FluX Density

  • Thread starter LandonV
  • Start date
  • #1
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Homework Statement



Loop like an "O" with current moving to the left
Diameter = 8cm
Current = 15A
FIND THE FLUX DENSITY OF THE CURRENT LOOP IN THE PLACE OF THE LOOP?

Homework Equations



B(flux density)=F(magnetic flux)/A(area)
B=H(magnetic field strength)*Permeability
H(magnetic field strength)=I(current)/A(area)
A(area)=Pi*r(radius)2
L(inductance)=Ioda(magnetic flux)/I(current)
Ioda(magnetic flux)=V(velocity)xT(time)
(permeability)=L(inductance)/d(diameter)

The Attempt at a Solution



Area = 50.24 = (3.14 *(42))
Magnetic Field Strangth = .299 or .3 = (15/50.24)

From this point I am lost because it appears to me that I am missing a key part to finish the equation. If I can find out the time or the charge then I can do some substitions and find the answer the equation. Can anyone point me in the right direction?

These are the possible answers:

1. 1.68 x 10^-5 Wb/meters squared
2. 2.36 x 10^-5 Wb/meters squared
3. 1.68 x 10^-4 Wb/meters squared
4. 2.36 x 10^-4 Wb/meters squared
5. 2.87 x 10 Wb/meters squared
 
Last edited:

Answers and Replies

  • #2
277
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That second equation might come in handy.
 
  • #3
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That en lies my problem...

For that equation, I need to find permeability!
To find permeability I need to find inductance
To find Inductance I need to find magnetic flux
To find magnetic flux I need to find velocity and time!

How am I to find 2 items to which I am unable to derive a possible answer for?
 
  • #4
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can no one show me what it is that I am missing in this problem?
 
  • #5
Dick
Science Advisor
Homework Helper
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Isn't this just the Biot-Savart law? That should be all you need. You have too many equations.
 
  • #6
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Well I was given no equations to work with because i missed this day in class and im playing catch up so I am lost. These were the "relevant" equations that I could find. Im gonna have to look up Biot...appreciate the possible direction point!
 
  • #7
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After doing some more investigation, I tried another avenue but to no avail.

Permeability = 4Pi x 10^-7 H/m (IN A VACUUM)

B(magnetic flux) = H(magnetic field strength) * permeability
B = .3*(4Pi x 10^-7)
B = 3.768 x 10^-7

But it is not the answer according to my instructor...he gave me a list of options as show above but I cant seem to get there. Anyone?
 
  • #8
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QUESTION AND EQUATION ANSWERED!!!!

B = permeability x current / 2 x radius

B = (4Pi x 10^-7) x 15A / 2 x (.04 meters)

B = 2.355 x 10^-4 (rounded off = 2.36 x 10^-4)
 
  • #9
277
1
B=H(magnetic field strength)*Permeability
H(magnetic field strength)=I(current)/A(area)
so
B= Permeability x Current / Area

B = permeability x current / 2 x radius
2 x radius looks strongly like a diameter rather than an area
 
  • #10
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Yeha im not exactly sure how any of this works because I missed the last 2 classes and im playing catch up and it sucks but the list of equations that I originally found were the ones I listed and they didnt help me in anyway but after doing some reading I found the equation that I used in my last post and that gave me the answer I was looking for so I have no clue how these equations supposedly tie in together but my instructor said my answer was correct so Im not complaining about that...just sucks that I can't figure out how this all ties in together
 

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