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What is the relative permeability of an air-core inductor, if its self-inductance increases from 0.5 mH to 50 mH due to the introduction of an iron core into it?
  • a)
    100
  • b)
    0.1
  • c)
    10000
  • d)
    0.0001
Correct answer is option 'A'. Can you explain this answer?
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What is the relative permeability of an air-core inductor, if its self...
Explanation:

Relative permeability is defined as the ratio of the permeability of a material to the permeability of free space (µ0). It is denoted by the symbol µr.

µr = µ/µ0

where µ is the permeability of the material and µ0 is the permeability of free space.

Given, self-inductance of air-core inductor (L1) = 0.5 mH
Self-inductance of iron-core inductor (L2) = 50 mH

The self-inductance of an inductor is given by the formula:

L = µr * µ0 * (N^2 * A) / l

where L is the self-inductance, N is the number of turns, A is the area of the coil, l is the length of the coil, µr is the relative permeability of the core material, and µ0 is the permeability of free space.

Since the number of turns, area, and length of the coil are constant, we can write:

L2/L1 = µr2/µr1

where µr2 is the relative permeability of the iron core and µr1 is the relative permeability of air.

Substituting the given values, we get:

50/0.5 = µr2/µr1

µr2/µr1 = 100

Therefore, the relative permeability of air-core inductor is 100.
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What is the relative permeability of an air-core inductor, if its self-inductance increases from 0.5 mH to 50 mH due to the introduction of an iron core into it?a)100b)0.1c)10000d)0.0001Correct answer is option 'A'. Can you explain this answer?
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