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A coil of 200 turns is wound uniformly over an iron ring of mean circumference of 80cm and cross-sectional area of 0.6cm2 . If current through the coil is 2 ampere Calculate: a) Magnetizing force b) Reluctance 3) Total flux 4) Flux density ir relative permeability of ring is 200?
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A coil of 200 turns is wound uniformly over an iron ring of mean circu...
Solution:
a) Magnetizing force:
Magnetizing force, H = (nI)/l
where n = number of turns = 200
I = current through the coil = 2 A
l = mean length of the magnetic path = mean circumference of the iron ring = 80 cm = 0.8 m
Therefore, H = (200 x 2)/0.8 = 500 A/m
b) Reluctance:
Reluctance, S = (l)/(μA)
where l = mean length of the magnetic path = mean circumference of the iron ring = 80 cm = 0.8 m
μ = relative permeability of the ring = 200
A = cross-sectional area of the ring = 0.6 cm2 = 6 x 10^-5 m^2
Therefore, S = (0.8)/(200 x 6 x 10^-5) = 6,667.67 A/Wb
3) Total flux:
Total flux, Φ = nI
where n = number of turns = 200
I = current through the coil = 2 A
Therefore, Φ = 200 x 2 = 400 Wb
4) Flux density or relative permeability of ring is 200?
Flux density, B = (μΦ)/(lA)
where μ = relative permeability of the ring = 200
Φ = total flux = 400 Wb
l = mean length of the magnetic path = mean circumference of the iron ring = 80 cm = 0.8 m
A = cross-sectional area of the ring = 0.6 cm2 = 6 x 10^-5 m^2
Therefore, B = (200 x 400)/(0.8 x 6 x 10^-5) = 66,666,667 T or 66.67 MT
The flux density is very high, indicating that the iron ring is highly magnetized. The high value of relative permeability (200) also confirms this. This means that the iron ring is a good magnetic material and can be used for various magnetic applications.
a) Magnetizing force:
Magnetizing force, H = (nI)/l
where n = number of turns = 200
I = current through the coil = 2 A
l = mean length of the magnetic path = mean circumference of the iron ring = 80 cm = 0.8 m
Therefore, H = (200 x 2)/0.8 = 500 A/m
b) Reluctance:
Reluctance, S = (l)/(μA)
where l = mean length of the magnetic path = mean circumference of the iron ring = 80 cm = 0.8 m
μ = relative permeability of the ring = 200
A = cross-sectional area of the ring = 0.6 cm2 = 6 x 10^-5 m^2
Therefore, S = (0.8)/(200 x 6 x 10^-5) = 6,667.67 A/Wb
3) Total flux:
Total flux, Φ = nI
where n = number of turns = 200
I = current through the coil = 2 A
Therefore, Φ = 200 x 2 = 400 Wb
4) Flux density or relative permeability of ring is 200?
Flux density, B = (μΦ)/(lA)
where μ = relative permeability of the ring = 200
Φ = total flux = 400 Wb
l = mean length of the magnetic path = mean circumference of the iron ring = 80 cm = 0.8 m
A = cross-sectional area of the ring = 0.6 cm2 = 6 x 10^-5 m^2
Therefore, B = (200 x 400)/(0.8 x 6 x 10^-5) = 66,666,667 T or 66.67 MT
The flux density is very high, indicating that the iron ring is highly magnetized. The high value of relative permeability (200) also confirms this. This means that the iron ring is a good magnetic material and can be used for various magnetic applications.
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A coil of 200 turns is wound uniformly over an iron ring of mean circumference of 80cm and cross-sectional area of 0.6cm2 . If current through the coil is 2 ampere Calculate: a) Magnetizing force b) Reluctance 3) Total flux 4) Flux density ir relative permeability of ring is 200?
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A coil of 200 turns is wound uniformly over an iron ring of mean circumference of 80cm and cross-sectional area of 0.6cm2 . If current through the coil is 2 ampere Calculate: a) Magnetizing force b) Reluctance 3) Total flux 4) Flux density ir relative permeability of ring is 200? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A coil of 200 turns is wound uniformly over an iron ring of mean circumference of 80cm and cross-sectional area of 0.6cm2 . If current through the coil is 2 ampere Calculate: a) Magnetizing force b) Reluctance 3) Total flux 4) Flux density ir relative permeability of ring is 200? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A coil of 200 turns is wound uniformly over an iron ring of mean circumference of 80cm and cross-sectional area of 0.6cm2 . If current through the coil is 2 ampere Calculate: a) Magnetizing force b) Reluctance 3) Total flux 4) Flux density ir relative permeability of ring is 200?.
A coil of 200 turns is wound uniformly over an iron ring of mean circumference of 80cm and cross-sectional area of 0.6cm2 . If current through the coil is 2 ampere Calculate: a) Magnetizing force b) Reluctance 3) Total flux 4) Flux density ir relative permeability of ring is 200? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A coil of 200 turns is wound uniformly over an iron ring of mean circumference of 80cm and cross-sectional area of 0.6cm2 . If current through the coil is 2 ampere Calculate: a) Magnetizing force b) Reluctance 3) Total flux 4) Flux density ir relative permeability of ring is 200? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A coil of 200 turns is wound uniformly over an iron ring of mean circumference of 80cm and cross-sectional area of 0.6cm2 . If current through the coil is 2 ampere Calculate: a) Magnetizing force b) Reluctance 3) Total flux 4) Flux density ir relative permeability of ring is 200?.
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