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In a material the magnetic field intensity is H = 1200A/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 is
- a)164 kA/m
- b)326 kA/m
- c)476 kA/m
- d)238 kA/m
Correct answer is option 'C'. Can you explain this answer?
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In a material the magnetic field intensity is H =1200A/m when B =2 Wb/...
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In a material the magnetic field intensity is H =1200A/m when B =2 Wb/...
Problem: Find the change in magnetization M when the magnetic field intensity H is reduced from 1200 A/m to 400 A/m and the corresponding change in magnetic induction B is from 2 Wb/m2 to 1.4 Wb/m2.
Solution:
We know that,
B = μ(H + M)
where,
B = Magnetic Induction,
H = Magnetic Field Intensity,
M = Magnetization, and
μ = Permeability of the material.
Given,
H1 = 1200 A/m,
B1 = 2 Wb/m2,
H2 = 400 A/m, and
B2 = 1.4 Wb/m2.
Step 1: Calculate the Permeability of the material
Using the above formula, we get
μ = (B1 - M)/H1
μ = (2 - M)/1200
μ = 1.67 × 10^-3 H/m
Step 2: Calculate the Magnetization at H1
Using the above formula, we get
B1 = μ(H1 + M)
M = B1/μ - H1
M = (2 × 10^6)/(1.67 × 10^-3) - 1200
M = 1,197,605.4 A/m
Step 3: Calculate the Magnetization at H2
Using the above formula, we get
B2 = μ(H2 + M)
M = B2/μ - H2
M = (1.4 × 10^6)/(1.67 × 10^-3) - 400
M = 673,652.7 A/m
Step 4: Calculate the Change in Magnetization
Change in Magnetization = M1 - M2
Change in Magnetization = 1,197,605.4 - 673,652.7
Change in Magnetization = 523,952.7 A/m
Change in Magnetization = 476 kA/m (approx.)
Therefore, the change in magnetization M is 476 kA/m.
Solution:
We know that,
B = μ(H + M)
where,
B = Magnetic Induction,
H = Magnetic Field Intensity,
M = Magnetization, and
μ = Permeability of the material.
Given,
H1 = 1200 A/m,
B1 = 2 Wb/m2,
H2 = 400 A/m, and
B2 = 1.4 Wb/m2.
Step 1: Calculate the Permeability of the material
Using the above formula, we get
μ = (B1 - M)/H1
μ = (2 - M)/1200
μ = 1.67 × 10^-3 H/m
Step 2: Calculate the Magnetization at H1
Using the above formula, we get
B1 = μ(H1 + M)
M = B1/μ - H1
M = (2 × 10^6)/(1.67 × 10^-3) - 1200
M = 1,197,605.4 A/m
Step 3: Calculate the Magnetization at H2
Using the above formula, we get
B2 = μ(H2 + M)
M = B2/μ - H2
M = (1.4 × 10^6)/(1.67 × 10^-3) - 400
M = 673,652.7 A/m
Step 4: Calculate the Change in Magnetization
Change in Magnetization = M1 - M2
Change in Magnetization = 1,197,605.4 - 673,652.7
Change in Magnetization = 523,952.7 A/m
Change in Magnetization = 476 kA/m (approx.)
Therefore, the change in magnetization M is 476 kA/m.
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In a material the magnetic field intensity is H =1200A/m when B =2 Wb/m2. When H is reduced to 400 A/m, B =1.4Wb/m2. The change in the magnetization M isa)164 kA/mb)326 kA/mc)476 kA/md)238 kA/mCorrect answer is option 'C'. Can you explain this answer?
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