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Homework Help: Line Integral/Ampere's Law: is my logic valid?

  1. Jun 1, 2015 #1
    This is a problem from a section on Line Integrals in my Calculus Textbook, I haven't studied any physics relating to E&M yet, and the solutions manual only gives solutions for odd numbered problems. Sorry, if I'm posting in the wrong forum, I hope I'm not.

    1. The problem statement, all variables and given/known data

    A steady current [itex]I[/itex] in a long wire produces a magnetic field [itex]\vec{B}[/itex] that is tangent to any circle that lies in the plane perpendicular to the wire and whose center is the axis of the wire.

    Ampere's Law relates the electric current to its magnetic effects and states that:

    [tex]\int_C \vec{B} \cdot d\vec{r} = \mu_0I[/tex]

    where [itex]I[/itex] is the net current that passes through any surface bounded by a closed curve C, and [itex]\mu_0[/itex] is a constant called the permeability of free space. By taking C to be a circle with radius r, show that the magnitude [itex]B = |\vec{B}|[/itex] of the magnetic field at a distance [itex]r[/itex] from the center of the wire is

    [tex]B = \frac{\mu_0I}{2\pi r}[/tex]

    2. Relevant equations

    [tex]\vec{A}\cdot \vec{B} = |\vec{A}||\vec{B}|\cos{\theta}[/tex]

    I also found this website that gives a different mathematical form for Ampere's Law


    [tex]\oint B ds = \mu_0I + \frac{1}{c^2}\int E\cdot dA[/tex]

    But I don't know what that second term is, so I am assuming that the form of Ampere's Law given me by the textbook is specific to closed circular paths around the wire. Is this correct? What is this second term all about?

    3. The attempt at a solution

    So [itex]\int_C \vec{B}\cdot d\vec{r}=\int_C \vec{B} \cdot \vec{T} ds[/itex] where [itex]\vec{T}[/itex] is a unit vector and so [itex]|\vec{T}| = 1[/itex]

    Also, [itex]ds = r dt[/itex] since the [itex]\vec{r} = \langle r\cos(t),r\sin(t)\rangle[/itex]

    Then, [itex]\vec{B} \cdot \vec{T} = |\vec{B}||\vec{T}|\cos{\theta}[/itex]. Obviously our unit tangent vector magnitude just becomes one, but here I also need to argue that [itex]\theta=0[/itex].

    Since it states that the magnetic field will be tangent to a circle at radius r from the wire, I feel comfortable stating that theta will be zero on the circular path.

    Here, one continues [itex]\int_C \vec{B} \cdot \vec{T} ds = \int_C |\vec{B}| ds = \mu_0I[/itex], provided C is a circle of radius r in the plane perpendicular to the wire/source and with the wire/source at the center of the circle.

    So [itex]\int_C |\vec{B}| ds = \int_0^{2\pi} |\vec{B}| r dt = |\vec{B}|r[t]_0^{2\pi}[/itex], but now I need to argue that [itex]|\vec{B}|[/itex] and [itex]r[/itex] are constants. This is no problem for [itex]r[/itex] as it is assumed to be a constant from the beginning of the exercise, but I am not 100% comfortable arguing that [itex]|\vec{B}|[/itex] is a constant in this case.

    I could argue that [itex]|\vec{B}|[/itex] is a constant because it is a function of [itex]r[/itex], and it states that in the setup to the problem itself, so I have no big issue with this intuitively. If we assume something to be so, then so be it...but...

    Is there anything in this mathematical statement of Ampere's Law

    [tex]\int_C \vec{B} \cdot d\vec{r} = \mu_0I[/tex]

    that tells us for certain that [itex]|\vec{B}|[/itex] is a function of [itex]r[/itex] and that if [itex]r[/itex] is constant then so is [itex]|\vec{B}|[/itex]? After all, it is a line integral that equals a constant value.

    If a line integral is a constant value, is it generally then the case that the magnitude of the vector field is a function only of the path through that field? Or only for closed paths? Is this what they mean by a conservative field? Or am I just veering completely into nonsense?
    Last edited: Jun 1, 2015
  2. jcsd
  3. Jun 1, 2015 #2


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    There is indeed an important and nontrivial point here, that unfortunately most books do not explain well (if at all). We have to use a symmetry argument to show that the magnitude of B cannot depend on the angle if we are at a fixed r. The argument is the following:

    Let's pick an arbitrary point ##P_1## at a distance r. There is a B field ##\vec{B}_1## there. Let's say that you close your eyes and I rotate the wire by an angle theta (around its axis of symmetry) in such a way that a second point ##P_2## (where there is a B field ##\vec{B}_2##) has being moved to where point ##P_1## was before. Now you open your eyes and you see the field ##\vec{B}_2## where you were seeing ##\vec{B}_1## before I rotated the wire. Now comes the key point: by symmetry, the wire looks exactly the same as what it was before I rotated it. In other words, if I don't tell you that I rotated the wire, you would not be able to see that the wire has been rotated. Therefore, since the wire looks exactly the same as it was just before I rotated it, the B field must also be the same as before! So we conclude that after the rotation, the B field ##\vec{B}_2 ## must be identical to the B field ##\vec{B}_1## that was there before we did the rotation (and by identical I mean both in magnitude and in direction). Since a rotation does not change the magnitudes of the B fields, we obtain immediately that the two B fields must have the same magnitude (note that it is crucial that we are considering two points P1 and P2 at the same distance r from the wire since other wise a rotation would not bring P2 to P1).

    Symmetry arguments are very powerful. For example, you can use them to show that the E field produced by an infinite uniform charge density has a direction necessarily perpendicular to the surface.
  4. Jun 1, 2015 #3


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    You left a time derivative out of the term involving the E field in the second term. A time varying electric field can also contribute to the magnetic flux. In your case, you don't have such a field.
  5. Jun 1, 2015 #4
    So then, will Ampere's Law as given by my textbook work just as well for any closed path around the wire?
  6. Jun 1, 2015 #5


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    Well, sure. Isn't that what I said? If there is no time varying electric field, you can leave that term out.
  7. Jun 2, 2015 #6
    No no no, I'm asking if the path around the wire has to be circular or not, in order for Ampere's equation to work.
  8. Jun 2, 2015 #7


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    It has to be a "closed curve" but not necessarily a circle or any specific shape. Since you posted this under "Calculus" rather than "Physics", I will point out that this is basically "Green's theorem" (more specifically, the "Kelvin-Stokes theorem", Green's theorem extended to three dimensions)- that, for F a differentiable function, and C a closed curve with A as its interior,
    [tex]\oint_C \vec{F}\cdot d\vec{r}= \int_A\int \nabla\times \vec{F}\cdot d\vec{S}[/tex]
  9. Jun 2, 2015 #8
    Ok. Now, this closed path line integral equals a constant, in this given case, because we're considering a steady current...and I've just read that for a conservative vector field the line integral of a closed path is zero, so this must mean that the magnetic field around a wire is NOT a conservative vector field?? Intuitively I would have thought it would be. What does it being a constant tell us mathematically?
  10. Jun 2, 2015 #9

    Ray Vickson

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    Certainly, symmetry arguments are more insightful and nicer all around, but it is satisfying to note that formal expressions can also give the same result. In this case, using the Biot-Savart Law (see, eg., http://en.wikipedia.org/wiki/Biot–Savart_law ) results in
    [tex] \vec{B}(\vec{r}) = \frac{\mu_0}{4 \pi} \int \frac{ d \vec{l} \times \vec{r}'}{|\vec{r}'|^3},[/tex]
    where ##\vec{r}' = \vec{r} - \vec{l}##. You end up with ##|\vec{B}(\vec{r})| = F(h)##, where ##h## is the (shortest) distance between the point at ##\vec{r}## and the wire (which the OP called ##r## in post #1).
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