전자기학
(Electromagnetics)
Chapter 6 Static Magnetic Fields
Geunchang Choi
School of Electrical and Electronics Engineering,
Chung-Ang University
Email: nightsky@cau.ac.kr
Chapter 6 Static Magnetic Fields
Boundary Conditions for Magnetostatic Fields
• Boundary conditions
Normal component
∆𝑆
𝐉𝑠
ර 𝐁 ∙ 𝑑𝐬 = 0
𝛻∙𝐁=0
∆ℎ
𝑆
Δℎ → 0
⨀⨀⨀
ර 𝐁 ∙ d𝒔 = 𝐁1 ∙ 𝐚𝑛2 Δ𝑆 + 𝐁2 ∙ −𝐚𝑛2 Δ𝑆 = 0
𝑆
𝐵1𝑛 = 𝐵2𝑛
Tangential component
𝛻 × 𝐁 = 𝜇𝐉𝑠
𝜇1 𝐻1𝑛 = 𝜇2 𝐻2𝑛
ර 𝐇 ∙ 𝑑𝒍 = 𝐼
𝐶
∆ℎ → 0,
න 𝐇 ∙ 𝑑𝒍 + න 𝐇 ∙ 𝑑𝒍 = 𝐽𝑠 ∆𝑤
𝑎𝑏
𝐇1 ∙ (Δ𝑤𝐚𝑎𝑏 ) + 𝐇2 ∙ −Δ𝑤𝐚𝑎𝑏 = 𝐽𝑠 ∆𝑤
𝑐𝑑
𝐻1𝑡 − 𝐻2𝑡 = 𝐽𝑠
𝐚𝑛2 × 𝐇1 − 𝐇2 = 𝐉𝑠
Chapter 6 Static Magnetic Fields
Example 6-12
Two magnetic media with permeabilities 𝜇1 and 𝜇2 have a common boundary. The
magnetic field intensity in medium 1 at the point 𝑃1 has a magnitude 𝐻1 and makes an
angle 𝛼1 with the normal. Determine the magnitude and the direction of the magnetic
field intensity at point 𝑃2 in medium 2.
𝐚𝑛2
𝑃1
𝑃2
𝜇1
𝜇2
𝐇2
𝛼1
𝐇1
𝛼2
Chapter 6 Static Magnetic Fields
Inductances and Inductors
• Two magnetically coupled loops
If a current 𝐼1 flows in 𝐶1 , a magnetic field 𝐁1
will be created. Some of the magnetic flux due
to 𝐁1 will link with 𝐶2 (will pass through the
surface 𝑆2 bounded by 𝐶2)
• Mutual flux (상호자속)
Φ12 = න 𝐁1 ∙ 𝑑𝒔2 (Wb)
𝑆2
• Biot-Savart law
𝜇0 𝐼
𝑑𝒍′ × 𝐚𝑅
𝐁=
ර
4𝜋 𝐶 ′ 𝑅2
𝐁∝𝐼
Φ12 = න 𝐁1 ∙ 𝑑𝒔2 ∝ 𝐼1
𝑆2
Φ12 = 𝐿12 𝐼1
Chapter 6 Static Magnetic Fields
𝐿12 (H): mutual inductance
between loops 𝐶1 and 𝐶2
Inductances and Inductors
• Two magnetically coupled loops
Φ12 = 𝐿12 𝐼1
𝐿12 (H): mutual inductance between loops
𝐶1 and 𝐶2
In case 𝐶2 has 𝑁2 turns,
flux linkage: Λ12 = 𝑁2 Φ12 = 𝐿12 𝐼1 (Wb)
Λ12
𝐿12 =
(H)
𝐼1
• Mutual inductance between loops 𝐶1 and 𝐶2
𝐿12 =
Λ12 𝑁2 Φ12 𝑁2
=
=
න 𝐁 ∙ 𝑑𝒔2 (H)
𝐼1
𝐼1
𝐼1 𝑆2 1
Chapter 6 Static Magnetic Fields
Inductances and Inductors
• Two magnetically coupled loops
Some of the magnetic flux produced by 𝐼1 links
only with 𝐶1 itself, and not with 𝐶2
• The total flux linkage with 𝐶1 caused by 𝐼1
Λ11 = 𝑁1 Φ11
• Self inductance of loops 𝐶1
𝐿11 =
Λ11 𝑁1 Φ11 𝑁1
=
=
න 𝐁 ∙ 𝑑𝒔1 (H)
𝐼1
𝐼1
𝐼1 𝑆1 1
• A conductor arranged in an appropriate shape to supply a certain amount of selfinductance is called an inductor
• Just as a capacitor can store electric energy, an inductor can storage magnetic
energy
Chapter 6 Static Magnetic Fields
Procedure for determining the self-inductance
Procedure for determining the self-inductance 𝐿11
1. Choose an appropriate coordinate system for the given geometry
2. Assume a current 𝐼 in the conducting wire.
3. Find 𝐁 from 𝐼 by Ampere’s circuital law,
𝜇0 𝐼
𝑑𝒍′ × 𝐚𝑅
ර 𝐁 ∙ 𝑑𝒍 = 𝜇0 𝐼,
𝐁=
ර
4𝜋
𝑅2
𝐶
𝐶′
4. Find the flux linking with each turn, Φ, from 𝐁 by integration
Φ = න 𝐁 ∙ 𝑑𝐬
𝑆
where 𝑆 is the area over which 𝐁 exists and links with the assumed
current.
5. Find the flux linkage 𝛬 by multiplying Φ by the number of turns.
6. Find 𝐿 by taking the ratio 𝐿 = 𝛬/𝐼
Procedure for determining the mutual inductance 𝐿12
1. Choose an appropriate coordinate system for the given geometry
2. Assume a current 𝐼1 3. Find 𝐁1 from 𝐼1
4. Find Φ12 , Φ12 = ‫𝐁 𝑆׬‬1 ∙ 𝑑𝐬2
5. Find the flux linkage 𝛬12 = 𝑁2 Φ12
2
6. 6Find
𝐿 by
taking Fields
the ratio 𝐿12 = 𝛬12 /𝐼1
Chapter
Static
Magnetic
Example 6-14
Assume that 𝑁 turns of wire are tightly wound on a toroidal frame of a rectangular
cross section with dimensions. Then, assuming the permeability of the medium to be
𝜇0 , find the self-inductance of the toroidal coil.
Chapter 6 Static Magnetic Fields
Example 6-15
Find the inductance per unit length of a very long solenoid with air core having 𝑛 turns
per unit length.
𝐵 = 𝜇0 𝑛𝐼
Φ = 𝐵𝑆 = 𝜇0 𝑛𝐼𝑆
𝛬 = 𝑛Φ = 𝜇0 𝑛2 𝐼𝑆
𝛬
𝐿 = = 𝜇0 𝑛2 𝑆
𝐼
Chapter 6 Static Magnetic Fields