# Magnetic field on an axis

1. Nov 1, 2009

### fluidistic

1. The problem statement, all variables and given/known data
I must calculate the magnetic field for any point $$P$$ on an axis passing through the middle of a square wire loop (whose radius is $$a$$ and current is $$I$$).
The situation is such that the square wire is on a plane that is perpendicular to the axis in which I must calculate the magnetic field. If the situation is unclear, please tell it to me so that I scan a sketch of the situation.

2. Relevant equations

Biot-Savart, despite the symmetry, I don't think that Ampère's law can be useful.

3. The attempt at a solution

I drew a sketch as follow : a horizontal axis (called x-axis) passing in the center of a square with radius $$a$$. I drew the square such that its sides coincide with up-bottom-left-right directions, for the sake of simplicity.
Let $$\theta (x)$$ be the angle measured from the point $$P$$, between the x-axis and the upper side of the square. I have that $$\theta (x)= \frac{a}{2} \cdot \frac{1}{x} \Rightarrow \theta = \arctan \left ( \frac{a}{2x} \right )$$.
I've found this angle because I wanted to know the angle within $$\vec l$$ and $$\vec r$$ in Biot-Savart law. This is where I'm stuck. Am I in the right direction?

From B-S law : $$\vec B =\frac{\mu _0}{4 \pi} I \int d\vec l \times \frac{d \vec r}{r^3}$$.

So I wanted to calculate the contribution of each side of the square on the magnetic field situated at point $$P$$.

2. Nov 1, 2009

### rl.bhat

Consider a current element dl at a distance l from the center of the conductor.
Can you find the distance r from dl to P?
If O is the center of the loop, draw a parallel line through O to the upper wire.If A is the position of dl from the center of the wire, Draw AB perpendicular to the parallel line. Now you can find the angle BAP which is needed in Biot-Savart Law

3. Nov 2, 2009

### fluidistic

Thank you for helping!

Hmm, on the axis passing through the center of the loop and by point P?
I guess yes, but I'm unsure where to place dl.
Waiting for a little clarification.

Anyway, here comes my sketch, in case of confusion :

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4. Nov 2, 2009

### rl.bhat

dl is a small element of any one wire at a distance l from the center of the wire.

5. Nov 2, 2009

### fluidistic

Ah then $$r=\sqrt{x^2+l^2}$$ where $$x$$ is the distance from the center of the conductor (see figure).

Edit: I reached $$\theta= \arccos \left ( \frac{a}{2 \sqrt{x^2+l^2}} \right )$$. Where $$\theta$$ is the angle BAP.

Last edited: Nov 2, 2009
6. Nov 2, 2009

### rl.bhat

The distance AP = sqrt[l^2 + x^2 + (a/2)^2]
And sin( BAP) = l/sqrt[l^2 + x^2 + (a/2)^2]
Direction of th field is not in the direction of the x-axis. You have to resolve them to get the net field along x-axis.

Last edited: Nov 2, 2009
7. Nov 2, 2009

### fluidistic

Ok thanks a lot. I will try to understand my error about the distance AP.

8. Nov 4, 2009

### fluidistic

Sorry to bother again, I still don't reach the angle BAP.
Here's the picture of how I understood how to plot everything. Please tell me if there's a confusion somewhere.
Which gave me $$\sin \theta = \sqrt{r^2- \frac{a^2}{4}} \cdot \frac{1}{r}$$ where $$r= \sqrt{l^2+x^2}$$.
I don't really know why you've searched for $$\sin \theta$$ instead of $$\cos \theta$$ since $$\vec l \times \vec r = lr\cos \theta$$.
At last I obtained $$B= \frac{\mu _0Ia^2}{8 \pi r^3}$$ which I almost know it's wrong. I should probably get something over $$r$$, not $$r^3$$.

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9. Nov 4, 2009

### rl.bhat

In the figure OB = l, OP = x and BP = sqrt(l^2 + x^2)
BA = a/2. So AP = r = sqrt[l^2 + x^2 + (a/2)^2]
AXB = ABsinθ.
Here sinθ = x/r.

10. Nov 4, 2009

### fluidistic

I understand all until you wrote "Here sinθ = x/r". So you've considered the angle θ as $$OPA$$ ? If so then everything's fine.
I still don't see why I needed $$\sin \theta$$ instead of $$\cos \theta$$.

11. Nov 4, 2009

### rl.bhat

If A' is the point on the wire vertically above O, to find the field at P, I want the angle A'AP and sine of that angle. To find the field we need the angle between current element I*Δl and r.

12. Nov 4, 2009

### fluidistic

Ah ok, I think I see a bit better. At least I understand why we search this angle $$\theta$$, but still not its sine.
However in this case I have that the distance between A' and P is $$\sqrt {\left ( \frac{a^2}{2} \right ) +x^2}$$ and not x. Thus $$\sin \theta = \sqrt {\left ( \frac{a^2}{2} \right ) +x^2} \cdot \frac{1}{r}= \sqrt {\left ( \frac{a^2}{2} \right ) +x^2} \cdot \sqrt{l^2+x^2}$$.

13. Nov 4, 2009

### rl.bhat

Yes. Your sinθ is correct. But what is l? dl is a small segment on the wire. So put OA = sqrt(OA'^2 + A'A^2). And write 1/r