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A ring having a cross-sectional area of 500 mm2, a circumference of and ϕ=800microWb 400 mm has a coil of 200 turns wound around it. Calculate the reluctance.
- a)1.68 * 10-4A/Wb
- b)1.68 * 104 A/Wb
- c)1.68 * 106 A/Wb
- d)1.68 * 10-6A/Wb
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A ring having a cross-sectional area of 500 mm2, a circumference of an...
From the given question:
Flux density= 800*10-6/500*106=1.6 Wb/m2.
Reluctance= 0.4/(380*4*pi*10-7*10-4*5)=1.68 * 106 A/Wb.
Flux density= 800*10-6/500*106=1.6 Wb/m2.
Reluctance= 0.4/(380*4*pi*10-7*10-4*5)=1.68 * 106 A/Wb.
Most Upvoted Answer
A ring having a cross-sectional area of 500 mm2, a circumference of an...
To calculate the reluctance of a ring with a coil wound around it, we need to understand the concept of reluctance and how it is related to magnetic flux.
Reluctance is a measure of how difficult it is for magnetic flux to flow through a material or a magnetic circuit. It is analogous to resistance in an electric circuit. The formula for reluctance is given by:
Reluctance (R) = Magnetic Flux (Φ) / Magnetomotive Force (MMF)
The magnetic flux (Φ) is given by:
Φ = B * A
where B is the magnetic field intensity and A is the cross-sectional area.
The magnetomotive force (MMF) is given by:
MMF = N * I
where N is the number of turns in the coil and I is the current flowing through the coil.
In this case, we are given the cross-sectional area (A) of the ring as 500 mm^2, the circumference of the ring as 400 mm, and the number of turns (N) as 200.
Step 1: Calculate the cross-sectional area of the ring
The cross-sectional area of the ring is given by the formula:
A = π * r^2
where r is the radius of the ring. Since the circumference of the ring is given as 400 mm, we can calculate the radius as:
C = 2 * π * r
400 = 2 * π * r
r = 400 / (2 * π) = 63.66 mm
Therefore, the cross-sectional area of the ring is:
A = π * (63.66)^2 = 3183 mm^2
Step 2: Calculate the magnetic flux (Φ)
The magnetic flux is given by the formula:
Φ = B * A
We are not given the magnetic field intensity (B), so we cannot directly calculate the magnetic flux.
Step 3: Calculate the magnetomotive force (MMF)
The magnetomotive force is given by the formula:
MMF = N * I
We are not given the current (I), so we cannot directly calculate the magnetomotive force.
Step 4: Calculate the reluctance (R)
The reluctance is given by the formula:
R = Φ / MMF
Substituting the formulas for magnetic flux and magnetomotive force, we get:
R = (B * A) / (N * I)
Since we do not have the values for B and I, we cannot directly calculate the reluctance.
Therefore, it is not possible to calculate the reluctance with the given information. The correct answer cannot be determined.
Reluctance is a measure of how difficult it is for magnetic flux to flow through a material or a magnetic circuit. It is analogous to resistance in an electric circuit. The formula for reluctance is given by:
Reluctance (R) = Magnetic Flux (Φ) / Magnetomotive Force (MMF)
The magnetic flux (Φ) is given by:
Φ = B * A
where B is the magnetic field intensity and A is the cross-sectional area.
The magnetomotive force (MMF) is given by:
MMF = N * I
where N is the number of turns in the coil and I is the current flowing through the coil.
In this case, we are given the cross-sectional area (A) of the ring as 500 mm^2, the circumference of the ring as 400 mm, and the number of turns (N) as 200.
Step 1: Calculate the cross-sectional area of the ring
The cross-sectional area of the ring is given by the formula:
A = π * r^2
where r is the radius of the ring. Since the circumference of the ring is given as 400 mm, we can calculate the radius as:
C = 2 * π * r
400 = 2 * π * r
r = 400 / (2 * π) = 63.66 mm
Therefore, the cross-sectional area of the ring is:
A = π * (63.66)^2 = 3183 mm^2
Step 2: Calculate the magnetic flux (Φ)
The magnetic flux is given by the formula:
Φ = B * A
We are not given the magnetic field intensity (B), so we cannot directly calculate the magnetic flux.
Step 3: Calculate the magnetomotive force (MMF)
The magnetomotive force is given by the formula:
MMF = N * I
We are not given the current (I), so we cannot directly calculate the magnetomotive force.
Step 4: Calculate the reluctance (R)
The reluctance is given by the formula:
R = Φ / MMF
Substituting the formulas for magnetic flux and magnetomotive force, we get:
R = (B * A) / (N * I)
Since we do not have the values for B and I, we cannot directly calculate the reluctance.
Therefore, it is not possible to calculate the reluctance with the given information. The correct answer cannot be determined.
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A ring having a cross-sectional area of 500 mm2, a circumference of and ϕ=800microWb 400 mm has a coil of 200 turns wound around it. Calculate the reluctance.a)1.68 * 10-4A/Wbb)1.68 * 104A/Wbc)1.68 * 106A/Wbd)1.68 * 10-6A/WbCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 ring having a cross-sectional area of 500 mm2, a circumference of and ϕ=800microWb 400 mm has a coil of 200 turns wound around it. Calculate the reluctance.a)1.68 * 10-4A/Wbb)1.68 * 104A/Wbc)1.68 * 106A/Wbd)1.68 * 10-6A/WbCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ring having a cross-sectional area of 500 mm2, a circumference of and ϕ=800microWb 400 mm has a coil of 200 turns wound around it. Calculate the reluctance.a)1.68 * 10-4A/Wbb)1.68 * 104A/Wbc)1.68 * 106A/Wbd)1.68 * 10-6A/WbCorrect answer is option 'C'. Can you explain this answer?.
A ring having a cross-sectional area of 500 mm2, a circumference of and ϕ=800microWb 400 mm has a coil of 200 turns wound around it. Calculate the reluctance.a)1.68 * 10-4A/Wbb)1.68 * 104A/Wbc)1.68 * 106A/Wbd)1.68 * 10-6A/WbCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 ring having a cross-sectional area of 500 mm2, a circumference of and ϕ=800microWb 400 mm has a coil of 200 turns wound around it. Calculate the reluctance.a)1.68 * 10-4A/Wbb)1.68 * 104A/Wbc)1.68 * 106A/Wbd)1.68 * 10-6A/WbCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ring having a cross-sectional area of 500 mm2, a circumference of and ϕ=800microWb 400 mm has a coil of 200 turns wound around it. Calculate the reluctance.a)1.68 * 10-4A/Wbb)1.68 * 104A/Wbc)1.68 * 106A/Wbd)1.68 * 10-6A/WbCorrect answer is option 'C'. Can you explain this answer?.
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