QI- The magnetic circuit shown is made of a cast-steel and has a coil of 1000 turns on its center limp. The center limp and each of the side limbs has a width of Scm and the thickness of the core t =5 cm. The means length of magnetic paths LABCD-LAFED= 60 cm, and the length of center limp=30 cm. Determine the current that the coil should carry to produce a flux of 2.5 mWb in the air-gap. Neglect magnetic leakage and fringing. The magnetization curve fore 1.2 steel is given B (lux density) Wb/ 0.2 0.5 0.7 cast- 1.0 by: mm (Amper-turn/ 300 540 650 900 1150 m)

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QI- The magnetic circuit shown is made of a cast-steel and has a coil of 1000 turns on its
center limp. The center limp and each of the side limbs has a width of 5cm and the
thickness of the core t =5 cm. The means length of magnetic paths
and the length of center limp=30 cm.
Determine the current that the coil should carry to produce a flux of 2.5 mWb in
the air-gap. Neglect magnetic leakage and fringing. The magnetization curve fore
1.2 steel is given
LABCD=LAFED= 60
cm,
B (flux density) Wb /
mm?
cast-
0.2
0.5 0.7 1.0
by:
H
(Amper-turn/ 300 540 650 900 1150
m)
0.1 cm
30 m
1000
Tru
Transcribed Image Text:QI- The magnetic circuit shown is made of a cast-steel and has a coil of 1000 turns on its center limp. The center limp and each of the side limbs has a width of 5cm and the thickness of the core t =5 cm. The means length of magnetic paths and the length of center limp=30 cm. Determine the current that the coil should carry to produce a flux of 2.5 mWb in the air-gap. Neglect magnetic leakage and fringing. The magnetization curve fore 1.2 steel is given LABCD=LAFED= 60 cm, B (flux density) Wb / mm? cast- 0.2 0.5 0.7 1.0 by: H (Amper-turn/ 300 540 650 900 1150 m) 0.1 cm 30 m 1000 Tru
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