Question
Download Solution PDFThe magnetic susceptibility of the oxygen gas at 20°C is 400 π × 10-11 H/m. The absolute and relative permeabilities are respectively
This question was previously asked in
KVS TGT WET (Work Experience Teacher) 8 Jan 2017 Official Paper
Answer (Detailed Solution Below)
Option 1 : 4.04 π × 10-7 H/m, 1.01
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Detailed Solution
Download Solution PDFConcept:
Susceptibility is defined as the ratio of the intensity of magnetization (I) to magnetizing force (H) i.e
K = I/H
RELATION BETWEEN B,H, I AND k
When a magnetic material (say iron) of cross-sectional area A, and relative permeability 𝜇ᵣ is placed in a uniform field of intensity H, two types of lines of induction pass through it: one due to the magnetizing field H and another due to the iron piece itself being magnetized by induction.
Thus total flux density B will be given by
B = μ0H + I
Thus total flux density B will be given by
B = μ0H + I
where,
μ0 → permeability of free space.
Now, absolute permeability is given by μa
\(μ_{a} = μ_{0}μ_{r}= \frac{B}{H} =\frac{μ_{0}H+I}{H}=μ_{0}+\frac{I}{H} = μ_{0} + K\)
Dividing either side by μ0, we get the relative permeability.
\(μ_{r} =1+\frac{K}{\mu_{0}} \)
Calculations:
Given;
K = 400 π × 10-11 H/m
μ0 → permeability of free space = 4π × 10-7
Then;
\(μ_{r} =1+\frac{K}{\mu_{0}} = 1+\frac{400\pi \times 10^{-11}}{4\pi \times 10^{-7}} = 1.01\)
and
\(\mu_{a} = \mu_{0}\mu_{r} = 4\pi \times 10^{-7}\times 1.01 = 4.04\pi \times 10^{-7}\)
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