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PROBLEM SOLUTIONS: Chapter 1
Problem 1-1
Part (a):
Rc = lc
µAc
=
lc
µrµ0Ac
= 0 A/Wb
Rg = g
µ0Ac
= 5.457× 106 A/Wb
Part (b):
Φ =
NI
Rc +Rg
= 2.437 × 10−5 Wb
Part (c):
λ = NΦ = 2.315× 10−3 Wb
Part (d):
L =
λ
I
= 1.654 mH
Problem 1-2
Part (a):
Rc = lc
µAc
=
lc
µrµ0Ac
= 2.419× 105 A/Wb
Rg = g
µ0Ac
= 5.457× 106 A/Wb
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Part (b):
Φ =
NI
Rc +Rg = 2.334 × 10
−5 Wb
Part (c):
λ = NΦ = 2.217× 10−3 Wb
Part (d):
L =
λ
I
= 1.584 mH
Problem 1-3
Part (a):
N =
√
Lg
µ0Ac
= 287 turns
Part (b):
I =
Bcore
µ0N/g
= 7.68 A
Problem 1-4
Part (a):
N =
√
L(g + lcµ0/µ)
µ0Ac
=
√
L(g + lcµ0/(µrµ0))
µ0Ac
= 129 turns
Part (b):
I =
Bcore
µ0N/(g + lcµ0/µ)
= 20.78 A
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instructor use. Not authorized for sale or distribution in any manner. This document may
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whole or part.
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Problem 1-5
Part (a):
Part (b):
Bg = Bm = 2.1 T
For Bm = 2.1 T, µr = 37.88 and thus
I =
(
Bm
µ0N
)(
g +
lc
µr
)
= 158 A
Part (c):
Problem 1-6
Part (a):
Bg =
µ0NI
2g
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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Bc = Bg
(
Ag
Ac
)
=
(
µ0NI
2g
)(
1− x
X0
)
Part (b): Will assume lc is “large” and lp is relatively “small”. Thus,
BgAg = BpAg = BcAc
We can also write
2gHg +Hplp +Hclc = NI ;
and
Bg = µ0Hg; Bp = µHp Bc = µHc
These equations can be combined to give
Bg =

 µ0NI
2g +
(
µ0
µ
)
lp +
(
µ0
µ
) (
Ag
Ac
)
lc

 =

 µ0NI
2g +
(
µ0
µ
)
lp +
(
µ0
µ
) (
1− x
X0
)
lc


and
Bc =
(
1− x
X0
)
Bg
Problem 1-7
From Problem 1-6, the inductance can be found as
L =
NAcBc
I
=
µ0N
2Ac
2g + µ0
µ
(lp + (1− x/X0) lc)
from which we can solve for µr
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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µr =
µ
µ0
=
L
(
lp + (1− x/X0) lc
)
µ0N2Ac − 2gL = 88.5
Problem 1-8
Part (a):
L =
µ0(2N)
2Ac
2g
and thus
N = 0.5
√
2gL
Ac
= 38.8
which rounds to N = 39 turns for which L = 12.33 mH.
Part (b): g = 0.121 cm
Part(c):
Bc = Bg =
2µ0NI
2g
and thus
I =
Bcg
µ0N
= 37.1 A
Problem 1-9
Part (a):
L =
µ0N
2Ac
2g
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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and thus
N =
√
2gL
Ac
= 77.6
which rounds to N = 78 turns for which L = 12.33 mH.
Part (b): g = 0.121 cm
Part(c):
Bc = Bg =
µ0(2N)(I/2)
2g
and thus
I =
2Bcg
µ0N
= 37.1 A
Problem 1-10
Part (a):
L =
µ0(2N)
2Ac
2(g + (µ0
µ
)lc)
and thus
N = 0.5
√√√√2(g + (µ0µ )lc)L
Ac
= 38.8
which rounds to N = 39 turns for which L = 12.33 mH.
Part (b): g = 0.121 cm
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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Part(c):
Bc = Bg =
2µ0NI
2(g + µ0
µ
lc0)
and thus
I =
Bc(g +
µ0
µ
lc)
µ0N
= 40.9 A
Problem 1-11
Part (a): From the solution to Problem 1-6 with x = 0
I =
Bg
(
2g + 2
(
µ0
µ
)
(lp + lc)
)
µ0N
= 1.44 A
Part (b): For Bm = 1.25 T, µr = 941 and thus I = 2.43 A
Part (c):
Problem 1-12
g =
µ0N
2Ac
L
−
(
µ0
µ
)
lc = 7.8× 10−4 m
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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Problem 1-13
Part (a):
lc = 2pi
(
Ri +Ro
2
)
− g = 22.8 cm
Ac = h(Ro − Ri) = 1.62 cm2
Part (b):
Rc = lc
µAc
= 0
Rg = g
µ0Ac
= 7.37× 106 H−1
Part (c):
L =
N2
Rc +Rg = 7.04× 10
−4 H
Part (d):
I =
BgAc(Rc +Rg)
N
= 20.7 A
Part (e):
λ = LI = 1.46× 10−2 Wb
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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Problem 1-14
See solution to Problem 1-13
Part (a):
lc = 22.8 cm
Ac = 1.62 cm
2
Part (b):
Rc = 1.37× 106 H−1
Rg = 7.37× 106 H−1
Part (c):
L = 5.94× 10−4 H
Part (d):
I = 24.6 A
Part (e):
λ = 1.46× 10−2 Wb
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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Problem 1-15
µr must be greater than 2886.
Problem 1-16
L =
µ0N
2Ac
g + lc/µr
Problem 1-17
Part (a):
L =
µ0N
2Ac
g + lc/µr
= 36.6 mH
Part (b):
B =
µ0N
2
g + lc/µr
I = 0.77 T
λ = LI = 4.40× 10−2 Wb
Problem 1-18
Part (a): With ω = 120pi
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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Vrms =
ωNAcBpeak√
2
= 20.8 V
Part (b): Using L from the solution to Problem 1-17
Ipeak =
√
2Vrms
ωL
= 1.66 A
Wpeak =
LI2peak
2
= 9.13× 10−2 J
Problem 1-19
B = 0.81 T and λ = 46.5 mWb
Problem 1-20
Part (a):
R3 =
√
(R21 +R
2
2) = 4.49 cm
Part (b): For
lc = 4l +R2 +R3 − 2h;
and
Ag = piR
2
1
L =
µ0AgN
2
g + (µ0/µ)lc
= 61.8 mH
Part (c): For Bpeak = 0.6 T and ω = 2pi60
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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λpeak = AgNBpeak
Vrms =
ωλpeak√
2
= 23.2 V
Irms =
Vrms
ωL
= 0.99 A
Wpeak =
1
2
LI2peak =
1
2
L(
√
2Irms)
2 = 61.0 mJ
Part (d): For ω = 2pi50
Vrms = 19.3 V
Irms = 0.99 A
Wpeak = 61.0 mJ
Problem 1-21
Part (a);
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
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Part (b):
Emax = 4fNAcBpeak = 118 V
part (c): For µ = 1000µ0
Ipeak =
lcBpeak```