Magnetic field from two wires distance

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Homework Statement




At one instant, the two conductors in a long
household extension cord carry equal 1.61 A
currents in opposite directions.
Find the magnetic field 2.4 cm away from
the middle of the straight cord, in the plane of
the two wires, if the centers of the two wires
are 2.92 mm apart.
Answer in units of µT.

At what distance is it one tenth as large?
Answer in units of cm.

The center wire in a coaxial cable carries current 1.61 A in one direction, and the sheath
around it carries current 1.61 A in the opposite direction. Consider a point P outside of
the sheath at a distance r from the middle of
the center wire. Assume the center wire without any sheath would create a magnetic field
of B at the point P.
What magnetic field does the cable and
sheath create at point P?
1. r B
2. None of these
3. More information is needed.
4. 2 B
5. 0.5√r B
6.√r B
7. 0.1 B
8.B/r
9. zero

Homework Equations



F = (mu/2(PI)*d)*(L)*(I1)*(I2)
B = (mu/2(PI))*(I/(d))


The Attempt at a Solution


I don't know how to do this problem at all, the equations I have don't even make since using because it is asking for the magnetic field. If I were to do the magnetic fields separately and add them would that give me the correct value? The only thing that matters since they are close together is the force that acts upon them correct? So if I were to do the wires separately:

d1 = 2.4cm - 2.92mm
d2 = 2.4cm + 2.92mm

B1 = (mu/2(PI))*(I/d1)
B2 = (mu/2(PI))*(I/d2)

B(at point A) = B1 + B2

This doesn't seem correct to me( I am studying for a test and this was a practice problem he gave us, hence I don't have the correct answer), is there any place I am going wrong about this?

On the third part I have no idea what it is asking at all...

Using the right hand rule I know that one of the magnetic fields is going into point A and one is going out of point A so would I subtract one from the other? I am lost, help is greatly appreciated.
 
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Answers and Replies

  • #2
jhae2.718
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If I were to do the magnetic fields separately and add them would that give me the correct value?
Yes. This is the principle of superposition.
 
  • #3
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All right then is the way I think of doing the fields correct?
 
  • #4
jhae2.718
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I think your expressions for the distances are incorrect. The 2.92 mm distance is the distance between the wires, and the 2.4 cm distance is from the middle of the wire, so if I'm intrepeting correctly I think the distances should be +/- 2.92/2 mm.

The magnetic fields of each wire should be given by [tex]\frac{\mu_0i}{2\pi r}[/tex] as you've said.
 
  • #5
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That is incorrect...

(4*PI e-7)/(2*PI) * (1.61/((2.4/100) - (0.00292)) + (4*PI e-7)/(2*PI) * (1.61/((2.4/100) + (0.00292)) = 27 micro T
 
  • #6
jhae2.718
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Ok, so if I'm wrong on the distance forget that. Otherwise your approach is correct.

Regarding the currents, since they are in different directions the magnetic fields of each wire will have a different sign, which you can determine by setting a coordinate system and using the right hand rule.

While it seems counterintuitive, magnetic fields (and electric fields, among others) combine linearly, so you can just add the contributions of each wire to find the total B-field.
 
  • #7
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Well I believe your distances are correct because I forgot that it said that the point is from the center of the extension cord. Could it be possible that one is negative and that I need to subtract it from the other? Why would there be a negative magnetic field?
 
  • #8
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I don't believe this is the right approach....
 
  • #9
jhae2.718
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The negative indicates direction; for example let's say that up is positive. Then the field that is up is positive, and the field that is down is negative.

If the total field is positive it is up under our coordinate system, and if it is negative it is down.
 
  • #10
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I understand that. I do not think this is correct: for one they don't tell me which conductor is going which direction current wise. This makes no sense....I'm not getting the correct answer at all...
 
  • #11
jhae2.718
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If you're looking for the magnitude, it doesn't matter. Otherwise you'll have to make an assumption. Do you have an answer for the problem? I get 1.64 μT.
 
  • #12
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Yes I keep getting 27 micro T
 
  • #13
jhae2.718
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The B-fields of the two wires will oppose each other, so the magnitude of the net B-field at the distance specified will be the absolute value of the difference of the two B-fields.

As for direction, that will result from the directions you assume for the currents.
 
  • #14
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Can you explain with equations how you got you answer?

I must be doing something wrong because if you used the same approach I did, then how did we get two different answers?
 
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  • #15
jhae2.718
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I assumed that the nearer wire had a current such that the B-field is positive (let's say up), which means the far wire has opposite current and B-field in the opposite direction.

From Ampere's law, the magnetic field is:
[tex]
B = \frac{\mu_0 i}{2\pi r}
[/tex]

Substituting in values:
[tex]
B_{net} = \frac{\mu_0 1.61\,\rm A}{2\pi (2.4\,\rm cm - 0.5 \cdot 2.92\,\rm mm)} - \frac{\mu_0 1.61\,\rm A}{2\pi (2.4\,\rm cm + 0.5 \cdot 2.92\,\rm mm)}
[/tex]
I used Wolfram Alpha to solve http://www.wolframalpha.com/input/?...+free+space*1.61+A)/(2pi(2.4+cm+++2.92mm/2))", and got ~1.6 μT.

Since I assumed the closer wire was positive and had a B-field pointing up, and since the net B-field is positive, the net B-field is also up.
 
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  • #16
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That is exactly what I did...where did I go wrong..

I forgot parenthesis around the 0.5*2.92mm part so it was adding the 0.5 to the centimeter measurement..

So for the next part I'd take the value of 1.6uT and multiply that by .1 to get 0.16uT and set that equal to the same equation but substitute the 2.4cm for a variable to solve for?
 
  • #17
jhae2.718
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That is exactly what I did...where did I go wrong..

I forgot parenthesis around the 0.5*2.92mm part so it was adding the 0.5 to the centimeter measurement..

So for the next part I'd take the value of 1.6uT and multiply that by .1 to get 0.16uT and set that equal to the same equation but substitute the 2.4cm for a variable to solve for?
Yes.
 
  • #18
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If you're still with me I got an answer of: 17.1270498 cm

*edit* It was wrong....

*edit 2* now I have 7.5768170098 cm
 
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  • #19
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If you're still with me I got an answer of: 17.1270498 cm

*edit* It was wrong....

*edit 2* now I have 7.5768170098 cm
Success it was correct! Thank you so much, now for the last question.
 
  • #20
jhae2.718
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Success it was correct! Thank you so much, now for the last question.
Good! The last one, if you think about it in terms of Ampere's Law, is pretty easy...
 
  • #21
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Is it possible to just do the last one as if we were doing the first just substitute the distance for r and the inner cable's magnetic field strength at point p for B?


integral(B dot ds) = u_0 * current through?
 
  • #22
jhae2.718
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Here's a hint: for a path outside of the coaxial cable, what is the enclosed current?
 
  • #23
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Here's a hint: for a path outside of the coaxial cable, what is the enclosed current?
If you include the cable in the enclosure wouldn't it be equal to the current of the cable, no matter how large it was/
 
  • #24
jhae2.718
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What is the total current of the cable (i.e. both center wire and outside wire)? Remember the signs...
 
  • #25
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Zero? But if that were the case then why didn't that happen in the earlier problems? Although the current is the same in opposite directions doesn't mean there is no magnetic field...
 
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