How to Calculate Magnetic Field on an Axis Through a Square Wire Loop?

[fluidistic]

Homework Statement

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.

Homework Equations

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

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.

 

[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

 

[fluidistic]

Thank you for helping!

Consider a current element dl at a distance l from the center of the conductor.

Hmm, on the axis passing through the center of the loop and by point P?

Can you find the distance r from dl to P?

I guess yes, but I'm unsure where to place dl.

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

Waiting for a little clarification.

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

 

[rl.bhat]

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

 

[fluidistic]

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

Ah then r=\sqrt{x^2+l^2} where x is the distance from the center of the conductor (see figure).
I will try to follow your instructions.

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

 

[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.

 

[fluidistic]

The distance AP = sqrt[l^2 + x^2 + (a/2)^2]
And sin( BAP) = sqrt(l^2 + x^2)/[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.

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

 

[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.

 

[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.

 

[fluidistic]

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.

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.

 

[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.

 

[fluidistic]

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.

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}.

 

[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