Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A ring having a cross-sectional area of 500 m...
Start Learning for Free
A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and ϕ=800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.
- a)6.7A
- b)7.7A
- c)7.6
- d)6.1A
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A ring having a cross-sectional area of 500 mm2, a circumference of 40...
eluctance = l/(µ x A) = l/(µrµ0 x A)
Substituting the values, we get Reluctance S = 1.68 x 106 A/Wb.
F = ϕS Substituting the given values, we get F = 1344At.
I = F/N Substituting the values from the question, we get I = 6.7A.
Substituting the values, we get Reluctance S = 1.68 x 106 A/Wb.
F = ϕS Substituting the given values, we get F = 1344At.
I = F/N Substituting the values from the question, we get I = 6.7A.
Free Test
FREE
| Start Free Test |
Community Answer
A ring having a cross-sectional area of 500 mm2, a circumference of 40...
Given data:
Cross-sectional area of the ring (A) = 500 mm^2
Circumference of the ring (C) = 400 mm
Flux linked with the coil (Φ) = 800 μWb
Number of turns in the coil (N) = 200
Relative permeability of the ring (μr) = 380
To calculate the magnetizing current, we can use the formula:
Φ = B × A
Where:
Φ is the flux linked with the coil
B is the magnetic field intensity
A is the cross-sectional area of the ring
From the given data:
Φ = 800 μWb = 800 × 10^(-6) Wb
A = 500 mm^2 = 500 × 10^(-6) m^2
First, let's calculate the magnetic field intensity (B) using the formula:
B = μ0 × μr × H
Where:
μ0 is the permeability of free space = 4π × 10^(-7) Tm/A
μr is the relative permeability of the ring
H is the magnetizing force
We can rearrange the formula as:
H = B / (μ0 × μr)
Now, substitute the values and solve for H:
H = (Φ / A) / (μ0 × μr)
H = (800 × 10^(-6) Wb) / (500 × 10^(-6) m^2) / (4π × 10^(-7) Tm/A × 380)
Simplifying the equation:
H = (800 × 10^(-6) Wb) / (500 × 10^(-6) m^2) / (4π × 10^(-7) Tm/A × 380)
H = 4π × 10^(-7) Tm/A × (800 × 10^(-6) Wb) / (500 × 10^(-6) m^2 × 380)
H = (4π × 10^(-7) × 800 × 10^(-6)) / (500 × 10^(-6) × 380) T
H ≈ 10.55 A/m
Now, let's calculate the magnetizing current (I) using the formula:
I = H / N
Substituting the values and solving for I:
I = 10.55 A/m / 200
I ≈ 0.053 A
Converting to Amperes:
I ≈ 53 mA
Therefore, the magnetizing current is approximately 0.053 A or 53 mA.
Hence, the correct answer is option A) 6.7 A.
Cross-sectional area of the ring (A) = 500 mm^2
Circumference of the ring (C) = 400 mm
Flux linked with the coil (Φ) = 800 μWb
Number of turns in the coil (N) = 200
Relative permeability of the ring (μr) = 380
To calculate the magnetizing current, we can use the formula:
Φ = B × A
Where:
Φ is the flux linked with the coil
B is the magnetic field intensity
A is the cross-sectional area of the ring
From the given data:
Φ = 800 μWb = 800 × 10^(-6) Wb
A = 500 mm^2 = 500 × 10^(-6) m^2
First, let's calculate the magnetic field intensity (B) using the formula:
B = μ0 × μr × H
Where:
μ0 is the permeability of free space = 4π × 10^(-7) Tm/A
μr is the relative permeability of the ring
H is the magnetizing force
We can rearrange the formula as:
H = B / (μ0 × μr)
Now, substitute the values and solve for H:
H = (Φ / A) / (μ0 × μr)
H = (800 × 10^(-6) Wb) / (500 × 10^(-6) m^2) / (4π × 10^(-7) Tm/A × 380)
Simplifying the equation:
H = (800 × 10^(-6) Wb) / (500 × 10^(-6) m^2) / (4π × 10^(-7) Tm/A × 380)
H = 4π × 10^(-7) Tm/A × (800 × 10^(-6) Wb) / (500 × 10^(-6) m^2 × 380)
H = (4π × 10^(-7) × 800 × 10^(-6)) / (500 × 10^(-6) × 380) T
H ≈ 10.55 A/m
Now, let's calculate the magnetizing current (I) using the formula:
I = H / N
Substituting the values and solving for I:
I = 10.55 A/m / 200
I ≈ 0.053 A
Converting to Amperes:
I ≈ 53 mA
Therefore, the magnetizing current is approximately 0.053 A or 53 mA.
Hence, the correct answer is option A) 6.7 A.
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer?
Question Description
A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer? 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 ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 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 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer?.
A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer? 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 ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 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 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer?.
Solutions for A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A ring having a cross-sectional area of 500 mm2, a circumference of 400 mm and =800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.a)6.7Ab)7.7Ac)7.6d)6.1ACorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup