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Homework Help: Magnetic Forces & Cylinders

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Hello, let me dish out the particulars straight away:
    "A large(and infinite) cylinder of the radius a, is fed a Current I through its surface; the same current, but travelling in the opposite direction exits the body through a smaller cylinder located within the greater one, and at a distance l from the center of the former, with the radius b(also infinite, both axes are parallel).
    Find the magnetic force of interaction:(Per unit length)"(Also see the attached diagram):

    2. Relevant equations
    Well, we would have to start with the Amperian Force:
    [tex]\vec{F}=I\int d \vec{B} \times \vec{l}[/tex]
    [tex]\frac{\partial \vec{F}}{\partial l} = I \int d \vec{B}[/tex]
    And of course, (and I assume there's no reason to prove it), the field of an infinite current, passing through a wire is:
    [tex]B = \frac{{\mu}_0I}{2 \pi r}[/tex] (r- being the radial distance from the current).
    3. The attempt at a solution
    Well, I've tried separating the current through some sort of surface distribution, i.e:
    [tex]dI'=\frac{I \cdot dS}{\pi (b^2-a^2)}[/tex], but as you might imagine, I've been rather unsuccessful, this where the resolution of this question jams; That is, after multiplying by B and then integrating over the remaining dr there, I am firstly uncertain as to how to set up the boundaries of the integral, and then, I recieve a logarithm, an unlikely relation(decrement), in problems of such kind.(More so, it doesn't match the advertised solution).

    What should I do next?

    I thank you for you time and attention,
    Hoping for your aid,
    Thanks!
    Daniel
     

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    Last edited: Sep 1, 2010
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  3. Sep 1, 2010 #2

    gabbagabbahey

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    Your first formula should be [itex]\textbf{F}=I\int \textbf{B}\times d\textbf{l}[/itex], and your second formula is simply nonsense. At best you can say [itex]d\textbf{F}=\textbf{B}\times d\textbf{l}[/itex]

    How is this relevant? It sounds like you have two cylinders carrying azimuthal surface currents.
     
    Last edited: Sep 1, 2010
  4. Sep 2, 2010 #3
    Thank you a for a very explicit response.
    Now, I concede that the proper formulae are perhaps these:
    [tex]\vec{F}=I\int \vec{B} \times \vec{dl}[/tex]
    And the Force per unit length is therefore F=I*B;
    But [tex]B = \frac{{\mu}_0I}{2 \pi r}[/tex] is tied with the fact that the current circulates here as in any other ordinary conductor, (at least so I think), and thus creates this radial magnetic field.
    If otherwise, I'd appreciate hearing some advice about framing any other field and its division of current, therein.
    Thanks again,
    Most grateful,
    Looking forward to your replies,
    Daniel
     
    Last edited: Sep 2, 2010
  5. Sep 2, 2010 #4

    gabbagabbahey

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    The question doesn't really make it clear how the current flows over the surface of each cylinder. Is it flowing parallel to the axis of the cylinder, or around its circumference?
     
  6. Sep 2, 2010 #5
    Hi again,
    If not sufficiently lucidly stated in the phraseology of the question itself, the current flows parallel to the axes of both cylinders, that is, the same value enters the larger one, and leaves, in the opposite direction(but still parallel, as always), through the smaller one; There's no need to reiterate that both are insulated from one another.

    What's next then?
    Thank you very much again,
    Daniel
     
  7. Sep 2, 2010 #6

    gabbagabbahey

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    First, the field of a cylinder carrying a uniformly distributed axial current (like each of the cylinders in this problem) isn't radial, it circles around the cylinder (it's azimuthal). Second, its magnitude is not [tex]B = \frac{{\mu}_0I}{2 \pi r}[/tex] everywhere, it can't be: the magnetic field is discontinuous whenever it crosses a surface current, so you know the field inside the cylinder has to be different from the field outsside the cylinder.
     
  8. Sep 3, 2010 #7
    Hi,
    If you mean that the current curls around the cylinders, then I beg to differ. according to the statement in the query, it obviously flows through the surface of each cylinder, or Pi*(R1^2-R2^2), respectively.
    The magnetic field therefore, outside the cylinder is 0, and it's also assumed that there are NO fringe effects at work at the intersection of both bodies, i.e. on the verge of the separation between them.

    How do you propose then, to compute, via Biot-Savart's (if it applies), or Ampere-Maxwell's law, the proper the magnetic field, as you say?
    Thanks again,
    Beholden,
    Daniel
     
  9. Sep 3, 2010 #8

    gabbagabbahey

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    No, I mean the magnetic field curls around the cylinders, not the current.

    Not if the current is flowing along the length of the cylinders (parallel to their axes). Why do you think otherwise?

    Think of each cylinder of being made up of an infinite number of thin (thickness [itex] ds[/itex] ), long straight wires, carrying current [itex] i = I\frac{ds}{2\pi a}[/itex] for the larger cylinder or [itex]i = I\frac{ds}{2\pi b}[/itex] for the smaller cylinder. You know that the field of each wire is [itex]\frac{\mu_0 i}{2\pi d}[/itex], where [itex]d[/itex] is the distance from the wire and it circles around the wire (right-hand rule). Integrate over the circumference of the cylinder to get the total field.
     
    Last edited: Sep 3, 2010
  10. Sep 3, 2010 #9
    Can I assume that you mean the circumference of the respective cylinders to be either 2*Pi*a, and 2*Pi*b?
    None the less, I thank you for it, you present a very coherent point. The only problem is, that I can't feature, in these attempts the given distance l between both centers. A try to combine that during integration yields a logarithm, that should not appear.

    What have I done wrong, therefore?
    Thanks again,
    Daniel
     
  11. Sep 3, 2010 #10

    gabbagabbahey

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    What are you getting for the magnetic field of the smaller cylinder (radius [itex]b[/itex])?
     
  12. Sep 4, 2010 #11
    Let me just post the process by which I integrated:
    Now, as you've said that ds would stand for the element of the circumference, we may thus inscribe:
    [tex]i=\frac{I ds}{2 \pi b}=\frac{I r d \phi}{2 \pi b}[/tex]
    Where as,
    [tex]dB = \frac{{\mu}_0 i}{2 \pi r}=\frac{{\mu}_0 I d \phi}{4 {\pi}^2 b}[/tex]
    And,
    [tex]B=\int^{2 \pi}_0 \frac{{\mu}_0 I d \phi}{4 {\pi}^2 b}= \frac{{\mu}_0 I}{2 \pi b}
    [/tex]

    And the same applies for the cylinder of the radius a, and the likes.

    What should I have done?
    Thank you!
    Daniel
     
  13. Sep 7, 2010 #12

    gabbagabbahey

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    You're forgetting that the magnetic field is a vector, you need to account for the direction of the field of each thin wire. The field from each wire points in a different direction and you will need to account for that when you integrate.
     
  14. Sep 8, 2010 #13
    Hey,
    Wouldn't you say though, that since [tex]\vec{B} \ \ || \ \ (\hat{I} \times \hat{r})[/tex], that in our coordinate system, the resultant will always be pointed in the [tex] \hat{z} [/tex] direction?
    If not, what angle should I incorporate?
    Thanks again,
    Daniel
     
  15. Sep 8, 2010 #14

    gabbagabbahey

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    No, the current is flowing in the z-direction (parallel to the axis of the cylinder, right?), so the resultant B will be perpendicular to the z-axis. (just as it is for each long straight wire)
     
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