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Homework Help: Induced Magnetic Field in a Non-Uniform Electric Field

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data

    An electric field is directed out of the page within a circular region of radius R = 3.00 cm. The field magnitude is [itex]E = (0.500 V/ms)(1 - \frac{r}{R})t[/itex], where t is in seconds and r is the radial distance (r≤R). What is the magnitude of the induced magnetic field at a radial distance of 2.00 cm?

    2. Relevant equations

    Maxwell - Ampere's Law
    [itex]\oint \vec{B} \cdot d\vec{s} = μ_0ε_0 \frac{d\Phi_B}{dt}[/itex]

    3. The attempt at a solution

    Since B and ds are parallel, their dot product will equal Bds, or since ds is a circle at radius r, [itex]2\pi rB[/itex]. I got caught up on the left-hand side of the equation. I know I can simplify it to:

    [itex]μ_0ε_0 \frac{d}{dt}\int \vec{E} \cdot d\vec{A}[/itex], which further simplifies to (since E and dA are parallel):

    [itex]μ_0ε_0 \frac{d}{dt}\int E dA[/itex]

    I'm lost as to how to simplify that integral. I think I could do a double integral to solve to surface integral, however, this is an introductory physics course only requiring Calculus 1 and 2, so I wouldn't have assumed such integrals would be necessary.

    Also, assuming I've simplified correctly, would I integrate, then take the time-derivative, or could I take the time derivative of E first, then integrate?

    Last edited: Nov 16, 2011
  2. jcsd
  3. Nov 16, 2011 #2
    Let E = Fxy(x,y,t)

    d/dt ∫ E dA
    = d/dt ∫ Fxy(x,y,t) dA
    = d/dt ∫x1x2y1y2 Fxy(x,y,t) dy dx
    = d/dt ∫x1x2 Fx(x,y2,t) - Fx(x,y1,t) dx
    = d/dt [ (F(x2,y2,t) - F(x1,y2,t)) - ((F(x2,y1,t) - F(x1,y1,t)) ]
    = (Ft(x2,y2,t) - Ft(x1,y2,t)) - (Ft(x2,y1,t) - Ft(x1,y1,t))
    = ∫x1x2 Fxt(x,y2,t) - (Fxt(x,y1,t) dx
    = ∫x1x2y1y2 Fxyt(x,y,t) dy dx
    = ∫x1x2y1y2 d/dt Fxy(x,y,t) dy dx
    = ∫x1x2y1y2 d/dt E dA
    = ∫ d/dt E dA

    You mean that E is everywhere normal to A right?

    The question is poorly worded (maybe it comes with a diagram), but if I'm understanding it correctly, the equation that you wrote for the electric field implies that it's constant across the entire surface of a sphere. Therefore, instead of numerically evaluating ∫E(t).dA, you can argue by symmetry that ∫E(t).dA = ∫|E(t)|.dA = |E(t)|∫dA = E(t)*4πr2.

    Saves you the trouble of actually evaluating the surface integral.
    Last edited: Nov 16, 2011
  4. Nov 16, 2011 #3
    So since the integral depends only on two variables, of which time is neither, the time-derivative can essentially be freely taken before or after integration?

    Thanks for the detailed proof as well.

    Yes, it does come with a diagram, but it's just a drawing of a circle so I didn't think it was all to vital to include. But I am saying that E is parallel to the area vector everywhere (it would be normal to the surface though). And since the electric field varies with radial distance from the center of the circular region, wouldn't it be inappropriate to remove it from an integral with respect to r? That is what I initially attempted, but it seemed incorrect for that reason.

    I think most of the ambiguity is coming from the problem description, and apologies there.
    Last edited: Nov 16, 2011
  5. Nov 16, 2011 #4
    There's no integration with respect to r. The integration is on the surface of the sphere, which means r is constant.

    To make this more intuitive, compare the above surface integral, which I shall rewrite as the following:

    E.dA = ∫∫ E.dS

    with this volume integral:

    ∫∫∫ E.dS r dr
    = ∫∫∫ F dV

    In words, after summing up all the elements which make up the surface area of the sphere, you integrate outward.

    *Note that the extra r term arises from the Jacobian.
    *Also note that I've written F for the volume integral since F=div(E).
    Last edited: Nov 16, 2011
  6. Nov 16, 2011 #5
    The surface is a circle, not a sphere. I found a picture of the region online, sorry for not including it originally. I'm finding the induced magnetic field at a radial distance from the center of this region; the electric field is coming out of the page in that picture, and its magnitude varies radially. So I assumed that to find the total flux through the circle, I would need to integrate across the radius.

    But in your volume example, I think I understand the logic. Since you're only integrating at one radius, it can be disregarded from the integration, correct?

    Attached Files:

  7. Nov 16, 2011 #6
    That makes more sense now. It's just like a cross section of a wire.

    The same logic that you recognized in your first post still stands. Since E is everywhere perpendicular to A, you don't have to worry about the dot product.

    ∫ E.dA = ∫ E dA

    (The dot product is there as a vector projection, kind of like finding the "effective" flux, but the projection of a vector E on another vector in the same direction An is just itself if both E and n are in the same direction.)

    Do the integral in polar coordinates:

    ∫ E dA = ∫ E r dr dθ

    You'll get your result immediately. Again, it doesn't matter if you take d/dt first or evaluate the integral first. I would recommend taking d/dt first.
  8. Nov 16, 2011 #7
    Excellent; thank you. I do apologize for the confusion in the beginning.

    Just out of curiosity, is it possible to solve the integral without resorting to a double integral? I don't see anyway to, since you have both A and E varying radially outwards (its not like A is constant, simplifying to just the product of A and E).
  9. Nov 16, 2011 #8
    There's no choice in this problem but to integrate with respect to r, since the electric field varies (continuously) according to the r.

    If you really want to avoid a double integral that badly, you can transform this into a one-dimensional problem by just deforming the circle into a rectangle with width 2π. Then the problem would be equivalent to:

    2π ∫ E r dr

    It isn't much different from the double-integral though, and I'd just do the double-integral to preserve the logic behind doing it.
  10. Nov 16, 2011 #9
    The double integral definitely makes the most sense. Like I stated in the original post, I was curious if it could be solved without the double integral since the course doesn't have multivariable calculus as a pre-req. Seemed strange to have problems with double integrals without a foundation in the area.

    Thanks again for your help (and the reminder that when switching to polar coordinates one needs to include r dr dθ, not just dr dθ!).
  11. Nov 16, 2011 #10
    I just noticed this:
    The problem with the logic here is that A is constant because it's a flat surface.

    ∫ dA = A

    In other words, a surface is f(x,y), or any function of two variables. If the surface you had wasn't A, but instead was f(x,y), then the surface isn't constant and the above relationship isn't true.

    The quantity that isn't constant is ∫ E dA. Think of E as that f(x,y) that distorts the area (which is equivalently a density function). It "causes" the flat surface to become a concave down surface, so the "new" area is deformed into something that isn't constant.
    Last edited: Nov 16, 2011
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