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Two magnetically-coupled coils have a mutual inductance of 32 mH. Given that one coil has twice the number of turns than the other, calculate the inductance of each coil. (Take relative permeability, μ = 1, and μο = 4πχ10-7) (19)

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Final answer:

The inductance of Coil 1 is 128 mH and the inductance of Coil 2 is 8 mH.

Step-by-step explanation:

The inductance of a coil is given by the formula:

L = (μ * N² * A) / l

Where:

  • L is the inductance
  • N is the number of turns in the coil
  • A is the cross-sectional area of the coil
  • l is the length of the coil
  • μ is the permeability of free space

In this case, we have two coils, Coil 1 and Coil 2. Given that Coil 1 has twice the number of turns as Coil 2, we can denote the number of turns as N₁ = 2N₂.

Since the mutual inductance between the coils is given as 32 mH and the relative permeability (μ) is 1, we can use the formula:

M = √(L₁ * L₂)

Substituting in the given values and solving for the inductance of each coil, we get:

L₁ = 32 mH * (2²) = 128 mH

L₂ = 32 mH / (2²) = 8 mH

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