Electromagnetic Field Lecture Notes PDF

Electromagnetic Field Dr. Wafaa Rady  Maxwell’s Equations 1- Gauss’s law for Electric The electric flux passing through any closed surface is equal to the total charge enclosed by that surface....

Electromagnetic Field Dr. Wafaa Rady  Maxwell’s Equations 1- Gauss’s law for Electric The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. 𝛻. 𝐷 2- Gauss’s law for Magnetic = 𝜌𝑣 The magnetic flux lines are closed and do not terminate on a “magnetic charge.” For this reason Gauss’s law for the magnetic field is 𝛻. 𝐵 = 0 3- Ampere's law Ampere’s circuital law states that the line integral of H about any closed path is exactly equal to the direct current enclosed by that path, ර 𝑯. 𝒅𝒍 = 𝐼𝑒𝑛𝑐  Magnetic Field 4- Faraday’s law Faraday try to prove that if a current could produce a magnetic field, then a magnetic field should be able to produce a current. A time-varying magnetic field produces an electromotive force (emf) that may establish a current in a suitable closed circuit. Magnetic Flux 𝜑 𝜑 = ඵ 𝑩. 𝒅𝒔  Magnetic Field Example 1:  Magnetic Field Example 2: Given the closed path having 0< x < 5t + 2t3 , 0< y < 20cm , z =0. Let B = 0.8x2az T. Find the electromotive force value at t = 0.4 s  Maxwell’s Equations Ampere's law ර 𝐻. 𝑑𝑙 = 𝐼𝑒𝑛𝑐 = Displacement current + Conduction current ර 𝐻. 𝑑𝑙 = ඵ 𝐽𝑑. 𝒅𝒔 + ඵ 𝐽𝑐. 𝒅𝒔 𝜕𝐷 𝐽𝑑 = 𝐽𝑐 = 𝜎𝐸 𝜕𝑡 𝜕 ර 𝐻. 𝑑𝑙 = ඵ 𝑫. 𝒅𝒔 + ඵ 𝝈𝑬. 𝒅𝒔 𝜕𝑡 By using Stock’s theory ර 𝐴. 𝑑𝑙 = ඵ 𝛻𝑥 𝐴. 𝑑𝑠 𝜕 𝜕𝐷 𝜕𝐸 ඵ 𝛻𝑥 𝐻. 𝑑𝑠 = ඵ 𝐷. 𝑑𝑠 + ඵ 𝜎𝐸. 𝑑𝑠 𝛻𝑥 𝐻 = + 𝜎𝐸 𝛻𝑥 𝐻 = ε + 𝜎𝐸 𝜕𝑡 𝜕𝑡 𝜕𝑡  Magnetic Field Example 3: If the electric field intensity is E = (106) cos(105 t)aρ V/m in a homogeneous material has conductivity σ = 10−5 S/m. find: a) displacement current density b) conduction current density.  Maxwell’s Equations Faraday’s law 𝑑𝜑 𝑒𝑚𝑓 = − Magnetic Flux 𝜑 𝑑𝑡 𝑑𝜑 ර 𝐸. 𝑑𝑙 = − 𝜑 = ඵ 𝑩. 𝒅𝒔 𝑑𝑡 𝑑 ‫𝑩 ׭‬. 𝒅𝒔 ර 𝐸. 𝑑𝑙 = − 𝑑𝑡 𝜕 ර 𝐸. 𝑑𝑙 = − ඵ 𝑩. 𝒅𝒔 𝜕𝑡 By using Stock’s theory ර 𝐴. 𝑑𝑙 = ඵ 𝛻𝑥 𝐴. 𝑑𝑠 𝜕 𝜕𝐵 𝜕𝐻 ඵ 𝛻𝑥 𝐸. 𝑑𝑠 = − ඵ 𝐵. 𝑑𝑠 𝛻𝑥 𝐸 = − 𝛻𝑥 𝐸 = −𝜇 𝜕𝑡 𝜕𝑡 𝜕𝑡  Maxwell’s Equations Maxwell’s Equations Maxwell’s Equations Integral Form Differential Form Faraday’s law 𝜕 𝜕𝐻 ර 𝐸. 𝑑𝑙 = − ඵ 𝑩. 𝒅𝒔 𝛻𝑥 𝐸 = −𝜇 𝜕𝑡 𝜕𝑡 Ampere's law 𝜕 𝜕𝐸 ර 𝐻. 𝑑𝑙 = ඵ 𝑫. 𝒅𝒔 + ඵ 𝝈𝑬. 𝒅𝒔 𝛻𝑥 𝐻 = ε + 𝜎𝐸 𝜕𝑡 𝜕𝑡 Gauss’s law for Electric 𝛻. 𝐷 = 𝜌𝑣 Gauss’s law for Magnetic 𝛻. 𝐵 = 0 By using Stock’s and Divergence theories  Magnetic Field Example 4: Using Maxwell equation 𝜕𝐷 𝛻𝑥 𝐻 = + 𝜎𝐸 𝜎=0 𝜕𝑡 𝜕𝐷 𝛻𝑥 𝐻 = 𝜕𝑡  Magnetic Field 𝜕𝐷 𝛻𝑥 𝐻 = 𝜕𝑡 Thank You