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B H Curves of various Magnetic Materials 1 A magnetic circuit is made up of steel laminations shaped as in Fig1 The width of the iron is 40 mm and the core is built up to a depth of 50 mm of which

B-H Curves of various Magnetic Materials
1. A magnetic circuit is made up of steel laminations shaped as in Fig.1. The width of the iron is
40 mm and the core is built up to a depth of 50 mm, of which 8 percent is taken up by insulation
between the laminations. The gap is 2 mm long and the effective area of the gap is 2500 mm 2.
The coil is wound with 800 turns. If the leakage factor (the ratio of the total flux linking the coil
over the air gap flux) is 1.2, calculate the magnetizing current required to produce a flux of
0.0025 Wb across the air gap. (Use the given B-H curve given above)
Fig. 1
2. Fig.2 shows a magnetic circuit with air gaps g 1 = g2 = g3 = 1 mm and coils N1 = 100 turns and
N2 = 200 turns. The cross sectional area A of the circuit is 200 mm. Assume the permeability of
the core material approaches infinity and the fringing effect is negligible. Calculate:
(a) the self and mutual inductances;
(b) the total magnetic energy stored in the system, if the currents in the coils are i 1 = i2 = 1 A;
(c) the mutual inductance between N1 and N2, if the air gap g3 is closed.
Fig. 2
3. It is desired to achieve a time varying magnetic flux density in the air gap of the magnetic
circuit of Fig. 3a of the form B g = B0 + B1sin(?t), where B0=0.5 T and B1=0.25 T. The dc field
B0 is to be created by a NdFeB permanent magnet, whereas the time varying field is to be
created by a time varying current. Assume the permeability of the iron is infinite and neglect the
fringing effect.
(a) For the air gap dimensions given in Fig. 3a, find the magnet length d if the magnet area Am
equals the air gap area Ag. Fig. 3b gives the demagnetization curve of NdFeB permanent
magnet.
(b) Find the excitation current required to achieve the desired time varying air gap flux density.
Fig. 3a
Fig. 3b
4. The Fig. 4 shows a magnetic circuit. Find the self and mutual inductances of the two windings
considering infinite relative permeability of the ferromagnetic material. Consider a thickness d
and 3d for the air gaps and a constant cross sectional area S for the branches of the magnetic
circuit. N1=120; N2=200; d=0.8mm; S=160cm2; µ0=1.257 10-6 H/m; i1=1.2A, i2=4A.
Fig. 4
5. The Fig. 5 show a magnetic circuit. Find the self and mutual inductances of the two windings
and the magnetic flux F. Consider a thickness d for the air gaps, a length L for the columns and
a constant iron normal section S. N 1=100; N2=200; d=1mm; S=100cm2; µ0=1.257 x 10-6 H/m,
µFe=2000, i1=2A, i2=3A.
F

Jul 27 2020 View more View Less

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