A ring has a mean diameter of 21 cm and cross sectional area of 10 cm2. the ring is made up of semi-circular sections of cast iron and cast steel with each joint having reluctance equal to an air gap of 0.2 mm. find the ampere turns required to produce a flux of 0.8 milli-wb. the relative permeability of cast steel and cast iron are 800 & 166 respectively. neglect fringing and leakage effects.
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# Answer- 1780.5 Ampere turns.
# Given-
Φ = 8×10^-4 Wb
A = 10^-3 m^2
# Solution-
Magnetic field is given by
B = Φ/A
B = 0.8 Wb/m^2
Air gap,
H = B/μ0
H = 6.366×10^5 AT/m
Total air gap length l = 4×10^-4 m
AT required = H×l
AT required = 254.64
Cast steel path, μr=800
H = B/μ0.μr
H = 795.77 AT/m
Length of cast steel path l = 0.3295 m
AT required = H×l
AT required = 262.20
Cast iron path, μ0=166
H = B/μ0.μr
H = 3835.05 AT/m
Length of cast iron path l = 0.3295 m
AT required = H×l
AT required = 1263.65
Total AT required = 254.64+262.2+1263
Total AT required = 1780.5 AT
Hope this was helpful...
# Answer- 1780.5 Ampere turns.
# Given-
Φ = 8×10^-4 Wb
A = 10^-3 m^2
# Solution-
Magnetic field is given by
B = Φ/A
B = 0.8 Wb/m^2
Air gap,
H = B/μ0
H = 6.366×10^5 AT/m
Total air gap length l = 4×10^-4 m
AT required = H×l
AT required = 254.64
Cast steel path, μr=800
H = B/μ0.μr
H = 795.77 AT/m
Length of cast steel path l = 0.3295 m
AT required = H×l
AT required = 262.20
Cast iron path, μ0=166
H = B/μ0.μr
H = 3835.05 AT/m
Length of cast iron path l = 0.3295 m
AT required = H×l
AT required = 1263.65
Total AT required = 254.64+262.2+1263
Total AT required = 1780.5 AT
Hope this was helpful...
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