# Flux through a circle not centered at origin.

• Kizaru
In summary, the magnetic flux through a circle in the xz plane of radius "a" due to a wire is given by: \Phi = k \iint _S \frac{1}{\rho} d\rho dz where k is a constant and the limits of integration over the inner integral are not constant.
Kizaru
I feel really dumb for this, but I keep getting strange answers so I may be forgetting something.

## Homework Statement

Find the magnetic flux through a circle in the xz plane of radius "a" due to a wire. . The center of the circle is (b,0,0).

## Homework Equations

Well I found the magnetic field to be proportional to
$$\frac{1}{\rho}$$

That field has only a component in the phi direction.

Where rho and phi are cylindrical coordinates.

## The Attempt at a Solution

I can't figure out the limits of integration for anything unless it's rectangular coordinates, but those become very messy and I obtain an integral which is clearly wrong. I KNOW the answer should reduce to something "nice."

I know I'm missing something stupid here.

Where is the wire? I guess it's the z-axis?

Yes, it is. It's an infinite wire carrying current I in the z-hat direction. Basically using the result of a previous problem, I know what the B field is. I just chose not to include the rest of the stuff for that field in this question.

According to wikipedia, $$\Phi _B=\iint _S \vec B \cdot d \vec S$$.
I'm not able to solve the problem so I'd love someone to help us.
If you have the expression of B in all the circle, I think it's easy to solve the integral.

I obtained B and I know what the definition of the magnetic flux is. I also know the vector element of area for cylindrical coordinates results in an integral over just $$\frac{1}{\rho}$$

IE, I have

$$\Phi _B= k \iint _S \frac{1}{\rho} d\rho dz$$
where k is a constant (having trouble texing the constants).
The issue is I am unsure of what to make the limits of integration.

Kizaru said:
I obtained B and I know what the definition of the magnetic flux is. I also know the vector element of area for cylindrical coordinates results in an integral over just $$\frac{1}{\rho}$$

IE, I have

$$\Phi _B= k \iint _S \frac{1}{\rho} d\rho dz$$
where k is a constant (having trouble texing the constants).
The issue is I am unsure of what to make the limits of integration.
Are you sure you have a $$d\rho dz$$ as a differential area? You said you were working cylindrical coordinates, I'm confused.

What? Phi is the angle that goes counterclockwise from the x axis. I would NOT be integrating over dphi because phi is constant = 0. The circle lies in the xz plane.

Draw the circle out on a piece of paper to see what I mean. Clearly rho does not go from 0 to a. Rho goes from (b-a) to (b+a) when z =0, and it goes from b to b when z = a. The limits of integration over the inner integral are clearly not constant.

Well, you could divide the circle into infinitesimal horizontal slices and calculate what is the area dA of a slice as a function of x (through trigonometry). On the circle, B is proportional to 1/x so the flux through a slice would be just B(x)(dA/dx)dx which you could integrate along the x axis. However, this gives a complicated integral which I'm not sure how to solve.

I tried a similar approach and obtained an integral which Mathematica integrated to have imaginary parts != 0. I am curious if I could shift my origin (make a substitution) and then convert to spherical coordinates (probably the other order). The spherical coordinates integral (after the substitution) for r would be from 0 to a, and for theta it would be 0 to pi. After the origin shift, my B field will still be in phi-hat direction only due to the symmetry.

My error, I thought it was in the x-y plane, sorry about that.

Kizaru said:
I tried a similar approach and obtained an integral which Mathematica integrated to have imaginary parts != 0.

It does that because it doesn't know the values of a and b, they might be such that the thing inside the square root is negative. If we assume that b>a>0, however, it's clear that the value of the integral is real. You have to use assumptions, though i don't know how to do that in Mathematica. Anyway, I got (assuming B(x) = B0/x)

$$\Phi = \int_{b-a}^{b+a} \frac{2 B_0 \sqrt{a^2 -(b-x)^2}}{x}dx = 2 B_0 a \int_{\frac{b}{a}-1}^{\frac{b}{a}+1} \frac{ \sqrt{1 -(\frac{b}{a}-x)^2}}{x}dx$$

which according to Maple is

$$2 B_0 a \,{\frac {\pi \, \left( 1+\frac{b}{a}\,\sqrt {-1+{\frac{b^2}{a^2}}}-{\frac{b^2}{a^2}} \right) }{ \sqrt {-1+{\frac{b^2}{a^2}}}}}$$

which can be simplified to $$2 \pi B_0 (b - \sqrt{b^2-a^2})$$

Which isn't that bad in the end (assuming I didn't make a mistake). I'm not sure how to calculate the integral by hand though.

Last edited:
YES! Thank you. That's exactly what the result should simplify to (it's very similar to that of a toroid, after all a toroid is roughly just N of those). Thank you. You were right, I didn't make the assumptions in Mathematica, but I'm still quite a novice and don't know how to do that yet.

Thank you :)

## 1. What is flux through a circle not centered at origin?

Flux through a circle not centered at origin refers to the amount of flow or movement of a vector field through a circle that is not located at the coordinate origin. It is a measure of how much of the vector field passes through the circle.

## 2. How is flux through a circle not centered at origin calculated?

The flux through a circle not centered at origin is calculated using the formula Φ = ∫∫F⋅dS, where Φ is the flux, F is the vector field, and dS is the infinitesimal surface area element of the circle.

## 3. Why is the circle not being centered at origin important for calculating flux?

The location of the circle is important because it affects the direction and magnitude of the vector field passing through it. If the circle is not centered at origin, the vector field passing through it will be influenced by the coordinates of its center, resulting in a different flux value.

## 4. What factors can affect the flux through a circle not centered at origin?

The flux through a circle not centered at origin can be affected by the magnitude and direction of the vector field, the size and location of the circle, and the orientation of the circle with respect to the vector field.

## 5. How is flux through a circle not centered at origin useful in science?

Flux through a circle not centered at origin is useful in many scientific fields, such as physics, engineering, and meteorology. It can help predict the flow of fluids, the movement of particles, and the behavior of electric and magnetic fields. It is also used in the study of fluid dynamics and electromagnetic radiation.

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