advertisement

Jose G. Chiriboga Alcivar TCET 2220 Prof. Viviana Homework 1 09-15-13 (6.1) The voltage between two parallel plates separated by a distance of 5 mm is 200 V. Determine the electric field intensity πΈ= π 200π = = 40000 π ⁄π π 5 π₯ 10−3 π (6.2) The Voltage between two parallel plates separated by a distance of 0.4 inch is 60 V. Determine the electric field intensity. x= 0.4 inch x 2.5cm = 1 cm = 0.01 m πππβ πΈ= π 60 π = = 6000 π ⁄π π 1 π₯ 10−2 π (6.3) The electric field intensity in the region between two parallel plates separated by a distance of 4 cm is 2 kV/m. Determine the voltage between the plates. 4 cm = 0.04m V = E x d = 2000 π ∗ 0.04 m = 80 V π (6.4) The electric field intensity in the region between two parallel plates separated by a distance of 8 mm is 200 V/mm. Determine the voltage between the plates. V V = E x d = 200 ∗ 8 mm = 1600V mm (6.5) A direct current of 5 A is flowing in a conductor. Determine the magnetic field intensity at a distance of 3 m from the conductor. H= I 5A = = 265.26 mA/m 2πd 2 ∗ π ∗ 3 m (6.6) A direct current of 40 mA is flowing in a conductor. Determine the magnetic field intensity at a distance of 5 ft. from the conductor. 5ft = 60 inches Chiriboga 1 x= 60 inch x 2.5cm = 150 cm = 1.5 m inch H= I 40 mA = = 4.244 mA/m 2πd 2 ∗ π ∗ 1.5 m (6.7) For the parallel plates of Problem (6.3), determine the electric flux density if the dielectric is polyethylene (∈π = π. ππ). ∈=∈π ∗∈π = 2.25 ∗ 8.842 ∗ 10−12 = 19.89 ∗ 10−12 F/m D =∈∗ E = 19.89 ∗ 10−12 F π ∗ 2000 = 39.78 nC/π2 m π (6.8) For the parallel plates of Problem (6.4), determine the electric flux density if the dielectric is air. ∈=∈π = 8.842 ∗ 10−12 F/m D =∈∗ E = 8.842 ∗ 10−12 F V ∗ 200 = 1.768 nC/mm2 m mm (6.9) For the current-carrying conductor of Problem (6.5), determine the magnetic flux density at a distance of 3 m from the conductor if the medium is air. μ = π0 = 4π ∗ 10−7 = 1.2566 ∗ 10−6 H/m B = μ ∗ H = 1.2566 ∗ 10−6 H mA ∗ 265.26 = 333.32 nWb/m2 m m (6.10) For the current-carrying conductor of Problem (6.6), determine the magnetic flux density at a distance of 5 ft from the conductor if the medium is air. μ = π0 = 4π ∗ 10−7 = 1.2566 ∗ 10−6 H/m B = μ ∗ H = 1.2566 ∗ 10−6 H mA ∗ 4.244 = 5.333 nWb/m2 m m (6.11) The electric flux density normal to a rectangular surface with dimensions 8 m x 75 cm is 4 ππͺ/ππ . Determine the value of the electric flux across the area. A = 8m ∗ 0.75m = 6 m2 ψ = A ∗ D = 6 m2 ∗ 4 ππΆ = 24 μC m2 Chiriboga 2 (6.12) The electric flux density normal to a circular surface with a diameter of 3m is 8 ππͺ/ππ . Determine the value of the electric flux across the area. A = ππ 2 = (π) ∗ 1.52 = 7.06 m2 ψ = A ∗ D = 7.06 m2 ∗ 8 ππΆ = 56.55 μC π2 (6.13) The magnetic flux density normal to a circular surface with a radius of 5m is 4 ππΎπ/ππ . Determine the value of the magnetic flux across the area. A = ππ 2 = (π) ∗ 52 = 78.539 m2 Ο = A ∗ B = 78.539 m2 ∗ 4 πππ = 314.159 nWb π2 (6.14) The magnetic flux density normal to a rectangular surface with dimensions 30 cm x 60 cm is 12 ππΎπ/ππ . Determine the value of the magnetic flux across the area. A = 0.30m ∗ 0.60m = 0.18 m2 Ο = A ∗ B = 0.18 m2 ∗ 12 πππ = 2.16 nWb π2 (6.15) A current of 8 A is uniformly distributed over a rectangular conductor with dimensions 5 mm x 4 mm. Determine the current density. A = 0.005m ∗ 0.004m = 0.00002 m2 J= I 8A A = = 400000 2 2 A 0.00002 m m (6.16) A current of 4 A is uniformly distributed over a circular conductor with a dimensions 3 cm. Determine the current density. A = ππ 2 = (π) ∗ (0.0152 ) = 706.85 ∗ 10−6 m2 J= I 4A kA = = 5.658 A 706.86 ∗ 10−6 m2 mπ (6.17) Assume that the conductivity for the conductor of Problem (6.15) is 5 MS/m. Determine the electric field intensity. E= J 400000 A⁄m2 = = 80 mV⁄m σ 5 MS/m Chiriboga 3 (6.18) Assume that the conductivity for the conductor of Problem (6.16) is π π πππ S/m. Determine the electric field intensity. J 5.658 kA⁄m2 E= = = 94.3 μ V⁄m σ 6 ∗ 107 S/m (6.19) The rms magnitude of the magnetic field of a plane wave in air is π―π = πππ ππ¨⁄π. Assuming that E is in the positive x-direction, determine the following for a circular surface of diameter 50 m in the x-y plane over which the fields are constant: (a)π¬π π0 = 120π = 377Ω πΈπ₯ = ππ»π¦ = 377 Ω ∗ 200 (b)π·π Pz = ππ΄ = 75.4 ππ/π π Ex 2 75.4 2 mV/m = = 15.08 μW/m2 n0 377 (c) Total power transmitted through area A = π ∗ r 2 = π ∗ 252 = 1.963 ∗ 103 m2 P = Pz ∗ π΄ = 15.08 μW ∗ 1.963 ∗ 103 m2 = 29.61 ππ m2 (6.20) The rms magnitude of the magnetic field of a plane wave in sea water (∈π = ππ) is π¬π = π π½/π. Assuming that H is in the positive y-direction, determine the following for a square surface with sides of 15 m each in the x-y plane over which the fields are constant: (a)π―π η= π»π¦ = (b)π·π π0 377 = = 42.15Ω √∈π √80 Ex 3 V/m = = 71.17 mA/m η0 42.15Ω Ex 2 32 V/m Pz = = = 126.46 W/m2 n0 71.17 mA/m (c) Total power transmitted through area Chiriboga 4 π΄ = 15m ∗ 15m = 225 m P = Pz ∗ π΄ = 126.46 W ∗ 225 m = 28.45 kW m2 (6.21) In a lossless dielectric medium, the rms electric and magnetic field intensities are π¬π = ππ ππ½/π and π―π = πππ ππ¨/π. Determine the following: (a) Intrinsic impedance η= πΈπ₯ 50 mV/m = = 500Ω π»π¦ 100 μA/m (b) Power density Pz = Ex ∗ Hy = 50 mV μA ∗ 100 = 5 μW/m2 m m (c) Dielectric constant η= ∈r = π0 377 = √∈π √∈π 3772 3772 = = 568.51 ∗ 10−3 η2 500Ω2 Chiriboga 5