Question

Q.3 Consider an infinitely long coaxial structure shown in the figure below. Inner conductor has a radius a and outer conducting shell has a radius b. Thickness of the outer conductor is ignored as it is very small. Between two conductors, there is a magnetic material with permeability () = Mo a Assume that the current I is distributed uniformly over the cross-section of the inner conductor whereas it flows on the surface of the outer conductor. a) Find the magnetic field intensity vector and the magnetic flux density vector B everywhere (r <a, a srsbr>b). b) Find the self-inductance per unit length of the coaxial structure. ur) b . HT)

Answer 1

* (11 for rea Applying ampere circuit law, фн-а Hidla ♡ I DOO Ir2 I ar? H.277 = 92 182 IY H 90 > I 2x8a2 2292 MIN B = MH 2792 By Magnetic flux density Mi permeability of inner wire Applying ampere circuit Low for aer<b> AK H. 227 = 1 I HE 227 a a<osb M(2) H = Mora a2 227 ad Magnetic flux density B= MH = B = Mo Ir ad 22q2 r for rob Applying ampere circuit Law Ь Ila H.267 = I-1 H=0 B = UH = 0 CSScanned with CamScanner

QWm (b) Magnetic energy density um = ELI? CY L 2 2 where Wm= B² dv 24 B12 dv 2 internal inducedance Lim See 2 I 24 MIT 229² where bi u magnetic flux density in innee wire, Bi= M²1²82 rdrd&dz 1²4 42²94 - Lin = SSI 22 9 M Ml 82 47²a4 dz fd da 83 dr where de un length of coaxial cable. M(M) 1² rdr dødz 2 B2 dv = . 12 Lext = 3. Saltar = IZ 24(0) 422g² where B2 in magnetic flux density between conducators, B2 = M(8) I 228 Mfr)= Mo 1² (M) 92 Lext - I za Slov*13 Mor2I2 rdrdd dz 4222222 22 do b Lext Mo 42292 4 Lext Mo ! (6²-927.27.d 42²92 Mo (6²-9²) I 4792 Total inductance L = Lin + Lext 44 87 + Mo (6²_9221 4292 M in the permeability of inner wire. CS Scanned with CamScanner