Homework 1 - City Tech OpenLab
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New York City College of Technology Fall 2013 Homework 1 Chapter 6 1 The voltage between two parallel plates separated by a distance of 5mm is 200V. Determine the electric field intensity. π½ 200 π = = ππ, πππ π½/π π 0.005 π π¬= 2 The voltage between two parallel plates separated by a distance for 0.4in is 60V. Determine the electric field intensity. 1in = 0. 254 m 0.4in = 0.01016 π¬= π½ 60 π = = π, πππ π½/π π 0.01016 π 3 The electric field intensity in the region between two parallel plates separated by a distance of 4cm is 2KV/m. Determine the voltage between the plates. π¬= π½ π = π½ = πΈ π₯ π = 2000 π₯ 0.04 = πππ½ 4 The electric field intensity in the region between two parallel plates separated by a distance of 8mm is 200V/mm. Determine the voltage between the plates. π¬= π½ = π π½ = πΈ π₯ π = 200 π₯ 8 = πππππ½ 5 A direct current of 5A is flowing a conductor. Determine the magnetic field intensity at a distance of 3m from the conductor. π―= π° 5π΄ = = πππππ¨/π ππ π 2π. 3 6 A direct current of 4mA is flowing in a conductor. Determine the magnetic field intensity at a distance of 5ft from the conductor. 1ft = 0.3048m 5ft = 1.524m π―= π° 0.004π΄ = = πππππ¨/π ππ π 2π. 1.524 7 For the parallel plates of Problem 6-3 determine the electric flux density if the dielectric is polyethylene (ππ = π. ππ). ∈=∈π π ∈° ∈= 2.25 π₯ (8.842π₯10−12 πΉ/π) = 19.895 π₯10−2 πΉ/π π« =∈ π π¬ = 19.895 π₯10−12 πΉ/π π₯ 2000π/ππ = ππ. ππ ππͺ/ππ 8 For the parallel plates of Problem 6-4, determine the electric flux density if the dielectric is air. π« =∈ π π¬ = 8.84 π₯10−12 πΉ 200π 1ππ π₯( π₯ ) = ππ. ππ ππͺ/ππ π π 0.001 9 For the current carrying conductor of Problem 6-5, determine the magnetic flux density at a distance of 3m from the conductor if the medium is air. π© = π π π― = (1.2566π₯10−6 )π(0.265π΄/π) = πππππͺ/ππ 10 For the current carrying conductor of Problem 6-6, determine the magnetic flux density at a distance of 5ft from the conductor if the medium is air. π© = π π π― = (1.2566π₯10−6 )π(0.0417/π) = ππ. πππͺ/ππ 11 The electric flux density normal to a rectangular surface with dimensions 8m x 75cm is 4μC/m2. Determine the value of the electric flux across the area. π = π«ππ¨ = 6π2 π₯ 4ππΆ/π2 = π. πππͺ 12 The electric flux density normal to a circular surface with a diameter of 3m is 8μC/m2. Determine the value of the electric flux across the area. π¨ = π π ππ = 3.14 π₯ (1.5)2 = 7.068 π = π«ππ¨ = 7.068π2 π₯ 8ππΆ/π2 = π. ππππͺ 13 the magnetic flux density normal to a circular surface with a radius of 5m is 4nWb/m2. Determine the value of the magnetic flux across the area. π¨ = π π ππ = 3.14 π₯ (5)2 = 78.54 π = π©ππ¨ = 4πππ/π2 π₯ 78.54π2 = π. πππππΎπ 14 The magnetic flux density normal to a rectangular surface with dimensions 30cm x 60cm is 12nWb/m2. Determine the value of the magnetic flux across the area. π = π π π = 30ππ π₯ 60ππ = 1800ππ2 ππ 0.18π2 π = π©ππ¨ = 12πππ/π2 π₯ 0.18π2 = π. ππππΎπ 15 A current of 8A is uniformly distributed over a rectangular conductor with dimensions 5mm x 4mm. Determine the current density. π = π π π = 5ππ π₯ 4ππ = 20ππ2 ππ 2π₯10−5 π2 π½= πΌ 8π΄ = = πππππππ¨/ππ π 2π₯10−5 π2 16 A current of 4A is uniformly distributed over a circular conductor with a diameter of 3cm. Determine the current density. π¨ = π π ππ = 3.14 π₯ (0.15)2 = 7.06π₯10−4 π2 π±= π° 4π΄ = = ππππ. πππ¨/ππ π 7.06π₯10−4 π2 17 Assume that the conductivity for the conductor of Problem 6-15 is 5MS/s. Determine the electric field intensity. π¬= π± 400π΄/π2 = = ππππ½/π πΉ 5π₯106 π/π 18 Assume that the conductivity for the conductor of Problem 6-16 is 6 x107 S/m. Determine the electric field intensity. π¬= π± 142.85π΄/π2 = = π. ππππ½/π πΉ 6π₯107 π/π 19 The rms magnitude of the magnetic field of a plane wave in air is π―π = πππππ¨/π. Assuming the E is in the positive x-direction, determine the following for a circular surface of diameter 50m in the x-y plane over which the field are constant. a) Ex π¬π =πΌ π―π πΈπ₯ = π 377Ω = = 1.885ππ/π π»π¦ 200ππ΄/π b) Pz ℘π = π¬π 1.885ππ/π = = 9425ππ/π2 πΌ 377Ω c) Total power transmitted through area π¨ = π π ππ = 3.14 π₯ (25π)2 = 78.58π2 π· = ℘π π π¨ = 9425ππ/π2 π₯ 78.54π2 = 740ππ 20 The rms magnitude of the electric field of a plane wave in a sea water (ππ = π. ππ) is Ex = 3V/m. Assuming that H is in the positive y-direction, determine the following for a square surface with sides of 15m each in the x-y plane over which the fields are constant: a) Hy π¬π =πΌ π―π π»π¦ = πΈπ₯ 3π/π = = 7.95ππ΄/π π 377Ω b) Pz ℘π = π¬π 3π/π = = 23.8ππ/π2 πΌ 377Ω c) Total power transmitted through area π¨ = π³π = (15π)2 = 225π2
