Suppose ABCD is square loop (conducting). There is a wire carrying a current $I$ into the plane of paper within the region of loop. So how to properly apply Ampere's Law in this case?

The closed line integral of $\vec{B}\cdot d\vec{l}$ about loop ABCD will be $\mu_0I$ surely. Up to this I understand. But then there is a statement in my book saying that due to symmetry integral of $\vec{B}\cdot d\vec{l}$ from B to C will be same as integral of $\vec{B}\cdot d\vec{l}$ from D to A. I could not understand this statement. Can someone please explain this statement? And also how do I find integral from A to B of $\vec{B}\cdot d\vec{l}$?


1 Answer 1


If the loop of wire is square and the second wire with a current passes directly through the center of the wire, then everything is symmetric. There is some $\vec{B}$ which you could find - it has magnitude $\frac{\mu_0 I}{2\pi r}$ in the right-handed direction.

Lets imagine you were being silly and ignored the symmetry of the problem. Instead you calculated every little detail of the problem, laboriously doing the integrals.

So, imagine point "A" is at $(-a,a)$ and point "B" is at $(a,a)$, "a" being the size of the square. The wire is at $(0,0)$ going into the plane. Then $$\vec{B}=\frac{\mu_0 I}{2\pi r}(\sin(\theta),-\cos{\theta})=\frac{\mu_0 I}{2\pi \sqrt{x^2+y^2}}(\frac{y}{\sqrt{x^2+y^2}},-\frac{x}{\sqrt{x^2+y^2}})$$

(The direction $(\sin(\theta),-\cos{\theta})$ is perpendicular to the radial vector $(\cos(\theta),\sin(\theta))$ in the clockwise sense. And I write out $\cos(\theta)=\frac{x}{r}$, $\sin(\theta)=\frac{y}{r}$, and $r=\sqrt{x^2+y^2}$)

We also have, along the path I choose, $d\vec{l}=(dx,0)$. Substitute $y=a$. Then...

$$\int_A^B \vec{B}\cdot d\vec{l}=\int_{-a}^a\frac{\mu_0 I}{2\pi \sqrt{x^2+a^2}}\frac{a dx}{\sqrt{x^2+a^2}}$$

If I plug this into Mathematica, I find:

$$\int_{-a}^a\frac{a dx}{2\pi (x^2+a^2)}=\frac{1}{4}$$

but this should have been obvious from the start. If you were to write out all of the other integrals, $\int_B^C \vec{B}\cdot d\vec{l}$, $\int_C^D \vec{B}\cdot d\vec{l}$, $\int_D^A \vec{B}\cdot d\vec{l}$, you'd find that you're evaluating exactly the same quantity four times! So if we know the sum of the four quantities is $\mu_0 I$, and we know each quantity equals every other quantity, each integral was bound to be $\frac{1}{4}\mu_0 I$ in the first place. The integration was wasted effort!

  • 1
    $\begingroup$ @SanchayanDutta Oh, and just in case I didn't make it clear: You should deduce that every integral is the same just from the fact that "every edge of the square looks like every other edge." You definitely should not parameterize $\vec{B}$ and try to integrate it before realizing every integral would be the same. $\endgroup$
    – user12029
    Jun 21, 2016 at 19:41

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