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I Ampere's Law: Double Negative Line Integral Help

  1. Feb 13, 2019 at 9:06 PM #1
    Hi all,

    I’m having some trouble finding a minus sign in a standard calculation I have been doing. I am trying to show that if there is no enclosed current around the example loop in the enclosed jpeg, the four piecewise paths add up to zero (for the line integral part of Amp’s law). For this question, I’m only concerned with the line integral for two of the paths in the loop in the enclosed jpeg that should cancel out. The main problem is that I cannot get the line integral of A to B to cancel out the line integral from C to D. I am picking up two minus signs from the line integral of C to D, one from the dot product of B and DL, the other minus sign from integrating from theta = 90 degrees (point C) to theta = 0 degrees (point D). Can someone help me reconcile my getting rid of my double negative? I put in purple in the jpegs the comments relevant to this question that I am addressing to the forum.
     

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  3. Feb 13, 2019 at 10:49 PM #2

    kuruman

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    You should not be using ##\oint## because A→B, B→C, etc. are not closed loop integrals. It should be clear from your diagram that the integral ##\int \vec B\cdot d\vec l## is positive for A→B (you add a whole lot of positive numbers) and negative for C→D (you add a whole lot of negative numbers). Formally,$$\vec B=\frac{\mu_0I}{2\pi r}\hat{\theta};~~~d\vec l=r d\theta ~\hat{\theta}~~\rightarrow~~\vec B\cdot d\vec l=\frac{\mu_0I}{2\pi}d\theta$$so that $$\int_{\theta_1}^{\theta_2}\vec B\cdot d\vec l=\frac{\mu_0I}{2\pi}({\theta_2-\theta_1}).$$If you reverse the limits of integration, you reverse the sign. For path C→D if you use ##\cos(180^o)## and reverse the limits of integration, you end up adding a whole bunch of positive numbers because you have introduced a double negative.
     
    Last edited: Feb 13, 2019 at 11:08 PM
  4. Feb 13, 2019 at 11:49 PM #3

    Delta2

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    You shouldn't introduce the negative from cos(180). That negative is included in ##d\theta## which is considered negative when you integrate for the path C to D, from ##\theta=\pi/2## to ##\theta=0##. When you integrate for the path A-B from ##\theta=0## to ##\theta=\pi/2## then ##d\theta ## is positive.

    To make you understand even more clearly what I am saying, take it we have a positive function ##f(x)>0## for all ##x\in[a,b]## where ##a<b##. If you integrate from b to a ,that is, consider the integral ##\int_b^af(x)dx## we know it will be negative (for example consider the integral ##\int_1^0 x^2dx=-1/3##. How come taking a (Riemann) sum of positive values ##f(x)dx ## comes out to a negative result? The answer is because ##dx## is considered to be negative when we take an integral for which the start limit of integration is greater than the end limit of integration
     
    Last edited: Feb 14, 2019 at 12:24 AM
  5. Feb 14, 2019 at 7:08 AM #4
    Hi all,

    I am reading this. I was going to reply just now but I am out of time and have to go. I will reply this evening. But, yes, I will post soon. I already appreciate what is here.
     
  6. Feb 14, 2019 at 6:10 PM #5
    First, yes, to kuruman in that I should have been writing ∫ and not the closed loop symbol of ∫. Yes, I agree the integral over A → B is all + numbers. I’m clear on the integral from A → B.

    It does seem clear to that dθ is negative. I had not thought of that and accept it. I am going in the opposite angular direction of A → B, no different than if you were on the x-axis and dx pointing left, dx would be negative.

    I didn't realize that when you reverse the limits of integration, the dx, dy, dθ, whatever your line element is, has an implicit negative embedded inside of it. Is the general topic of what I am asking about called “signed integration”? I looked on the web some more to see what you are all saying.

    So, to complete what Delta2 was saying, if I do the integral of x^2 from 0 to 1, that integral is evaluated to be 1/3. dx is +. If I now flip the limits to, from 0 to 1, the integral becomes -1/3. The flipping of limits contains the information that the sign of the line element has switched. Thus, for this case, since dθ was integrated from 0 to 90, if I flip the limits from 90 to 0, I see how it is with the x^2 analogy that the sign on dθ doesn’t flip. This does make sense to me, the flipping of integral limits and correct sign for this part of it.

    I am still struggling with one part of this, though: the dot product of two vectors producing a minus sign. Let us assume that there is no integral for a moment and that I take the dot product of B counterclockwise and dθ clockwise on path C à D. Would this entity by itself produce a minus sign? I know that for the Work due to friction integrated from x1 to x2, the friction force and dx point in opposite directions and the overall Work is negative. Or is this a different situation from what I am talking about and I am mixing apples and oranges?
     
  7. Feb 14, 2019 at 9:59 PM #6

    Delta2

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    Yes of course two vectors can have a negative dot product. It is just that in our case the negativity (or positivity) of the dot product is included in the sign of ##d\theta##. Looking back at what @kuruman wrote at post #2 we can see that the dot product of B and dl is ##\frac{\mu_0}{2\pi}Id\theta## and it is negative (or positive) when ##d\theta## is negative (or positive)
     
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