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A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.?
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A coil is wound uniformly with 300 turns over a steel ring of relative...
Solution:
(i) Calculation of m.m.f.
m.m.f. = N × I
Where N is the number of turns and I is the current passing through the coil.
Given that, N = 300 and I = 25 A
Therefore, m.m.f. = 300 × 25 = 7500 ampere-turns (AT)
(ii) Calculation of reluctance of the ring
Reluctance (S) is defined as the opposition offered by the magnetic circuit to the flow of magnetic flux.
S = l / (μA)
Where l is the mean length of the magnetic path, A is the cross-sectional area of the magnetic path, and μ is the relative permeability of the steel ring.
Given that, l = mean circumference = 40 mm
A = cross-sectional area = 50 mm^2
μ = 900
Therefore, S = 40 / (900 × 50 × 10^-6) = 0.00089 H^-1
(iii) Calculation of flux
Flux (Φ) is defined as the total magnetic field passing through a given area.
Φ = m.m.f. / S
Given that, m.m.f. = 7500 AT and S = 0.00089 H^-1
Therefore, Φ = 7500 / 0.00089 = 842696.62 Wb (Weber)
Conclusion:
Thus, the m.m.f. of the coil is 7500 AT, the reluctance of the ring is 0.00089 H^-1, and the flux passing through the ring is 842696.62 Wb.
(i) Calculation of m.m.f.
m.m.f. = N × I
Where N is the number of turns and I is the current passing through the coil.
Given that, N = 300 and I = 25 A
Therefore, m.m.f. = 300 × 25 = 7500 ampere-turns (AT)
(ii) Calculation of reluctance of the ring
Reluctance (S) is defined as the opposition offered by the magnetic circuit to the flow of magnetic flux.
S = l / (μA)
Where l is the mean length of the magnetic path, A is the cross-sectional area of the magnetic path, and μ is the relative permeability of the steel ring.
Given that, l = mean circumference = 40 mm
A = cross-sectional area = 50 mm^2
μ = 900
Therefore, S = 40 / (900 × 50 × 10^-6) = 0.00089 H^-1
(iii) Calculation of flux
Flux (Φ) is defined as the total magnetic field passing through a given area.
Φ = m.m.f. / S
Given that, m.m.f. = 7500 AT and S = 0.00089 H^-1
Therefore, Φ = 7500 / 0.00089 = 842696.62 Wb (Weber)
Conclusion:
Thus, the m.m.f. of the coil is 7500 AT, the reluctance of the ring is 0.00089 H^-1, and the flux passing through the ring is 842696.62 Wb.
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A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.?
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A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.?.
A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.?.
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A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.?, a detailed solution for A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.? has been provided alongside types of A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40 mm and cross-sectional area of 50 mm2 . If a current of 25 amps is passed through the coil, find (i) m.m.f. (ii) reluctance of the ring and (iii) flux.? theory, EduRev gives you an
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