Magnetic field on an axis

In summary, the student is trying to solve for the magnetic field at a point on a wire loop. He is stuck because he does not understand Ampère's law.
  • #1
fluidistic
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Homework Statement


I must calculate the magnetic field for any point [tex]P[/tex] on an axis passing through the middle of a square wire loop (whose radius is [tex]a[/tex] and current is [tex]I[/tex]).
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 [tex]a[/tex]. I drew the square such that its sides coincide with up-bottom-left-right directions, for the sake of simplicity.
Let [tex]\theta (x)[/tex] be the angle measured from the point [tex]P[/tex], between the x-axis and the upper side of the square. I have that [tex]\theta (x)= \frac{a}{2} \cdot \frac{1}{x} \Rightarrow \theta = \arctan \left ( \frac{a}{2x} \right )[/tex].
I've found this angle because I wanted to know the angle within [tex]\vec l[/tex] and [tex]\vec r[/tex] in Biot-Savart law. This is where I'm stuck. Am I in the right direction?

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

So I wanted to calculate the contribution of each side of the square on the magnetic field situated at point [tex]P[/tex].
 
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  • #2
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
Thank you for helping!


rl.bhat said:
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 :
 

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  • #4
dl is a small element of anyone wire at a distance l from the center of the wire.
 
  • #5
rl.bhat said:
dl is a small element of anyone wire at a distance l from the center of the wire.

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

Edit: I reached [tex]\theta= \arccos \left ( \frac{a}{2 \sqrt{x^2+l^2}} \right )[/tex]. Where [tex]\theta[/tex] is the angle BAP.
 
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  • #6
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.
 
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  • #7
rl.bhat said:
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.
 
  • #8
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 [tex]\sin \theta = \sqrt{r^2- \frac{a^2}{4}} \cdot \frac{1}{r}[/tex] where [tex]r= \sqrt{l^2+x^2}[/tex].
I don't really know why you've searched for [tex]\sin \theta[/tex] instead of [tex]\cos \theta[/tex] since [tex] \vec l \times \vec r = lr\cos \theta[/tex].
At last I obtained [tex]B= \frac{\mu _0Ia^2}{8 \pi r^3}[/tex] which I almost know it's wrong. I should probably get something over [tex]r[/tex], not [tex]r^3[/tex].
 

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  • #9
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
rl.bhat said:
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 [tex]OPA[/tex] ? If so then everything's fine.
I still don't see why I needed [tex]\sin \theta[/tex] instead of [tex]\cos \theta[/tex].
 
  • #11
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
rl.bhat said:
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 [tex]\theta[/tex], but still not its sine.
However in this case I have that the distance between A' and P is [tex]\sqrt {\left ( \frac{a^2}{2} \right ) +x^2}[/tex] and not x. Thus [tex]\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}[/tex].
 
  • #13
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
 

1. What is a magnetic field on an axis?

A magnetic field on an axis refers to the strength and direction of the magnetic force at a specific point along an imaginary line passing through the center of a magnet, also known as the axis. This axis is usually considered to be the central line of a bar magnet or the imaginary line connecting the north and south poles of a magnet.

2. How is the strength of a magnetic field on an axis measured?

The strength of a magnetic field on an axis is measured in units of magnetic flux density, also known as Tesla (T). This measurement is obtained by using a device called a magnetometer, which can detect the presence and strength of a magnetic field.

3. What factors affect the strength of a magnetic field on an axis?

The strength of a magnetic field on an axis is affected by the distance from the magnet, the angle at which the magnet is placed, and the strength of the magnet itself. The closer the point is to the magnet, the stronger the magnetic field will be. Additionally, the stronger the magnet, the stronger the magnetic field will be at any given point.

4. How does the direction of a magnetic field on an axis change?

The direction of a magnetic field on an axis can change based on the position of the magnet. At the north pole of a magnet, the magnetic field lines are directed away from the magnet, while at the south pole, they are directed towards the magnet. Along the axis, the direction of the magnetic field will change as the magnet rotates or moves.

5. How is a magnetic field on an axis used in everyday life?

Magnetic fields on an axis have many practical applications in everyday life. They are used in compasses for navigation, in motors and generators for converting energy, and in speakers and headphones for producing sound. They are also used in medical imaging technology such as MRI machines to create detailed images of the body's internal structures.

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