The figure below shows a simplified diagram of a relay composed of a c-shaped fixed element, a movable armature [the horizontal bar in the middle], an air gap, and a coil of wire with N = 300 turns and i = 2 amperes. Upper segment Lower segment air --- gap fixed element N-turn winding movable armature not to scale The fixed element and the movable armature are made of the same material, with relative permeability of 2500. To allow the armature to move freely, there are spacers [yellow rectangles] where the fixed element and the armature brush against each other. For this problem, assume that we can neglect the effect of the spacers in terms reluctance. The mean length of the flux path along the upper segment of the fixed element is 50 cm [blue dashed arrows]. The same holds for the lower segment [red dashed arrows]. At the instant shown, the mean length of the flux path along the armature is 20 cm and the air gap is 0.1 cm long. The cross-sectional area (that is, the area normal to the flux path) for both the upper and lower segments of the fixed element is 25 cm², while the cross-sectional area for both the armature and the air gap is 50 cm². For this problem, use μ = 4 x 10-7 H/m. Hint: You may want to convert to Sl units (e.g., meters) before beginning your calculations.
The figure below shows a simplified diagram of a relay composed of a c-shaped fixed element, a movable armature [the horizontal bar in the middle], an air gap, and a coil of wire with N = 300 turns and i = 2 amperes. Upper segment Lower segment air --- gap fixed element N-turn winding movable armature not to scale The fixed element and the movable armature are made of the same material, with relative permeability of 2500. To allow the armature to move freely, there are spacers [yellow rectangles] where the fixed element and the armature brush against each other. For this problem, assume that we can neglect the effect of the spacers in terms reluctance. The mean length of the flux path along the upper segment of the fixed element is 50 cm [blue dashed arrows]. The same holds for the lower segment [red dashed arrows]. At the instant shown, the mean length of the flux path along the armature is 20 cm and the air gap is 0.1 cm long. The cross-sectional area (that is, the area normal to the flux path) for both the upper and lower segments of the fixed element is 25 cm², while the cross-sectional area for both the armature and the air gap is 50 cm². For this problem, use μ = 4 x 10-7 H/m. Hint: You may want to convert to Sl units (e.g., meters) before beginning your calculations.
Introductory Circuit Analysis (13th Edition)
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Question
Calculate the reluctance of the air gap. Answer in ampere-turns/weber
Calculate the total reluctance of the system Answer in ampere-turns/weber
![The figure below shows a simplified diagram of a relay composed of a c-shaped fixed element, a movable armature [the horizontal bar in the
middle], an air gap, and a coil of wire with N = 300 turns and i = 2 amperes.
Upper
segment
Lower
segment
air
gap
fixed element
N-turn winding
movable armature
not to scale
The fixed element and the movable armature are made of the same material, with relative permeability of 2500. To allow the armature to
move freely, there are spacers [yellow rectangles] where the fixed element and the armature brush against each other. For this problem,
assume that we can neglect the effect of the spacers in terms reluctance.
The mean length of the flux path along the upper segment of the fixed element is 50 cm [blue dashed arrows]. The same holds for the lower
segment [red dashed arrows]. At the instant shown, the mean length of the flux path along the armature is 20 cm and the air gap is 0.1 cm
long.
The cross-sectional area (that is, the area normal to the flux path) for both the upper and lower segments of the fixed element is 25 cm²,
while the cross-sectional area for both the armature and the air gap is 50 cm². For this problem, use μ = 4π × 10-7 H/m.
Hint: You may want to convert to Sl units (e.g., meters) before beginning your calculations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0975fa50-6543-481e-b78a-4d0bf136934b%2F1fc6d26b-1fbf-4edb-a0e0-70be3e0924ca%2Flucb29_processed.png&w=3840&q=75)
Transcribed Image Text:The figure below shows a simplified diagram of a relay composed of a c-shaped fixed element, a movable armature [the horizontal bar in the
middle], an air gap, and a coil of wire with N = 300 turns and i = 2 amperes.
Upper
segment
Lower
segment
air
gap
fixed element
N-turn winding
movable armature
not to scale
The fixed element and the movable armature are made of the same material, with relative permeability of 2500. To allow the armature to
move freely, there are spacers [yellow rectangles] where the fixed element and the armature brush against each other. For this problem,
assume that we can neglect the effect of the spacers in terms reluctance.
The mean length of the flux path along the upper segment of the fixed element is 50 cm [blue dashed arrows]. The same holds for the lower
segment [red dashed arrows]. At the instant shown, the mean length of the flux path along the armature is 20 cm and the air gap is 0.1 cm
long.
The cross-sectional area (that is, the area normal to the flux path) for both the upper and lower segments of the fixed element is 25 cm²,
while the cross-sectional area for both the armature and the air gap is 50 cm². For this problem, use μ = 4π × 10-7 H/m.
Hint: You may want to convert to Sl units (e.g., meters) before beginning your calculations.
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VIEWStep 2: Calculation of reluctance of the air gap.
VIEWStep 3: Calculation of reluctance of the upper segment of the fixed element.
VIEWStep 4: Calculation of reluctance of the lower segment of the fixed element.
VIEWStep 5: Calculation of reluctance of the armature.
VIEWStep 6: Calculation of total reluctance of the system.
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