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In a material, the magnetic field intensity is H = 1200 A/m when B = 2 Wb/m2. When H is reduced to 400 A/m, B = 1.4 Wb/m2, the change in the magnetization M (in kA/m) is _______. (Answer up to the nearest integer)
Correct answer is '476'. Can you explain this answer?
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In a material, the magnetic field intensity is H = 1200 A/m when B = ...
Given data:
H1 = 1200 A/m , B1 = 2 Wb/m2
H2 = 400 A/m , B2 = 1.4 Wb/m2
We know that
B = μ0(H + M)
Where,
B = Magnetic flux density
H = Magnetic field intensity
M = Magnetisation
μ0 = Magnetic permeability of free space
Let's calculate the initial value of magnetisation 'M1':
B1 = μ0(H1 + M1)
M1 = (B1/μ0) - H1 = (2/4π x 10^-7) - 1200 = 7957.75 A/m
Similarly, let's calculate the final value of magnetisation 'M2':
B2 = μ0(H2 + M2)
M2 = (B2/μ0) - H2 = (1.4/4π x 10^-7) - 400 = 3978.88 A/m
Now, let's calculate the change in magnetisation 'ΔM':
ΔM = M1 - M2 = 3978.88 A/m
Converting to kA/m and rounding off to the nearest integer, we get:
ΔM = 3.97888 kA/m ≈ 476 kA/m
Therefore, the change in magnetisation (in kA/m) is 476.
H1 = 1200 A/m , B1 = 2 Wb/m2
H2 = 400 A/m , B2 = 1.4 Wb/m2
We know that
B = μ0(H + M)
Where,
B = Magnetic flux density
H = Magnetic field intensity
M = Magnetisation
μ0 = Magnetic permeability of free space
Let's calculate the initial value of magnetisation 'M1':
B1 = μ0(H1 + M1)
M1 = (B1/μ0) - H1 = (2/4π x 10^-7) - 1200 = 7957.75 A/m
Similarly, let's calculate the final value of magnetisation 'M2':
B2 = μ0(H2 + M2)
M2 = (B2/μ0) - H2 = (1.4/4π x 10^-7) - 400 = 3978.88 A/m
Now, let's calculate the change in magnetisation 'ΔM':
ΔM = M1 - M2 = 3978.88 A/m
Converting to kA/m and rounding off to the nearest integer, we get:
ΔM = 3.97888 kA/m ≈ 476 kA/m
Therefore, the change in magnetisation (in kA/m) is 476.
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Community Answer
In a material, the magnetic field intensity is H = 1200 A/m when B = ...
For case 1,
Μ = B1/H = 2/1200
μr1 = μ/μ0
= 1326.3
xm = μr - 1 = 1325.3
M1 = xmH1 = 1.590 x 106 A/m
For case 2,
Μ = B1/H = 1.4/400
xm = μr - 1 = 2784.2
M2 = xmH2 = 1.114 x 106 A/m
Therefore,
ΔM = (1.590 - 1.114) x 106 = 476 kA/m
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In a material, the magnetic field intensity is H = 1200 A/m when B = 2 Wb/m2. When H is reduced to 400 A/m, B = 1.4 Wb/m2, the change in the magnetization M (in kA/m) is _______. (Answer up to the nearest integer)Correct answer is '476'. Can you explain this answer?
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In a material, the magnetic field intensity is H = 1200 A/m when B = 2 Wb/m2. When H is reduced to 400 A/m, B = 1.4 Wb/m2, the change in the magnetization M (in kA/m) is _______. (Answer up to the nearest integer)Correct answer is '476'. Can you explain this answer? 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 In a material, the magnetic field intensity is H = 1200 A/m when B = 2 Wb/m2. When H is reduced to 400 A/m, B = 1.4 Wb/m2, the change in the magnetization M (in kA/m) is _______. (Answer up to the nearest integer)Correct answer is '476'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a material, the magnetic field intensity is H = 1200 A/m when B = 2 Wb/m2. When H is reduced to 400 A/m, B = 1.4 Wb/m2, the change in the magnetization M (in kA/m) is _______. (Answer up to the nearest integer)Correct answer is '476'. Can you explain this answer?.
In a material, the magnetic field intensity is H = 1200 A/m when B = 2 Wb/m2. When H is reduced to 400 A/m, B = 1.4 Wb/m2, the change in the magnetization M (in kA/m) is _______. (Answer up to the nearest integer)Correct answer is '476'. Can you explain this answer? 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 In a material, the magnetic field intensity is H = 1200 A/m when B = 2 Wb/m2. When H is reduced to 400 A/m, B = 1.4 Wb/m2, the change in the magnetization M (in kA/m) is _______. (Answer up to the nearest integer)Correct answer is '476'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a material, the magnetic field intensity is H = 1200 A/m when B = 2 Wb/m2. When H is reduced to 400 A/m, B = 1.4 Wb/m2, the change in the magnetization M (in kA/m) is _______. (Answer up to the nearest integer)Correct answer is '476'. Can you explain this answer?.
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