전자기학솔루션

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전자기학 솔루션입니다 수업 과제에 많은 도움을 줄것입니다.

목차

chap10.pdf
chap11.pdf
chap1a.pdf
chap1b.pdf
chap1c.pdf
chap2.pdf
chap3a.pdf
chap3b.pdf
chap4o.pdf
chap5o.pdf
chap6.pdf
chap7.pdf
chap8a.pdf
chap8b.pdf
chap9.pdf

본문내용

Chapter 10 – Odd-Numbered
10.1. In Fig. 10.4, let B = 0.2 cos 120πt T, and assume that the conductor joining the two ends
of the resistor is perfect. It may be assumed that the magnetic field produced by I(t) is
negligible. Find:
a) Vab(t): Since B is constant over the loop area, the flux is Φ = π(0.15)2B = 1.41 ×
10−2 cos 120πt Wb. Now, emf = Vba(t) = −dΦ/dt = (120π)(1.41 × 10−2) sin 120πt.
Then Vab(t) = −Vba(t) = −5.33 sin 120πt V.
b) I(t) = Vba(t)/R = 5.33 sin(120πt)/250 = 21.3 sin(120πt) mA
10.3. Given H = 300 az cos(3 × 108t − y) A/m in free space, find the emf developed in the general
(중략)
and the energy is now
WH =

1
0

2π
0

a
c
μ0I2(ρ2 − c2)2
8π2ρ2(a2 − c2)2 ρdρdφdz =
μ0I2
4π(a2 − c2)2

a
c

ρ3 − 2c2ρ +
C4
ρ

dρ
=
μ0I2
4π(a2 − c2)2

1
4
(a4 − c4) − c2(a2 − c2) + c4 ln
a
c


J/m
The internal inductance is then
Lint =
2WH
I2 =
μ0
8π

a4 − 4a2c2 + 3c4 + 4c4 ln(a/c)
(a2 − c2)2

H/m

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압축파일 내 파일목록

chap10.pdf
chap11.pdf
chap1a.pdf
chap1b.pdf
chap1c.pdf
chap2.pdf
chap3a.pdf
chap3b.pdf
chap4o.pdf
chap5o.pdf
chap6.pdf
chap7.pdf
chap8a.pdf
chap8b.pdf
chap9.pdf