $/ $$ b. 'he value of the magnetic field strength )hen r is "00 is a*-und // A2' -* 14// A2'
EXERCISE 3$, Page 8
2
1. ;art of a magnetic circuit is made from steel of length 120 mm$ cross-sectional area 1" cm and relative !ermeability 900% ,alculate a. the reluctance and b. the absolute !ermeability of the steel%
l
=
0%12
900 ×1" ×10 −4 × a. Relu+!an+e, S = = 74$8/ 2H − µ = µ0 ×µ r = 4π×10 × 900 = 1 'H2' b. A)s-lu!e e*'ea)l!"$
µ0 µr A
4π× 10
−
or 74$8/ A2()
%. A mild steel closed magnetic circuit has a mean length of " mm and a cross-sectional area of 2 320%2 mm % A current of 0%40 A flo)s in a coil )ound uniformly around the circ uit and the flux
!roduced is 200 µWb% +f the relative !ermeability of the steel at this value of current is 400 find 5 6-0n #*d Pu)ls0ed )" Ta"l-* and F*an+s
a. the reluctance of the material and b. the number of turns of the coil% l
a. Relu+!an+e, S = m%m%f% / + = Φ Φ b. < =
µ0 µ r A
=
0%0" 4 π×10
−
× 400 × 320%2 ×10−# = /// 2H <Φ +
from )hich$ nu')e* -& !u*ns, N =
=
4##000 × 200 × 10−# 0%40
= %33
EXERCISE 3, Page 84 2
1. A magnetic circuit of cross-sectional area 0%4 cm consists of one !art 3 cm long$ of material having a relative !ermeability 1200$ and a second !art 2 cm long of material having a relative !ermeability "0% With a 100 turn coil carrying 2 A$ find the value of flux existing in the circuit%
<1 = eluctance of !art 1$
<2 = eluctance of !art 2$ 'otal reluctance$
<' =
m%m%f% Φ
=
l1 µ0 µ r1 A1
l2 µ 0 µ r2 A 2
=
3 ×10 −2 4π× 10− ×1200 × 0%4 ×10−4
=
2 × 10
4π × 10− × "0 × 0%4 ×10−4
= "30"1# 7
<' = <1 + <2 = 4&3"& + "30"1# = 1029" 7
/+ Φ
= 4&3"& 7
−2
Φ= from )hich$ &lux$
/ + <'
=
100 × 2 1029"
= /.14$ '() 2
%. a. A cast steel ring has a cross-sectional area of #00 mm and a radius of 2" mm% etermine the m%m%f% necessary to establish a flux of 0%9 mWb in the ring% :se the (- curve for cast steel sho)n on !age 93% b. +f a radial air ga! 1%" mm )ide is cut in the ring of !art a. find the m%m%f% no) necessary to maintain the same flux in the ring% −3 −# −3 2 a. c%s%a%$ A = #00 × 10 m $ length of magnetic circuit$ l = 2 πr = 2π × 2" ×10 m $ Φ = 0%9 ×10 Wb
(=
Φ
A
=
0%9 × 10−3 #00 × 10−# = 1%33333 ' = 1%33 '$ correct to 2 decimal !laces%
5rom the gra!h of cast steel on !age 93$ )hen ( = 1%33 '$ = 120 A7m −3 '.'.&. = × l = 120 × 2π × 2" × 10 = %7/ A
( b. 5or the air ga!$
= µ0
( from )hich$ =
µ0
=
1%33333 4π× 10−
= 10#1030 A
−3
m%m%f% = × l = 10#1030 × 1%" ×10 = 1"&2 5 6-0n #*d Pu)ls0ed )" Ta"l-* and F*an+s
T-!al '.'.&. = 20 > 1"&2 = 18/ A$ correct to 4 significant figures
3. A closed magnetic circuit made of silicon iron consists of a 40 mm long !ath of cross-sectional area &0 mm2 and a 1" mm long !ath of cross-sectional area 0 mm 2% A coil of "0 turns is )ound around the 40 mm length of the circuit and a current of 0%3& A flo)s% 5ind the flux density in the 1" mm length !ath if the relative !ermeability of the silicon iron at this value of magnetising force is 3000% F-* !0e / '' l-ng a!0
40 × 10 −3
l1 eluctance <1 =
µ0µr A1 =
4π×10 − .3000.&0 ×10 −# . = 119&2%""7
F-* !0e 1$ '' l-ng a!0
1" × 10−3
l2 eluctance <2 =
µ 0µ r A 2
− −# π× × -4 10 .-30000.-0 10 . = "#941%0"7 =
'otal circuit reluctance < = <1 > <2 = 119&2%"" > "#941%0" = 1433%#7 m%m%f% <=
Φ
m%m%f % i%e% Φ =
<
"0 × 0%3& /+ = < = 1433%# = 1%11# 10 -4 Wb Φ
1%11# × 10−4
5lux density in the 1" mm !ath$ ( = A =
0 × 10−# = 1.$4 T
. 5or the magnetic circuit sho)n belo)$ find the current + in the coil needed to !roduce a flux of 0%4" mWb in the air-ga!% 'he silicon iron magnetic circuit has a uniform cross-sectional area of 3 cm2 and its magnetisation curve is as sho)n on !age 93 of textboo*%
Φ 5or the silicon iron core$ ( = A
=
0%4" ×10−3 3 ×10−4
= 1%" '
5rom the magnetisation curve on !age 93.$ = 3"00 A7m
5 6-0n #*d Pu)ls0ed )" Ta"l-* and F*an+s
(
µ $ = µo since = 1 r
5or the air ga!
( from )hich$
'otal m%m%f$
µo = 4π x 10- = 11&3##2 A7m
=
m%m%f% '
= m%m%f%ring > m%m%f%ga! = l
r l l
1%"
= 3"00
ring
> l
ring
× 20 × 10
ga!
−2
= 00 A
−3 = 11&3##2 ×1%" × 10 = 1&0%" A
ga!
m%m%f%'
= 00 > 1&0%" =
24&0%" A
24&0%"
3000 = /.83 A
+u**en!, I =
$. A ring forming a magnetic circuit is made from t)o materials@ one !art is mild steel of mean 2 length 2" cm and cross-sectional area 4 cm $ and the remainder is cast iron of mean length 20 cm 2 and cross-sectional area %" cm % :se a tabular a!!roach to determine the total m%m%f% reuired to cause a flux of 0%30 mWb in the magnetic circuit% 5ind also the total reluctance of the circuit% :se the magneti6ation curves sho)n on !age 93%
4
− 8ild steel l = 0%2" m$ c%s%a%$ A = 4 ×10 m
2
2 −4 ,ast iron l = 0%20 m$ c%s%a%$ A = %" ×10 m
Pa*! -& +*+u!
8ild steel ,ast iron
2
9():
A9m : −3
0%30 ×10 −3 0%30 ×10
4 ×10
−4
%" × 10−4
# 9T:
H 9A2':
l 9':
''& ; H
9&*-' g*a0:
0%"
1000
0%2"
2"0
0%40
1"00
0%20
300
T-!al
m%m%f % T-!al *elu+!an+e, S =
Φ
=
$$/ A
""0 0%30 × 10−3 =
1.83
1/ 2 H
. 'he diagram belo) sho)s the magnetic circuit of a relay% When each of the air ga!s are 1%" mm )ide find the m%m%f% reuired to !roduce a flux density of 0%" ' in the air ga!s% :se the (- curves sho)n on !age 93%
5 6-0n #*d Pu)ls0ed )" Ta"l-* and F*an+s
l
2
Pa*! -& +*+u!
A9m :
# 9T:
l 9':
''& ; H
0%"
H 9A2': 9&*-' g*a0: ""00
,ast iron 8ild steel
0%9 × 10 −4 0%9 × 10 −4
0%20
1100
0%"
1000
0%09
90
Air
0%9 × 10 −4
0%"
0%" 4π×10
−
= "Σ
2 ×1%" ×10 −3 T-!al
5 6-0n #*d Pu)ls0ed )" Ta"l-* and F*an+s
1&0 %47/ A
l
