Solutions to ENEE 381 First Examination: Electromagnetic Waves and Circuits - Prof. Christ, Exams of Electrical and Electronics Engineering

Download Solutions to ENEE 381 First Examination: Electromagnetic Waves and Circuits - Prof. Christ and more Exams Electrical and Electronics Engineering in PDF only on Docsity! H 0.065= H-field amplitude in A/m εr 10:= µr 5:= Z µ0 µr⋅ ε0 εr⋅ := Z 266.391= E 2 Z⋅ Savg⋅:= E 20.591= E-field amplitude in V/m H 2 Savg⋅ Z := H 0.077= H-field amplitude in A/m With a 45 degree phase shift Hnew H e i 45⋅ deg⋅ ⋅:= 45 deg⋅ 0.785= radians Hnew  is complex conjugate Savg 0.5 Re E Hnew  ⋅( )⋅:= Savg 0.563= ENEE 381 First Examination Solutions P 1025:= Power of point source in W(1) R 1012:= Savg P 4 π⋅ R2⋅ := Average Poynting vector calculated using surface area of sphere Savg 0.796= W/m2 ε0 8.854 10 12− ⋅:= µ0 4 π⋅ 10 7− ⋅:= Z0 µ0 ε0 := Characteristic impedance of free space Z0 376.734= Savg E( )2 2 Z0⋅ = Savg Z0 H( )2 2 ⋅= E 2 Z0⋅ Savg⋅:= E 24.487= E-field amplitude in V/m H 2 Savg⋅ Z0 := The current in the wire is I 1 Z := I 0.866 0.866i−= H I 2 π⋅ r⋅ := The real current is Ireal I:= Ireal 1.225= Hreal Ireal 2 π⋅ r⋅ := From Amperes Law Hreal 19.492= Note that ohmic dissipation in wire is P 0.5 Re 1 I  ⋅( )⋅:= P 0.433= Poynting vector flux into wire PS 2 π⋅ r⋅ 0.5⋅ Re 1 H  ⋅( )⋅:= PS 0.433= same as ohmic dissipation (2) σ 3 107⋅:= r 10 2−:= ν 10 109⋅:= Frequency δ 1 π µ0⋅ ν⋅ σ⋅ := δ 9.189 10 7−×= Skin depth in meters Rs 1 σ δ⋅ := Rs 0.036= surface resistance in ohms Resistance of wire R Rs 2 π⋅ r⋅ := ωL R:= Z R i ωL⋅+:= Z 0.577 0.577i+= Impedance of 1m of wire in ohms