Understanding Steady Magnetic Fields and Their Laws
Explore the concept of steady magnetic fields, the generation of magnetic fields by charged particles, the prediction of magnetic fields using Biot-Savart law, and the application of Ampere's law in differential form. Discover how current distributions impact magnetic fields and learn about integral formulas like Biot-Savart and Coulomb's law.
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Presentation Transcript
1 Chapter 7. Steady magnetic field EMLAB
Generation of magnetic field 2 A charged particle in motion generates magnetic field nearby. n : number of charges in unit volume v : velocity of charges A : cross-section area of the wire e: amount of charges in an electron ? = ???? In the same way, currents generate magnetic field nearby. EMLAB
Prediction of magnetic field : Biot-Savart law 3 ?? =??? ? 4??2 Current segment ??? ? ? = ? ? Direction of H-field ? The magnetic field can be predicted by Biot-Savart s law with known current distribution. EMLAB
Biot-Savart law : integral form 4 ??? ? 4??2 Line current ??? ? = ; ? ? ? ??? 4??2 surface current = ; ? ? ??? 4??2 ? = ; Volume current ? ??? ??? EMLAB
Prediction of magnetic field 5 In practice, current distributions on objects are unknown and are apt to change with external magnetic field. EMLAB
EM problems 6 ?2? ??2 1 ?2? ??2= 0 ?2 Current/charge distributions are known only at the point of sources. If surface currents are known everywhere, integral formulas such as Biot-Savart s and Coulomb s law are useful. EMLAB
8.3 Curl (Amperes law in a differential form) 7 ? ?? = ? ? = ? ? From the integral form, we will derive the differential form of Ampere s law. ? ?? ? ?? ? ?? Line integrals from these currents add up to zero. ?? ? Line integrals over a closed path is equal to the sum of line integrals over infinitesimally small loops. EMLAB
Amperes law over an infinitesimally small closed path 8 ? (?,?,?) ? ? 2 3 4 1 ? ?? = ???? + ???? + ???? + ???? ?? 1 2 3 4 2 3 4 1 ?? ? + ? ?? ? + ? ?? ? ? ?? ? ? = ?? + ?? + ?? + ?? 2 2 2 2 1 2 3 4 = ?? ? + ? ?? ? ? ? ?? ? ? ?? ? + ? ? 2 2 2 2 ??? ?? ??? = ? ? = ?? ( ?) ? ? ? ?? ??? ?? ? ? ??? ?? ??? ?? = Lim ? 0 = ? ?= ?? ?? ??? ?? ??? ( ?)? = ?? ?? EMLAB
9 ??? ?? ??? ? ?? = ?? = ?? ; Ampere s law in a differential form ?? ? In the same way, Jx, Jy can be derived and are shown as follows. ??? ?? ??? ??? ?? ??? ??? ?? ??? = ?? ?? ??? ?? ??? = ???+ ???+ ???= ? ??? ?? ??? ??? ?? ??? + ? ? + ? = ?? ?? ?? ?? ?? = ?? ?? Because the expression for line integrals are different for each coordinate systems, x operator is used. ??? ?? ??? ??? ?? ??? ??? ?? ??? + ? ? = ? + ? = ? ?? ?? ?? EMLAB
Curl operator in other orthogonal coordinates 10 Rectangular coordinate : ??? ?? ??? ??? ?? ??? ??? ?? ??? + ? ? = ? + ? = ? ?? ?? ?? Circular cylindrical coordinate : ??? ?? ??? ??? ?? ??? ?(???) ?? ??? ?? 1 ? + ?1 ? = ? + ? ?? ?? ? Spherical coordinate : ?(??sin?) ?? ??? ?? ?(???) 1 ??? ?? + ?1 1 + ?1 ?(???) ?? ??? ?? ? = ? ?sin? ? sin? ?? ? EMLAB
11 8.4 Stokes theorem Line integrals can be transformed to surface integrals by curl operations. ? ?? = ? ? = ? ? ? ?? = ( ?) ?? ? ? ? ?? = ? ? ? ?? ? ?? = ? ?? ( ?) ? ?? ? ? ? EMLAB
Example 8.3 12 Evaluate the line integral over the specified path. ? ?? ? ? = ?6?sin? + ?18?sin?cos? ? (1) Explicit line integral ?? = ??? + ???? + ??sin? ?? Since r =4 on the contour, r component becomes zero. ? ?? = 18?2sin2?cos??? Over the path 1 and 3, the integrand becomes zero in that d = 0. 0.3? 0.3? 18?2sin2?cos??? = 18 42sin20.1? ? ?? = cos??? =22.2 ? 0 0 EMLAB
13 (2) Stokes theorem ?sin?(36?sin?cos? cos?) ? +1 1 1 sin?6?cos? 36?sin?cos? ? ? = ? ?? = ??2sin????? 0.3? 0.1? (36?2sin?cos? cos?)???? = 22.2 ? ?? = ? 0 0 EMLAB
Important vector identities 14 1) ( : arbitrary scalar function) ? = 0 2 1 2?? ???? + 1?? ???? ? ?? = ? ?? = ? ?? + ? ?? = ? ? 1 2 1 2 = ?(?2) ?(?1) + ?(?1) ?(?2) = 0 2) (A: arbitrary vector function) ( ?) = 0 ( ?)?? = ( ?) ?? = ? ?? + ? ?? = 0 ? ? ?1 ?1 ?1 ?1 EMLAB
8.5 Magnetic flux and magnetic flux density 15 Magnetic flux = ? ?? ? EMLAB
Magnetic field and magnetic flux density 16 With the same incident magnetic field, magnitudes of B changes with a magnetic permeability of surrounding medium. ??>> 1 ? = ?? = ???0? Magnetic flux density Permeability ?0= 4? 10 7[H/m] EMLAB
DC solution of Maxwells equations 17 (1) Scalar potential ? = ?? ??= 0 Because the curl of E is always equal to zero, E can be represented by a gradient of an arbitrary scalar function. ? = ? +?? ??= ? ? = ? ? = 0 ? ? ? = ( ?) = 0 ? ?? = 0 ? ?? = ? ?? = 0 ? ? ? ? = 0 ? = ? EMLAB
18 (2) Vector potential ? = ?? ??= 0 Because the divergence of B is always equal to zero, B can be represented by a curl of an arbitrary vector function. ? = ? +?? ??= ? ? = ? ? = 0 ? ? ? = ? = 0 EMLAB
Potential equations 19 2? = ? ? = 0 ? = ? (1) ? ? = ? ? ? = ? ? = ? 1 ? ? = ? ? = ?? (2) ? = 0 ? = ? ? = ? 2? 2? = ?? Steady state Maxwell s equations ( ? = 0) 2? = ? ??? 4?? ? =1 ? ? ? A solution of Poisson equation for free space. 2? = ?? ??? 4?? ? = ? ? EMLAB
Laplacian operator in other orthogonal coordinates 20 The differential equation for source-free region becomes a Laplace equation. (rectangular coordinate) 2? =?2? ??2+?2? ??2+?2? ??2 (cylindrical coordinate) ?2? ??2+?2? 2? =1 ? ?? ??? ?? +1 ?2 ??2 ? (spherical coordinate) ?2? ??2 1 ?2 ? ?? ?2?? 1 ? sin??? 1 2? = + + ?2sin? ?2sin2? ?? ?? ?? EMLAB
Solution of a Poisson equation for a point charge 21 2? = ? Free space ? ?2? ??2 1 ?2 ? ?? ?2?? 1 ? sin??? 1 2? = + + ?2sin? ?2sin2? ?? ?? ?? Suppose there is only one point charge in the space, the potential field will have spherically symmetric distribution. 1 ?2 ? ?? ?2?? = ?(?) ?? G : Green s function (1) Homogeneous solution 1 ?2 ? ?? ?2?? ? =?1 ?2?? = 0 ??= ?1 ?+ ?2 ?? EMLAB
22 ?2= 0 ?( ) = 0 To find the value of ?1, volume integrals of both sides are needed. ?1 ?2 2? ?? = ?2sin????? = 4??1= 1 ? ?? = ? ? ? 1 ?1= 4? 1 ?(?,? ) = 4?? If the position of the source is changed to r , the expression becomes, 1 ? = ? ? ? = 4??, EMLAB
23 The charge density ?(?) can be expressed as a weighted integral of delta function, ? ? ? ? ? ?? ? ? = ? The voltage ?(?) can also be expressed as a weighted integral of Green function, ? 2? = ? ?(? ) ? ?(? ) ? 1 ?(?,? )?? = 4? ? ? ?? ? = ? ? ? ?(? ) ? ? ? ? ? ?,? ?? = ?(?) 2? = 2?(?,? )?? = ? ? ? For proof, a point charge with q coulomb at origin yields Coulomb s law. ?(?) = ??(?) ? =?? ? +1 ?? ?? 1 ?? ?? ? ? + ? ? ?? ? ?sin? ? = ? = 4???2 ? ? ? = ? ? ? = 4???2, EMLAB
Equation for the vector potential 24 2? = ?? 2??= ??? 2??= ??? 2??= ??? 2? = ?(? ? ) ??? 4??= ? (?? ? + ?? ? + ?? ?)?? ? = ? 4?? ? ? ??? 4??= ??? 4?? ? =1 =1 1 ? ??? +1 ? (??? ) ? ? = 1 4? ? 4? ? ? ? ??? ? ?2 = EMLAB
25 For an infinitesimally small current segment with current I and length dL at origin, ?(?) = ??(? ? ) ? = ???? ? = ? ? 4??, ??? ? ?2 ? =1 ??? 4?? 1 1 ? ??? +1 1 ? ? = = 4? ? (???) = 4? EMLAB
Example 26 Calculate the magnetic field inside the coaxial cable sheath carrying current I. ?2?? ??2+?2?? 2??=1 ? ?? ???? ?? +1 ??2= ??? ?2 ? ? ? ???? ??= ?1 1 ? ? ?? ???? ?? ??= ?1ln? + ?2 = 0 ??(?) = ?1ln? + ?2= 0 ? =1 ?? = ?1 ? = ? = ?1 ?? ? ? ? 2? ? ?? = ? =1 ?1 ?? ? ??? ? = 2??1 ?? = ? ? 0 ? ? ? +? ? = 2??1 ?1= ?? ? 2? ? ? 2?? ? ? = EMLAB
