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 8, Problem 8.9P

A current of-100az A/m flows on the conducting cylinder p = 5 mm, and +500az A/m is present on the conducting cylinder p = 1 mm. Find the magnitude of the total force per meter length that is acting to split the outer cylinder apart along its length.

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Figure 1 shows a ferromagnetic core with a relative permeability of 1850, the depth of the core is 10 cm. The air gap on the core is 0.2 cm with effective area 5 % larger than their physical size due to fringing effects. Given the number of turns N = 500 and current i = 2 x, where x is the last digit of your student ID (example: EEE1705590, then i = 20 A): (a) Find the total reluctance of the core and air gap. [CLO1-PL01:C2] (b) Find the flux density of the air gap. [CLO1-PLO1:C2] 13 cm 23 cm 9 cm 5.6 сm Air gap 31 cm N turns 7 cm Figure 1
A magnetic core with an air gap is shown below. For a relative permeability is 9500, and the space free permeability is 4Ttx107 H/m. Find the reluctance of the left leg and the top side of the core. Ignore the gap in your calculations. 10 cm 0.1 cm gap 20 cm 5 cm 5 cm 40 cm 5 cm core depth = 10 cm Select one: a. Rjeft = 3.770 KA.t./Wb , Rtop = 4.607 KA.t/Wb b. None c. Rieft = 377.0 A.t./Wb , Rtop = 460.7 A.t/Wb d. Rjeft = 325.1 A.t./Wb , Rtop = 402.3 A.t/Wb ----------
> Figure 1 shows a ferromagnetic core whose mean path length is 40 cm. There is a small gap of 0.05 cm in the structure of the otherwise whole core. The cross-sectional area of the core is 12cm2, the relative permeability of the core is 4000 and the coil of wire on the core has 400 turns. Assume that fringing in the air gap increases the effective cross-sectional area of the air gap by 5 percent. Given this information. Find the total reluctance of the flux path (iron plus air gap) and the current required to produce a flux density of 0.5 T in the air gap. N=400 tums 0.05 cm A=12 cm? le-40 cm Figure 1

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