Finding Inductance: Solving Homework Statement

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Discussion Overview

The discussion revolves around a homework problem related to calculating inductance using the formula L = μ N² A / l. Participants express confusion about unit conversions and the application of the formula, as well as the significance of different units for inductance.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states the formula for inductance but is unsure why their calculations yield an incorrect answer.
  • Another participant suggests that the area A must be in square meters and that the length L should also be in meters for the calculations to be correct.
  • A participant mentions their calculation for area using the diameter but later realizes they made a mistake in unit conversion, confusing square millimeters with square meters.
  • There is a discussion about the meaning of the symbol "μ," with some participants clarifying its use as both permeability and a prefix for micro (1.0E-6).
  • One participant expresses confusion about how to convert units and seeks advice on remembering unit conversions.
  • Another participant corrects a misunderstanding regarding the area calculation, emphasizing the need to square the conversion factor when changing units.
  • There is a clarification about the notation used for permeability and its context in the problem, highlighting potential confusion with the symbol "μ" in different contexts.

Areas of Agreement / Disagreement

Participants generally agree on the need for correct unit conversions but have differing views on the application of the formula and the interpretation of the symbol "μ." The discussion remains unresolved regarding the exact calculations and the proper use of units.

Contextual Notes

Limitations include unresolved mathematical steps related to unit conversions and the dependence on definitions of symbols used in the context of the problem.

Josh225
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Homework Statement


See attached image.

Homework 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?

Thank you in advanced!

The Attempt at a Solution



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

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Josh225 said:

Homework Statement


See attached image.

Homework 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?

Thank you in advanced!

The Attempt at a Solution



L= (4 Pi x 10-7) (1002) (.25) / 1" = 3.14 x 10-3[/B]
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.
 
Thanks!
 
Last edited:
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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?
 
## \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.
 
Charles Link said:
## \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.

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 can't 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
 
Josh225 said:
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 can't 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
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.
 
Charles Link said:
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.

Oh dang haha. Quite a difference. So it would be 31.7 μm2... Which would actually equal 3.17E-5 m, right?
 
Josh225 said:
Oh dang haha. Quite a difference. So it would be 31.7 μm2... Which would actually equal 3.17E-5 m, right?
You need to call it 3.17 E-5 "m^2", (square meters), but otherwise correct.
 
  • #10
Charles Link said:
You need to call it 3.17 E-5 "m^2", (square meters), but otherwise correct.

Strange.. In my book it's expressed as 31.7 μm2.
 
  • #11
Josh225 said:
Strange.. In my book it's expressed as 31.7 μm2.
That part is ok . I was simply correcting the missing "2" on your m^2 (at the end of post #8).
 
  • #12
Oh! I see. Thank you for the info!
 
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  • #13
Josh225 said:
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)
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##.
 
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  • #14
NascentOxygen said:
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##.
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.
 

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