Find flux density

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  • #1
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an iron ring of mean length 5 cma has an air gap of 1mm and a winding of 200 turns. If the permeability of iron is 300 when 1A current flows through the coil find the flux density.


its answer is 94.2 mWb/m^3

kindly help how I would drive it
 

Answers and Replies

  • #2
berkeman
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an iron ring of mean length 5 cma has an air gap of 1mm and a winding of 200 turns. If the permeability of iron is 300 when 1A current flows through the coil find the flux density.


its answer is 94.2 mWb/m^3

kindly help how I would drive it

Welcome to the PF. Try listing the Relevant Equations for this type of problem. There is an analogy with resistive circuits, where the current is determined by the driving voltage and the sum of the resistances in the circuit. Can you state whqt the analogous equations are for a magnetic circuit like this one?
 
  • #3
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hi dear,

there is no person to help me in this
 
  • #4
berkeman
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hi dear,

there is no person to help me in this

Then use wikipedia and other learning resources to help you:

http://en.wikipedia.org/wiki/Magnetic_circuit

We can offer tutorial hints here (as I have done above), but we do not do the work for you.

What class is this for? What learning resources do you have from the class?
 
  • #5
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hi dear,


MMF = Ni


B= [tex]\mu[/tex]H

[tex]\mu[/tex] = [tex]\mu[/tex]r[tex]\mu[/tex]o


[tex]\mu[/tex]NA


H = Ni/Length
 
  • #6
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I am not to sure what is meant by the air gap of 1mm. Also the mean length of the iron ring? It that the diameter. Man this is poorly worded.

First all what does flux density mean? How do you think you are going to calculate using the information you are given?
 
  • #7
berkeman
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I am not to sure what is meant by the air gap of 1mm. Also the mean length of the iron ring? It that the diameter. Man this is poorly worded.

That's how this type of air-gap transformer/inductor problem is usually worded. The mean diameter is basically down the middle of the ring, so it's a 5cm ring with an air gap of 1mm (missing magnetic material).
 
  • #8
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hi dear,

this is as as same worded as I have
 
  • #9
125
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That makes sense now.
 
  • #10
berkeman
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hi dear,

this is as as same worded as I have

So now you can solve your problem. Use the link I provided, and write out the equations for the reluctances and the other parts of the problem. Please show us your work as you solve for the flux...
 
  • #11
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4
an iron ring of mean length 5 cma has an air gap of 1mm and a winding of 200 turns. If the permeability of iron is 300 when 1A current flows through the coil find the flux density.


its answer is 94.2 mWb/m^3

kindly help how I would drive it

1) Find reluctances in the air gap and in the core
2) Find emf (or F)
3) Find flux
4) Find the flux denisty

I would leave it to you for finding all those formulas and relationships between reluctance, F, flux, and flux density.

Hint: You should already know this that you can build a electric circuit from above and treat F as V, flux as I and reluctance as R
 
  • #12
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thank you root x I am at work at this moment as soon as Igo back I would perform it and then tell you thank you
 
  • #13
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hi dear,

kindly guide me that here we found mmf or emf
 
  • #14
berkeman
Mentor
61,341
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hi dear,

kindly guide me that here we found mmf or emf

Use the link in post #4, and rootX's hints in post #11, and show us your work in detail.

We will not do this problem for you. You have all that you need to work out the problem. Now show us your full work.
 
  • #15
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Hi dear,,

how are you kindly help me the equation you tell me to drive for reluctance in this the area is not given and I have not know equation to drive it kidly help
 
  • #16
ideasrule
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Hi dear,,

how are you kindly help me the equation you tell me to drive for reluctance in this the area is not given and I have not know equation to drive it kidly help

Call the area "A" and proceed with the calculation.
 
  • #17
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hi dears but how can we put 2 unknown in a same equation for example for reluctance we need area S= l/ mu * A.


and same for flux weber= NI / l/munot *mur*A
 
  • #18
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hi dear,

sory for late reply I was out of city as per your quidence I had put the area as A


now I have

total reluctance 24.3 A(area) AT/wb

H = 40.81 * 1o^2 AT/m


F = .5020 A(area) AT/m


now if I go to fine the E.M.F as per your guidence

i have this formula current*resistance=e.m.f

to find resistance I have R=conductance l/A

but still I have the problem of Area kindly help\me now with the equation from which I can find out the area


Kindly help me quickly I needed it urgent
 
  • #19
15
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hi dears,


sorry for late reply acctualy I was out of city


as per your instructions I had put the area as a A


and I find out these total reluctance 24.3 A(area)AT/wb


magnetising force H = 40.81*10^2 AT/m


F = .5020 A(area) AT/wb

now to find out the flux or resistance agian I need the value of area which I have a problem

kndly help me with the equation with which I can find the area
 
  • #20
2
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GIVEN:
L of Core=.5 m
Air Gap=10^3 m
N=200 turns
Permeability of iron(uFe)=300
I=1A
Uo=4pix10^-7
find B?

from uNI/L=B ;since B=Hu
therefore we have,NI=BL/u
NI(total)=NI(AIR)+NI(IRON)
(200)(1)=B(.5)/uFexU0 + B(10^3)/u(air)xUo
solve for B;
B=.094247779 Wb/m^2

hope you learn it (from heart)
from Philippines
 
  • #21
1
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GIVEN:
L of Core=.5 m
Air Gap=10^3 m
N=200 turns
Permeability of iron(uFe)=300
I=1A
Uo=4pix10^-7
find B?

from uNI/L=B ;since B=Hu
therefore we have,NI=BL/u
NI(total)=NI(AIR)+NI(IRON)
(200)(1)=B(.5)/uFexU0 + B(10^3)/u(air)xUo
solve for B;
B=.094247779 Wb/m^2

hope you learn it (from heart)
from Philippines

what is the value of u(air)...?
 

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