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Calculate the inductance, LL, for the following long coils: (Note: 1 m=100 cm1\text{ m}=100\text{ cm} and 1 m2=10,000 cm21\text{ m}^2=10,000\text{ cm}^2.) a. air core, 2020 turns, area 3.14 cm23.14\text{ cm}^2, length 25 cm25\text{ cm}. b. same coil as step a with ferrite core having a μr\mu_\text{r} of 50005000. c. air core, 200200 turns, area 3.14 cm23.14\text{ cm}^2, length 25 cm25\text{ cm}. d. air core, 2020 turns, area 3.14 cm23.14\text{ cm}^2, length 50 cm50\text{ cm}. e. iron core with μr\mu_\text{r} of 20002000, 100100 turns, area 5 cm25\text{ cm}^2, length 10 cm10\text{ cm}.

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Answered 1 year ago
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In this exercise, we need to find the inductance LL.

In part (a)

  • μr=1,\mu_r=1, the permeability of an air-core.
  • N=20N=20, number of turns in the coil.
  • A=3.14cm2A=3.14\,\mathrm{cm^2}, area of each turn.
  • l=25cml=25\,\mathrm{cm}, length of coil.

In part (b)

  • μr=5000,\mu_r=5000, the permeability of a ferrite-core.
  • N=20N=20, number of turns in the coil.
  • A=3.14cm2A=3.14\,\mathrm{cm^2}, area of each turn.
  • l=25cml=25\,\mathrm{cm}, length of coil.

In part (c)

  • μr=1,\mu_r=1, the permeability of a ferrite-core.
  • N=200N=200, number of turns in the coil.
  • A=3.14cm2A=3.14\,\mathrm{cm^2}, area of each turn.
  • l=25cml=25\,\mathrm{cm}, length of coil.

In part (d)

  • μr=1,\mu_r=1, the permeability of a ferrite-core.
  • N=20N=20, number of turns in the coil.
  • A=3.14cm2A=3.14\,\mathrm{cm^2}, area of each turn.
  • l=50cml=50\,\mathrm{cm}, length of coil.

In part (e)

  • μr=2000,\mu_r=2000, the permeability of a iron-core.
  • N=100N=100, number of turns in the coil.
  • A=5cm2A=5\,\mathrm{cm^2}, area of each turn.
  • l=10cml=10\,\mathrm{cm}, length of coil.

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