## Related questions with answers

Question

Calculate the inductance, $L$, for the following long coils: (Note: $1\text{ m}=100\text{ cm}$ and $1\text{ m}^2=10,000\text{ cm}^2$.) a. air core, $20$ turns, area $3.14\text{ cm}^2$, length $25\text{ cm}$. b. same coil as step a with ferrite core having a $\mu_\text{r}$ of $5000$. c. air core, $200$ turns, area $3.14\text{ cm}^2$, length $25\text{ cm}$. d. air core, $20$ turns, area $3.14\text{ cm}^2$, length $50\text{ cm}$. e. iron core with $\mu_\text{r}$ of $2000$, $100$ turns, area $5\text{ cm}^2$, length $10\text{ cm}$.

Solution

VerifiedAnswered 1 year ago

Answered 1 year ago

Step 1

1 of 8**Information**

In this exercise, we need to find the inductance $L$.

In part (a)

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

In part (b)

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

In part (c)

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

In part (d)

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

In part (e)

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

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