Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9780078028151
Author: Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher: Mcgraw-hill Education,
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Chapter 9, Problem 9.2P
To determine

The emf at any time, t.

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Consider a ferromagnetic ring (nucleus) that has a mean circumference of 40 cm, 300coiled copper turns and a sectional area of 5 cm2. A current of 2 is appliedA to winding, producing a flux of 1.5 mWb.Determine the permeability of the core material.Determine the relative permeability of the core material.If the ferromagnetic ring has an air gap of 2 mm. Find the required MMFso that the flux density in the gap is 0.5 T.
Biot-Savart law expresses the magnetic flux intensity produced by a dc electrical current Select one: O a. True b. None of the above O c. False
Figure 1 shows a ring formed with two different materials – cast steel and mild steel. Mild steel has a mean length of 400 mm and an area of 500 mm². Cast steel has a mean length of 300 mm and an area of 312.5 mm2. The magnetisation curve for mild steel and cast steel is shown in Figure 2. Find the total MMF required to cause a flux of 500 µWb in the magnetic circuit. Determine also the total circuit reluctance. 1.8 Cașt steel Cast-steel 1.6 Silicon iron 1.4 Mild steel 1.2 1.0 0.6 Mild stel Cast iron 0.6 Figure 1: Ring arrangement 0.4 0.2 3000 4000 Magnetic field strength, H(A/m) 1000 2000 5000 600ỘI 7000 Figure 2: Magnetization Curve MMF = AT Reluctance = x 10* ΑT/Wb Flux density, B(T)
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Publisher:Mcgraw-hill Education,