# Finding Inductance

1. Jul 17, 2016

### Josh225

1. The problem statement, all variables and given/known data
See attached image.

2. Relevant equations
The only equation that I have learned to find inductance is : L= μ N2 A / ι

However, when I do this, I get the incorrect answer. I am not so sure what is going on in the given solution and was wondering if somebody could explain that to me.The last step of the solution shows the equation that I stated, but how come you arent able to just plug numbers in directly?

3. The attempt at a solution

L= (4 Pi x 10-7) (1002) (.25) / 1" = 3.14 x 10-3

#### Attached Files:

• ###### 20160717_131526-1.jpg
File size:
33.4 KB
Views:
42
2. Jul 17, 2016

You need to work completely in MKS. You need the area A in square meters-I think you simply put in the diameter in inches. You also need the length L in meters. Your answer will then be in Henrys.

3. Jul 17, 2016

### Josh225

Thanks!

Last edited: Jul 17, 2016
4. Jul 17, 2016

### Josh225

Also, when I plug all the correct numbers in and run the equation through, it comes out to .01568 and I have to move the decimal 3 places to the right (like the answer is in meters and I need to convert to millimeters).
I am honestly not so hot wih unit conversion, so I was wondering what an easy way would be for me to remember to move the decimal in this case.

Also, what is the significance of writing the answer in μH as opposed to just H?

5. Jul 17, 2016

$\mu$ means 1.0 E-6. I'd like to see your calculations, particularly for area $A$ to see what you got. The arithmetic here should be pretty routine. You might need a little practice with exponents, etc.

6. Jul 17, 2016

### Josh225

Hmmm I was under the impression that "μ" meant permeability and the letter following it represented a specific number (μo = 4 pi x 10-7) (still cant find that pi button on here!).

I kinda "cheated" and looked up some of the conversion.. I really do need to practice them, but to find "A" I did:
Pi (d)2 / 4
Pi (6.35 mm)2 /4
= 31.7 m2

7. Jul 17, 2016

Those are millimeters you have there. 1mm=1.0E-3 m. This will give you a 1.0E-6 (m^2) when it gets squared. As area goes, this cylinder has a small area (cross section). Meanwhile 31.7 m^2 is about as big as an average living room.

8. Jul 17, 2016

### Josh225

Oh dang haha. Quite a difference. So it would be 31.7 μm2... Which would actually equal 3.17E-5 m, right?

9. Jul 17, 2016

You need to call it 3.17 E-5 "m^2", (square meters), but otherwise correct.

10. Jul 17, 2016

### Josh225

Strange.. In my book it's expressed as 31.7 μm2.

11. Jul 17, 2016

That part is ok . I was simply correcting the missing "2" on your m^2 (at the end of post #8).

12. Jul 17, 2016

### Josh225

Oh! I see. Thank you for the info!

13. Jul 18, 2016

### Staff: Mentor

It depends on context. Certainly, $\mu$ is the symbol for permeability, both as $\mu_{\scriptsize 0}$ and relative permeability, $\mu_r$. But you have also seen $\mu$ denoting the coefficient of friction of a surface; and the most common encounter is in association with units as a dimensionless multiplier meaning ×10-6 and written as a prefix pronounced "micro", e.g., $7\ \mu V$.

14. Jul 18, 2016

Very good comment. For the OP: In the second part of the above problem, they use the letter $\mu_r$ for the permeability of the iron inside the inductor. To avoid any confusion, it would be nice if they could use a different letter for both the magnetic $\mu_o=4 \pi E-7$ and the magnetic $\mu_r =2000$, because the $\mu$ in $\mu m^2$ and $\mu H$ means 1.0E-6, but in the magnetic theory, $\mu$ (in this problem with a subscript) is the letter that the science people have chosen for these magnetic parameters.