Question

Magnetic Circuits y Part A - Calculate reluctances Learning Goal: To understand how magnetic structures can be analyzed by drawing an equivalent circuit, and to use the equivalent circuit to calculate magnetic fluxes and coll currents. When analyzing magnetic structures, the geometry is often complex enough that using the fundamental rules can be very difficult without numerical methods. However, there are approximate methods that are often sufficient for engineering calculations. When the magnetic field is mostly contained within cores of high permeability and small air gaps, the structure can often be converted into an equivalent magnetic circuit. These circuits can then be analyzed using standard techniques for electric circuits The role of a voltage source is taken by coils wrapped around the core. The magnetomotive force is the analogue of the voltage. A coil with N turns carrying a current i has a magnetomotive force F=Ni Using the information given in the introduction, calculate the reluctances for the equivalent circuit shown A turns Express your answers to three significant figures in ► View Available Hint(s) Wb 190 AED ut vec ? R₂, R₂, Rio A-tums Submit - Part B - Calculate the flux density The cores and air gaps all have a reluctance, which plays a similar role to the resistance in an electrical circuit. The reluctance of a path is R where L is the mean length of μΑ the path is the magnetic permeability along the path, and is the cross-sectional area of the path. We use the mean length because the cross-dimensions of magnetic structures are often significant, so a single length is not well defined. The permeability of air is taken to be the permeability of free space H = 4 x 10-7 Wb Am For magnetic materials, the permeability is usually given as a relative permeability such Both coils have N = 500 turns and the current is i = 26 A. What is the flux density in the air gap? Express your answer to three signficant figures in T. ► View Available Hint(o) VO AXD! vec ? The magnetic core shown has dimensions d = 8 cm and g 0.5 cm. The relative permeability of the core is p = 400. What are the reluctances in the equivalent magnetic circuit? The Iron core has a cross sectional area of 5.2 cm and the effective area of the air gap due to fringing is 8.3 cm B

Answer 1

da 8cm : 0:08 g=0.5cm 0.005 2 E. Mra N = Soo 400 5.2X164 m2, A2= 8.3x109 m2 ALE SO! Ry I Mo Mę A, 3d Molle A, R 3x0.08 40x107x 400x 5.2 X10-4 918201.5948 R2 = le la Moller A, + 0.08 40x167 x 400 x 5.2x164 0.5 x 16? 40X167x 8,3x104 Mo Az R2= 306067.1983 + 4793823.5873 5099890.785 R3 Riz 918201·5948 from diagrama kel at point А R3=R1 A, R2 A F-A Ri E-A + R3 ale F R2 4 2 (F-A) R, A R2 QE Ali+me) RI A = QF R2 2 R₂ + Ri FR₂ Rat Ri 2 NI flux in R2= A Rat R 5099890785+ 918201.5948 2 € 500x 2.6 5558991.583 2.33855' x104 Wb.

B dag Pag dag Aau 0.281753 T. gap Az 8.3x104 (0) When g= 0.3 Cm. Now 0:08 R2 아 urx107x 400 x 5:2x164 0.003 4 01x107x8.37104 R2= 306067.1983+ 2876294,152 3182361:351 BAS NI 11 from e. Ag (R2+ R 1) Az 0.28 1753 = Soo I 3641462.148 X 8.3x104 'I = 1.70315 Amp