2. A current of I = 3 [A] flows over the N = 200 winding wire as given in the figure below. The cross-sectional area of the core is S = 10-3 [m²] and its relative magnetic permeability is given as , = 5000. Calculate each magnetic flux.
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- A ferromagnetic core is shown in Figure PI-2. The depth of the core is 5 cm. The other dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of 0.005 Wb. With this current, what is the flux density at the top of the core? What is the flux density at the right side of the core? Assume that the relative permeability of the core is 1000. -10 cm- - 20 cm- 15 cm 400 turns 15 cm 15 cm Core depth 5 cmA toroidal core with a mean circumference of 100 cm and a cross-sectional area of 10 cm2 is wound with 500 turns of wire. What current would be required to generate a flux of 1 mWb in the core. Assume the core has a relative permeability of 800A ferromagnetic core is shown in Figure Pl-2. The depth of the core is 5 cm. The other dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of 0.005 Wb. With this current, what is the flux density at the top of the core? What is the flux density at the right side of the core? Assume that the relative permeability of the core is 800. 1-5. 10 cm- 5em 20 em 15 cm 15 cm 15 cm Coe depth - Scm SOLUTION There are three regions in this core. The top and bottom form one region, the left side forms a second region, and the right side forms a third region. If we assume that the mean path length of the flux is in the center of each leg of the core, and if we ignore spreading at the corners of the core, then the path lengths are I, = 2(27.5 cm) = 55 cm, I, = 30 cm, and /, = 30 cm. The reluctances of these regions are:
- A ferromagnetic core is shown below. The depth of the core is 5 cm. The other dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of 0.003 Wb. With this current, what is the flux density at the top of the core? What is the flux density at the right side of the core? Assume that the relative permeability of the core is 1000. 1. - 10 cm--- - 20 cm - 15 cm 500 tums 15 cm 15 cm [1.21 A, 0.4 T, 1.2 T]An iron circuit with a small0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I = 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm2, and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (u, = 800 for iron). Figure 1For the magnetic circuit shown in the figure, the iron core with N = 500 turns -Magnetic permeability is 200. its average length is 10 cm -The diameter of the cross-sectional area is 1 cm. Find the current that must pass through the winding to produce 0.5 mWb of flux in a 1mm air-gap magnetic circuit?
- A magnetic circuit has a length of 100 cm and cross-sectional area of 5 square cm. The total flux is 10 x 104 webers. The coil has 100 turns and current of 4 amperes. What is the magnetic strength? a. 500 AT/m b. 400 AT/m c. 350 AT/m d. 450 AT/mAn iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I = 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm², and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (ur = 800 for iron).An iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I = 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm2, and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (4, = 800 for iron).
- An iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I = 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm?, and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (4, = 800 for iron). Figure 1B. For the following magnetic circuit, the flux passing through the core is 1.32 mWb, the cross section of the core is 3 cm by 4 cm, the laminated section has a stacking factor of 0.9, and the gap is 1 mm. Determine the flux density in each section. Neglect fringing d N turns Cast iron Air gap Laminated sheet steelIn a certain cast-steel core series magnetic circuit with a 400 turns, 170 At/m, mean length of 0.16 m and 0. 002 square meters cross-sectional area. the value of current to develop a 0.0004 Wb is