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Infinite wire carrying current, magnetic field

  • Thread starter scholio
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  • #1
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Homework Statement



an infinite wire carries current into the paper as shown. (see attach.) compute [integral(B dI)] along the closed path indicated

Homework Equations



electric potential V = IR where I is current, R is resistance

magnetic field B = mu_0(I)/2pi(r) = mu_0(I)/dr where mu_0 is constant = 4pi*10^-7, r is radius, I is current, dr is change in radius

The Attempt at a Solution



the path is basically a part of a small inner circle of radius r_1, and 'part' of a larger circle of radius r_2. to find the change in radius dr, i need to find the path which equals r_net ( see 2nd attach).

does the fact that the current is carried into the paper just affect the sign of the current? or does it have a greater affect?

i'm not sure whether dr = path length, if it is not, to find dr am i supposed to just do (r_2 - r_1 = dr)?

how do i determine change in current I, dI? am i missing the correct equation?
 

Attachments

Answers and Replies

  • #2
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couldn't edit the original post, i am supposed to get " mu_0(I) " for the final answer, where mu_0 is a constant = 4pi*10^-7, and I is the current(a constant)

*******problem is to solve [integral(Bdl)] not [integral(BdI)], path length not current --> thanks calef/mindscrape
 
Last edited:
  • #3
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dr is the infinitesimal path length. This is a hairy problem if you aren't familiar with vector calculus.

Are you using Biot-Savart?
 
  • #4
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err, are you sure you aren't supposed to compute Bdl, as in, the integral of Magnetic Field with respect to differential length (as opposed to current?)?

Cause if so, all this problem is doing is trying to illustrate ampere's law.
 
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  • #5
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calef, i think you may be right, i think i should actually be solving for [integral(Bdl)] as opposed to [integral(BdI)].

even still, i may have written/used the wrong terms on the attachment to indicate path length, i used radii. anyways, did i determine the path length correctly? i used simple geometry.

do i just treat the magnetic field B as a constant, take it out of the integral, what would be by integrating limits?

no once i've found the total path,
 
  • #6
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calef, i think you may be right, i think i should actually be solving for [integral(Bdl)] as opposed to [integral(BdI)].

even still, i may have written/used the wrong terms on the attachment to indicate path length, i used radii. anyways, did i determine the path length correctly? i used simple geometry.

do i just treat the magnetic field B as a constant, take it out of the integral, what would be by integrating limits?

no once i've found the total path,
Well, I think you're finding too much, actually. The point of ampere's law is that the quantity I*mu_0 is independent of the path you take for the integral. So no matter what your path is, be it a circle, or that horrible geometric mess they've given you, the integral of magnetic field with respect to length along that path will always be the current within the loop times mu_0.
 
  • #7
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Ampere's law? What kind of loop would you use?

I'm pretty sure he is stuck with Biot-Savart and what seems like not enough information.

Edit:Ohhh, I see what the problem is getting at. Yeah sure, it follows from Maxwell's equations that [itex]\int \mathbf{B} \cdot d\mathbf{l} = \mu_0 I[/itex]. But as far as finding B goes... oof.
 
  • #8
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Ampere's law? What kind of loop would you use?

I'm pretty sure he is stuck with Biot-Savart and what seems like not enough information.

Edit:Ohhh, I see what the problem is getting at. Yeah sure, it follows from Maxwell's equations that [itex]\int \mathbf{B} \cdot d\mathbf{l} = \mu_0 I[/itex]. But as far as finding B goes... oof.
Doesn't matter, all that matters is that the loop is closed.
 
  • #9
dynamicsolo
Homework Helper
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Doesn't matter, all that matters is that the loop is closed.
Quite so! As a suggestion to the OP, consider that Ampere's Law is a magnetic analog to Gauss' Law. There, the flux integral, integral(E · dA) is given by the charge enclosed by the surface divided by epsilon_0 . It makes no difference how lumpy the surface is, as long as it is closed and continuous; you don't need to know the total surface area or the shape of the surface. Similarly here, integral(B · dL) itself will not depend on the arclength of the closed path or its shape.

(Tip for "psyching out" problems in introductory physics courses: if the problem posed seems to have a portion which would require incredibly complicated calculation, you are generally being asked to recall and apply a straightforward concept...)
 
  • #10
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ohhh, so basically i am just solving maxwell's eq, so do i just use the net length on the attachment i came up with and just multiply it by mu_0 = 4pi*10^-7, to get the answer i'm looking for?
 
  • #11
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The point is that you don't have to solve anything, it is already solved for you. You just recite the equation

[tex]\int \mathbf{B} \cdot d\mathbf{l} = \mu_0 I[/tex]
 

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