The relative permeability of certain silicon steel is 4500. A certain magnetic loop consists of silicon steel of 0.0005 m square, 0.007 m long and an air gap of 0.0007 m. What is the reluctance of the magnetic circuit? gilbert per maxwell
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compute the reluctance of the magnetic circuit
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- A mmf is supplied by a current of 1 A through 100 turns. The magnetic unit consists of a steel core of 800 relative permeability, length of 10 cm and cross-sectional area of 4 cm2 and an air gap of 1 cm long. What is the total reluctance of the magnetic circuit in AT/Wb?An air gap has a length of 0.1 cm. What length of iron core has the same reluctance as the air gap? The relative permeability of the iron is 5000. Assume that the cross-sectional areas of the gap and the core are the same.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 1
- 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 1An iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current 1 = 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 (t, = 800 for iron).The relative permeability of a certain silicon steel is 4500. A certain magnetic loop consists of a silicon steel of 10 cm square, 20 cm long and an air gap of 0.25 cm. Find the reluctance of the magnetic circuit.
- The property of a material which opposes the production of - magnetic flux in it is known as mmf O reluctance permeability permittivity OExplain the following. A. Magnetic Field Strength B. Boundary ConditionsFor 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?
- show thatvthis is a magnetic flux density: then prove and interpret it with reasons on each step(3.4)The length of a magnetic circuit in a moving iron instrument is 300mm. The coil around the soft-iron core has 360 turns and takes a current of 1.75 A. The core is square in section with sides of 20 mm. Take the relative permeability of soft iron as 1100. 3.4.1) Determine the magnetomotive force in the core 3.4.2)Determine the field strength 3.4.3) Determine the total fluxThe length of a magnetic circuit in a moving iron instrument is 300mm. The coil around the soft iron core has 360 turns and takes a current of 1.75 A. The core is square in section with sides of 20mm. Take the relative permeability of soft iron as 1100. Determine the total flux