CH.8 Time Varying Field
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CHAPTER (8) TIME VARYING ELECTRIC AND MAGNETIC FIELDS 47 AAST Prof. Darwish Faraday’s law: U Due to the first experiment of Faraday, we can say that a time-varying magnetic field produces an electromotive force (emf) which may establish a current in a suitable closed circuit. Faraday’s law is stated as ๐๐๐ (๐) ๐๐๐ = − ๐๐ ๐๐ (๐) [๐ฝ] (๐) A non zero value of ๏ฟฝ ๐ ๏ฟฝ may result from any of the ๐๐ following situations: • A time-changing flux linking a stationary closed path. • Relative motion between a steady flux and a closed path. • A combination of the two previous situations. Note: 1- The minus sign is an indication that the emf is in such a direction as to produce a current whose flux, if added to the original flux, would reduce the magnitude of the emf. This statement that the induced voltage acts to produce an opposing flux is known as Lenz’s law. U 2- For N – turns filamentary conductor closed path ๐๐๐ (๐) ๐๐๐ = − ๐ต ๐๐ (๐) 48 AAST Prof. Darwish The emf is considered as the voltage about a specific closed path, and it is defined as ๐๐๐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ ๏ฟฝโ ๐. ๐. ๐ = ๏ฟฝ ๐ฌ . ๐ ๐ = − (๐) ๐๐ In electro-statics, equation (3) must lead to zero potential difference about a closed path. In time varying fields, the line integral leads to an emf or a potential difference. Equation (3) can be written as ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ = − ๏ฟฝ๏ฟฝโ . ๐ ๐ ๐. ๐. ๐ = ๏ฟฝ ๐ฌ ๐๐๐ ๐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ (๐) ๏ฟฝ๏ฟฝโ . ๐ ๐ = − ๏ฟฝ๐ฉ ๐๐ ๐๐ In general, Faraday’s law manifests itself in either or both a stationary circuit linked by a time varying magnetic flux, such as a transformer, or the magnetic flux may be static, but the circuit is moving relative to the flux in such a way as to produce a time varying flux enclosed by the circuit. A rotating machine generates an emf by the latter mechanism. ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐ ๐๐๐๐๐๐๐๐ Applying Stock’s theorem on equation (4) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ = ๏ฟฝ ๐๐๐ฌ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ = − ๏ฟฝ๏ฟฝโ . ๐ ๐ ๏ฟฝ๏ฟฝโ . ๐ ๐ ๐. ๐. ๐ = ๏ฟฝ ๐ฌ ๐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๏ฟฝโ . ๐ ๐ ๏ฟฝ๐ฉ ๐๐ So, ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐ ๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๏ฟฝ๏ฟฝโ ๐๐ฉ ๏ฟฝ๏ฟฝโ = − ๐๐๐ฌ (๐) ๐๐ 49 AAST Prof. Darwish And, ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๏ฟฝ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๐ฌ ๐ ๐ = ๏ฟฝ ๐๐๐ฌ ๐ ๐ = − ๐ ๏ฟฝ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๐ฉ ๐ ๐ ๐๐ Displacement current density ๐ฑโ๐ (๐) U It is shown from ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐๐๐๐๐๐๐๐๐ in differential form (5) ๏ฟฝ๏ฟฝโ = − ๐๐๐ฌ ๏ฟฝ๏ฟฝโ ๐๐ฉ ๐๐ that a time varying magnetic field produces an electric ๏ฟฝ๏ฟฝโ. From the definition of the curl, the electric field field ๐ฌ ๏ฟฝ๏ฟฝโ has the special property of circulation; its line integral ๐ฌ about a general closed path is not zero. Now we will consider the time varying electric field. From Curl law as it applies to steady magnetic fields, where ๏ฟฝ๏ฟฝโ = ๐ฑโ๐ ๐ ๐ ๏ฟฝ๐ฏ (๐) ๐๐๐ค๐ข๐ง๐ ๐ญ๐ก๐ ๐๐ข๐ฏ๐๐ซ๐ ๐๐ง๐๐ ๐จ๐ ๐๐จ๐ญ๐ก ๐ฌ๐ข๐๐๐ฌ ๐จ๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง (๐), ๏ฟฝ๏ฟฝโ = ๐ . ๐ฑโ๐ ๐ . ๐ ๐ ๏ฟฝ๐ฏ (๐) Mathematically, the left hand side of equation (8) ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐), while the right hand side of equals zero (๐. ๐ ๐ ๐ฏ equation (8), from continuity equation (conservation of charge) equals to ๐๐๐ โ ๐ . ๐ฑ๐ = − (๐) ๐๐ AAST Prof. Darwish 50 So, equation (8) becomes ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ . ๐ฑโ๐ = − ๐ .๐ ๐ ๐ฏ ๐๐๐ ๐๐ (๐๐) Since in equation (10), the L.H.S = 0, but the R.H.S ≠ 0, so there is a missing term that appear in time varying field as that ๏ฟฝ๏ฟฝโ ๐๐ฉ appear in first Maxwell’s equation๏ฟฝ− ๏ฟฝ, in order to fulfill that ๐๐ R.H.S = L.H.S. This missing term that must be added to the ๐๐ L.H.S is ๏ฟฝ+ ๐ ๏ฟฝ. So ๐๐ ๏ฟฝ๏ฟฝโ = − ๐ . ๐ ๐ ๏ฟฝ๐ฏ ๐๐๐ ๐๐๐ + ๐๐ ๐๐ (๐๐) ๏ฟฝ๏ฟฝโ = ๐๐ , equation (11) Using the divergence law, ๐ . ๐ becomes ๏ฟฝ๏ฟฝโ = − ๐ . ๐ ๐ ๏ฟฝ๐ฏ So, ๐๐๐ ๐ ๏ฟฝ๏ฟฝโ ๏ฟฝ + ๏ฟฝ๐ . ๐ ๐๐ ๐๐ ๏ฟฝโ ๐ ๏ฟฝ๐ ๐๐๐ ๏ฟฝ๏ฟฝ๏ฟฝโ − ๐ . ๏ฟฝ๐ ๐ ๐ฏ ๏ฟฝ= − = ๐ . ๐ฑโ๐ ๐๐ ๐๐ ๏ฟฝโ ๐ ๏ฟฝ๐ ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฑโ๐ + ๐๐๐ฏ ๐๐ ๏ฟฝ๏ฟฝโ = ๐ฑโ๐ + ๐ฑโ๐ ๐ ๐ ๏ฟฝ๐ฏ Jc , Jd = 0 Jd , Jc = 0 E ๐ฑโ๐ ๐๐ ๐๐๐ ๐๐๐ ๐๐๐๐๐๐ ๐๐๐๐๐๐๐ ๐ ๐๐๐๐๐๐ ๐ฑโ๐ ๐๐ ๐๐๐ ๐ ๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐ ๐ ๐๐๐๐๐๐ AAST Prof. Darwish A 51 So, ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐ ๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฑโ๐ + ๐ฑโ๐ ๐๐๐ฏ (๐๐) And, ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๏ฟฝโ ๐ ๏ฟฝ๐ ๏ฟฝ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ ๏ฟฝ๐ฏ ๐ ๐ = ๏ฟฝ ๐ ๐ ๏ฟฝ๐ฏ ๐ ๐ = ๏ฟฝ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๐ ๐ + ๏ฟฝ ๐ฑโ๐ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๐ ๐ (๐๐) ๐๐ 3rd Maxwell's equation: PU UP rd 0 3 Maxwell's equation in Integral form ⋅ = → B d S ∫ Applying the divergence theorem ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝโ ๐ ๐ = ๐ ๏ฟฝ ๏ฟฝ๐ฉ ๐ ๐ = ๏ฟฝ ๐ . ๏ฟฝ๐ฉ So, ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐ ๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๏ฟฝโ = ๐ ๐ . ๏ฟฝ๐ฉ (๐๐) And, ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝโ ๐ ๐ = ๐ ๏ฟฝ ๏ฟฝ๐ฉ ๐ ๐ = ๏ฟฝ ๐ . ๏ฟฝ๐ฉ (๐๐) 52 AAST Prof. Darwish 4th Maxwell’s equation: PU UP Applying the divergence theorem on the electric flux ๏ฟฝ๏ฟฝโ, we have density ๐ซ ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝโ ๐ ๐ = ๏ฟฝ ๐๐ ๐ ๐ ๏ฟฝ ๏ฟฝ๐ซ ๐ ๐ = ๐ธ๐๐ = ๏ฟฝ ๐ . ๏ฟฝ๐ซ So, ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐ ๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๏ฟฝ๏ฟฝโ = ๐๐ ๐ .๐ซ (๐๐) And, ๐๐๐ ๐ด๐๐๐๐๐๐′ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝโ ๐ ๐ = ๏ฟฝ ๐๐ ๐ ๐ ๏ฟฝ ๏ฟฝ๐ซ ๐ ๐ = ๐ธ๐๐ = ๏ฟฝ ๐ . ๏ฟฝ๐ซ Summary: (๐๐) U Maxwell's equation Differential form ๏ฟฝโ = − ๐๐๐ฌ ๏ฟฝ๏ฟฝโ ๐๐ฉ ๐๐ Integral form ๏ฟฝ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ ๏ฟฝ๐ฌโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๐ ๐ = ๏ฟฝ ๐๐๐ฌ ๐ ๐ = − ๏ฟฝ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ ๏ฟฝ๐ฏ ๐ ๐ = ๏ฟฝ ๐ ๐ ๏ฟฝ๐ฏ ๐ ๐ = ๏ฟฝ ๏ฟฝ๏ฟฝโ = ๐ฑโ๐ + ๐ฑโ๐ ๐ ๐ ๏ฟฝ๐ฏ ๏ฟฝ๏ฟฝโ = ๐๐ ๐ .๐ซ ๐ ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ ๏ฟฝ๐ฉ ๐ ๐ ๐๐ ๏ฟฝโ ๐๐ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๐ ๐ + ๏ฟฝ ๐ฑโ๐ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๐ ๐ ๐๐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ธ๐๐ = ๏ฟฝ ๐ . ๐ซ ๏ฟฝ๏ฟฝโ . ๐ ๐ ๏ฟฝ๏ฟฝโ ๐ ๐ = ๏ฟฝ ๐๐ ๐ ๐ ๏ฟฝ๐ซ ๏ฟฝ๏ฟฝโ = ๐ ๐ .๐ฉ ๏ฟฝโ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝโ ๐ ๐ = ๐ ๏ฟฝ ๏ฟฝ๐ฉ ๐ ๐ = ๏ฟฝ ๐ . ๏ฟฝ๐ฉ 53 AAST Prof. Darwish Jd = ε ∂E ∂D = 0 ∂t ∂t ∇ ⋅ Jc + ∂ρ v = 0 ∂t Id = J d ⋅A 54 AAST Prof. Darwish Example: U A coaxial cable of ε r = 4, the inner conductor radius = 1mm and the outer conductor radius = 5mm. Find the displacement current between the two conductors per meter if the voltage difference equal 100 cos (12π.106t) volt. R R P P Solution: U 1- Using circuit concept U For a coaxial cable : 2πε 2π ∗ 8.85 ∗ 10 −12 * 4 = C= ๏ฃซb๏ฃถ ๏ฃซ5๏ฃถ ln๏ฃฌ ๏ฃท ln๏ฃฌ ๏ฃท ๏ฃญa๏ฃธ ๏ฃญ1๏ฃธ Ιd = C 1 5 dV dt dV = 100 sin(12π ∗ 10 6 t ) ∗ 12π ∗ 10 6 dt 2π ∗ 8.85 ∗ 10−12 * 4 100 sin(12π ∗ 106 t ) ∗ 12π ∗ 106 A/m Ιd = ๏ฃซ5๏ฃถ ln๏ฃฌ ๏ฃท ๏ฃญ1๏ฃธ 2- Using electro-magnetic field concept U The displacement current density ๏ฟฝ๏ฟฝโ ๐๐ซ from ๐ฑโ๐ = , after determining ๐๐ Gauss’s law as follows: ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ธ๐๐ ๏ฟฝโ . ๐ ๐ 1) โฎ ๏ฟฝ๐ซ 2) ๐๐๐๐๐๐ ๐๐ ๐ฎ๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐ 3) ๐ธ๐๐ = ๐๐ ๐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐๐ ๐๐ ๐ ๐ซ ๏ฟฝโ . ๐ ๐ 4) โฎ ๏ฟฝ๐ซ 5) ๐๐ ๐๐ ๐ ๐ซ = ๐๐ ๐ ๐ ๏ฟฝโ = ๐ ๏ฟฝ๐๐ ๐ ๐ช๏ฟฝ ๐ 6) ๏ฟฝ๐ซ ๐๐ ๐ ๐ AAST ๏ฟฝโ = ๐ ๏ฟฝ ๐๐ 7) ๏ฟฝ๐ฌ ๐ ๐๐ ๐๐ ๐บ๐ ๐บ๐ ๐๐ ๐ต๏ฟฝ ๐ช Prof. Darwish ๐ฑโ๐ can be determined ๏ฟฝ๐ซ ๏ฟฝโ by applying the 55 In order to determine ๐๐ in terms of the potential difference V, the potential V is determined as ๐ ๐ฝ๐๐ = ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๏ฟฝโ . ๐ ๐ ๐ฝ๐๐ = − ๏ฟฝ ๐ฌ ๐ ๐ ๐๐ ๏ฟฝ ๐๐ − ∫๐ ๐ ๐๐ ๐บ๐ ๐บ๐ ๐๐ ๏ฟฝ๐๐ ๐ ๐๐ .๐ ๐๐ ๐๐๐ ๐๐จ๐ฌ (๐๐๐ . ๐๐ ๐) = ๐๐ (๐) ๐๐ ๐บ๐ ๐บ๐ ๐ ๐๐๐ ๐ ๐บ๐ ๐บ๐ ๐๐จ๐ฌ (๐๐๐ . ๐๐๐ ๐) ๐๐ = ๐๐ (๐) So, ๏ฟฝ๏ฟฝโ = ๐ ๏ฟฝ ๐๐ ๐ซ ๐๐ ๐๐๐ ๐ ๐บ๐ ๐บ๐ ๐๐๐ (๐๐๐ . ๐๐๐ ๐) ๏ฟฝ ๐๐ = ๐ ๐๐ ๐๐ ๐๐ ๐๐ ๐๐ (๐) ๏ฟฝ๏ฟฝโ ๐๐๐ ๐ ๐บ๐ ๐บ๐ ๏ฟฝ๐๐๐ . ๐๐๐ ๏ฟฝ๐๐๐ (๐๐๐ . ๐๐๐ ๐) ๐๐ซ ๏ฟฝ ๐๐ =−๐ ๐ฑโ๐ = ๐๐ ๐๐ ๐๐ (๐) ๐๐ ๐ ๐ ๐ฐ๐ = ๏ฟฝ ๐ฑโ๐ . ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๐๐ ๐ฐ๐ = ๏ฟฝ ๐ ๐ ๏ฟฝ ๐๐ ๏ฟฝ −๐ ๐ ๐๐๐ ๐ ๐บ๐ ๐บ๐ (๐๐๐ . ๐๐๐ )๐๐๐ (๐๐๐ . ๐๐๐ ๐) ๏ฟฝ ๐๐ ๐๐ ๐ ๐๐ ๐ .๐ ๐๐ ๐๐ ๐๐ (๐) ๐๐๐ ๐ ๐บ๐ ๐บ๐ (๐๐๐ . ๐๐๐ )๐๐๐ (๐๐๐ . ๐๐๐ ๐) ๐ ๐ฐ๐ = ๐๐ (๐) ๐๐๐ ๐ ๐บ๐ ๐บ๐ (๐๐๐ . ๐๐๐ )๐๐๐ (๐๐๐ . ๐๐๐ ๐) ๐ฐ๐ = ๐จ/๐ ๐ ๐๐ (๐) 56 AAST Prof. Darwish Inductance: U When dealing with time – varying current and thus time – varying magnetic fields, the inductance of the inductor as a circuit concept becomes quite important and is defined as I I N ๐ณ = ๐ฒ ๐ฐ [๐ฏ] φm Where: ๐ฒ (lambda) is the total flux linkage of the inductors and I is the current flowing in the inductor. A flux linkage of one exists when 1 Wb of flux links one turn of the inductor. So, if all the flux link all the turns, the total flux linkage is equal to [๐๐ ๐ญ๐ฎ๐ซ๐ง๐ฌ] ๐ฒ = ๐ ๐๐ฆ Where, ๐๐ฆ is the total flux produced by the inductor L is the self inductance If the coil has N turns: ๐ฒ ๐ ๐๐ฆ ๐ณ = = ๐ฐ ๐ฐ 57 AAST Prof. Darwish Example: U Find the self inductance of the solenoid shown in figure. Solution: U ๐ฒ ๐ ๐๐ฆ = ๐ฐ ๐ฐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ ๐ = ๐ ๏ฟฝ๐๐๐ ๏ฟฝ ๏ฟฝ๏ฟฝโ . ๐๐ฌ ๐๐ฆ = โฌ ๐ ๐ณ = ๏ฟฝ๏ฟฝโ = ๐๐จ ๐๐ซ ๐ ๏ฟฝ๏ฟฝโ ๐ ๏ฟฝ๏ฟฝโ can be determined by The magnetic field intensity ๐ applying Ampere’s circuital law as follows: ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฐ๐๐ ๏ฟฝ๏ฟฝโ . ๐๐ฅ (1) โฎ ๐ (2) ๐ช๐๐๐๐๐ ๐๐ ๐จ๐๐๐๐๐๐๐ ๐๐๐๐ (3) ๐ฐ๐๐ = ๐ต ๐ฐ ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฏ ๐ ๏ฟฝ๏ฟฝโ . ๐๐ฅ (4) โฎ ๐ ๏ฟฝ๏ฟฝโ = ๐ต ๐ฐ ๐๏ฟฝ๐ณ (5) ๐ ๐ So, ๏ฟฝ๐ ๏ฟฝโ = ๐๐จ ๐๐ซ ๐ ๏ฟฝ๏ฟฝโ = ๐๐จ ๐๐ซ ๐ต ๐ฐ ๐๏ฟฝ๐ณ ๐ ๐ต๐ฐ ๐๐ฆ = ๐ ๏ฟฝ๐๐ ๏ฟฝ = ๐๐จ ๐๐ซ ๏ฟฝ๐๐๐ ๏ฟฝ ๐ ๐ ๐ณ = ๐ฒ ๐ฐ = ๐ ๐๐ฆ ๐ฐ = ๐ ๐จ ๐๐ซ ๐ต๐ ๐ ๏ฟฝ๐๐๐ ๏ฟฝ [๐] 58 AAST Prof. Darwish Example: U Find the self inductance of the torrid shown. Solution: ๐ฒ ๐ ๐๐ฆ ๐ณ = = ๐ฐ ๐ฐ ๏ฟฝ๏ฟฝโ can be The magnetic field intensity ๐ determined by applying Ampere’s circuital law as follows: ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฐ๐๐ ๏ฟฝ๏ฟฝโ . ๐๐ฅ (1) โฎ ๐ (2) ๐ช๐๐๐๐๐ ๐๐ ๐จ๐๐๐๐๐๐๐ ๐๐๐๐ (3) ๐ฐ๐๐ = ๐ต ๐ฐ ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฏ ๐๐ ๐๐ ๏ฟฝ๏ฟฝโ . ๐๐ฅ (4) โฎ ๐ ๏ฟฝ๏ฟฝโ = ๐ต ๐ฐ ๐๏ฟฝ๐ (5) ๐ U So, ๐๐ ๐๐ ๏ฟฝ๐ ๏ฟฝโ = ๐๐จ ๐๐ซ ๐ ๏ฟฝ๏ฟฝโ = ๐๐จ ๐๐ซ ๐ต๐ฐ ๐๐ ๐๐ ๐๏ฟฝ๐ Average path radius ๐๐|๐๐๐๐๐๐๐ = ๐ = ๏ฟฝ๏ฟฝโ๐๐ = ๐ = ๐ ๏ฟฝ๐ ๐ฉ ๐ฆ ๐+๐ ๐ ๐๐ฆ = ๐๐๐ฏ ๐ ๐−๐ ๐ ๐ = ๐๏ฟฝ ๏ฟฝ ๐ ๐ต๐ฐ ๐−๐ ๐ ๐๐ฆ = ๐๐จ ๐๐ซ ๐๏ฟฝ ๏ฟฝ ๐๐ ๐ ๐ ๐ต ๐ฐ(๐ − ๐)๐ ๐๐ฆ = ๐๐จ ๐๐ซ ๐(๐ + ๐) AAST ๐ต๐ (๐ − ๐)๐ ๐ฒ ๐ ๐๐ฆ ๐ณ = = = ๐ ๐จ ๐๐ซ ๐ฏ๐๐๐๐๐ ( ) ๐ฐ ๐ ๐+๐ ๐ฐ Prof. Darwish 59 Example: U Find the self inductance per unit length of a T.L. (coaxial cable). Solution: U ๐ฒ ๐ ๐๐ฆ ๐ณ = = ๐ฐ ๐ฐ ๐ต=๐ I I a b L ๏ฟฝ๏ฟฝโ can be determined by The magnetic field intensity ๐ applying Ampere’s circuital law as follows: ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฐ๐๐ ๏ฟฝ๏ฟฝโ . ๐๐ฅ (1) โฎ ๐ (2) ๐ช๐๐๐๐๐ ๐๐ ๐จ๐๐๐๐๐๐๐ ๐๐๐๐ (3) ๐ฐ๐๐ = ๐ฐ ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฏ ๐๐ ๐๐ ๏ฟฝ๏ฟฝโ . ๐๐ฅ (4) โฎ ๐ ๏ฟฝ๏ฟฝโ = ๐ฐ ๐๏ฟฝ๐ (5) ๐ ๐๐ ๐๐ So, ๏ฟฝ๐ ๏ฟฝโ = ๐๐จ ๐๐ซ ๐ ๏ฟฝ๏ฟฝโ = ๐๐จ ๐๐ซ ๐ฐ ๐๏ฟฝ๐ ๐๐ ๐๐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ = ∫๐ฅ ∫๐ ๐๐จ ๐๐ซ ๐ฐ ๐๏ฟฝ๐ . ๐๏ฟฝ๐ ๐๐๐ ๐ ๐ ๏ฟฝ๏ฟฝโ . ๐๐ฌ ๐๐ฆ = โฌ ๐ ๐ ๐ ๐๐ ๐ ๐๐ฆ = AAST ๐ณ = ๐ฒ ๐ฐ ๐๐จ ๐๐ซ ๐ฐ๐ = ๐๐ ๐๐ฆ ๐ฐ ๐ฅ๐ง(๐๐ )๐๐ = = ๐๐จ ๐๐ซ ๐ ๐๐ Prof. Darwish ๐๐จ ๐๐ซ ๐ฐ๐ ๐๐ ๐ ๐ฅ๐ง ๏ฟฝ ๏ฟฝ ๐ ๐ ๐ ๐ฅ๐ง ๏ฟฝ ๏ฟฝ ๐ [๐] 60 Example: ๐ ๐๐จ ๐ ๐ซ ๐ = ๐ฅ๐ง ๏ฟฝ ๏ฟฝ ๐ ๐๐ ๐ [๐/๐ฆ] U Find the inductance per meter for two parallel wires. Solution: U I d a I ๐ฒ ๐ ๐๐ฆ ๐ณ = = ๐ฐ ๐ฐ ๐ต=๐ ๏ฟฝ๏ฟฝโ can be determined by The magnetic field intensity ๐ applying Ampere’s circuital law as follows: ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฐ๐๐ ๏ฟฝ๏ฟฝโ . ๐๐ฅ (1) โฎ ๐ (2) ๐ช๐๐๐๐๐ ๐๐ ๐จ๐๐๐๐๐๐๐ ๐๐๐๐ (3) ๐ฐ๐๐ = ๐ฐ ๏ฟฝ๏ฟฝ๏ฟฝโ = ๐ฏ ๐๐ ๐๐ ๏ฟฝ๏ฟฝโ . ๐๐ฅ (4) โฎ ๐ ๏ฟฝ๏ฟฝโ = ๐ฐ ๐๏ฟฝ๐ (5) ๐ So, ๐๐ ๐๐ ๏ฟฝ๐ ๏ฟฝโ = ๐๐จ ๐๐ซ ๐ ๏ฟฝ๏ฟฝโ = ๐๐จ ๐๐ซ AAST ๐ฐ ๐๐ ๐๐ ๐๏ฟฝ๐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ = ∫๐ฅ ∫๐−๐๐ ๐๐จ ๐๐ซ ๐ฐ ๐๏ฟฝ๐ . ๐๏ฟฝ๐ ๐๐๐ ๐ ๐ ๏ฟฝ๏ฟฝโ . ๐๐ฌ ๐๐ฆ = โฌ ๐ ๐ ๐ ๐๐ ๐ Prof. Darwish ๐ 61 ๐๐ฆ = ๐๐จ ๐๐ซ ๐ฐ๐ ๐๐ ๐ฅ๐ง(๐๐ )๐−๐๐ = ๐ ๐๐จ ๐๐ซ ๐ฐ๐ ๐๐ ๐ฅ๐ง ๏ฟฝ ๐−๐๐ ๐ ๐ฒ ๐๐ฆ ๐๐จ ๐๐ซ ๐ ๐ − ๐๐ ๐ณ = = = ๐ฅ๐ง ๏ฟฝ ๏ฟฝ ๐๐ ๐ฐ ๐ฐ ๐ ๏ฟฝ [๐] Mutual inductance: U Mutual inductance exists between two magnetic circuits that share a common flux linkage. So, the mutual inductance is due to the influence of one circuit on another and vice versa. ๐ด๐๐ = ๐ต๐ ๐๐ฆ๐๐ ๐ฒ๐๐ = ๐ฐ๐ ๐ฐ๐ Where, ๐ฒ๐๐ is the linkage of circuit 2 produced by ๐ฐ๐ in circuit 1. For a linear magnetic medium, it can be shown that ๐ด๐๐ = ๐ด๐๐ . ๐ด๐๐ = ๐ฒ๐๐ ๐ฐ๐ = ๐ต๐ ๐๐ฆ๐๐ ๐ฐ๐ 62 AAST Prof. Darwish Example: U Find M 12 . R R Solution: U ψ 12 μ μ Ν I (c − a ) = 0 r 1 1 4(c + a ) N 2ψ 12 Μ 12 = 2 I1 μ μ Ν N I (c − a ) μ μ Ν N (c − a ) ∴ M 12 = 0 r 1 2 1 = 0 r 1 2 4 I1 (c + a ) 4(c + a ) 2 2 63 AAST Prof. Darwish Magnetic Circuits: U Example: U ≡ NI = Rψ m R = Reluctance = The impedance of magnetic material for flow of magnetic flux R= l µ0 µ r S l1 R1 = → l = magnetic path length , S = cross sectional area µ 0 µ r S1 R2 = R3 = R4 = , l2 µ0 µr S2 , l3 µ 0 µ r S3 , l 41 µ0 µr S4 ∴ψ m (R1 + R2 + R3 + R4 ) = ΝΙ 64 AAST Prof. Darwish ∴ψ m = ΝΙ = Β⋅ Α = Β⋅S R1 + R2 + R3 + R4 Example: U I Rg = lg µ0 S2 ∴ψ m (R1 + R2 + R3 + R4 + Rg ) = ΝΙ ψm = ΝΙ = B.S (R1 + R2 + R3 + R4 + Rg ) 65 AAST Prof. Darwish Example: U ≡ ψ m3 = ψ m1 + ψ m 2 Rt = Rg + R3 + R1R2 R1 + R2 ψ m 3 ⋅ Rt = ΝΙ R2 ∴ψ m1 = ψ m3 . R1 + R2 ψ m2 ψ m3 = → ΝΙ Rt , R1 = ψ m3. R1 + R2 66 AAST Prof. Darwish
