# Magnetic flux (and flux in general)

The general interpretation of flux as I understand it (and please correct me if I'm wrong) is that it represents how much something is going through another (surface or volume (and perhaps lines?)), I'll quote Khanacademy :

Magnetic flux is a measurement of the total magnetic field which passes through a given area.
Source

Considering that magnetism is a force, I very well understand that we only want the force that is pushing in the direction of the infinitesimal surface and keeping in mind the definition given before, it seems much logical to me to use this : $$\iint \frac{\mathbf{B}\cdot\mathbf{dS}}{|\mathbf{dS}|}$$ We find the direction with the dot product but take off the surface and then we sum up the force. I probably am misunderstanding the flux definition and hope someone would have the kindness to clear this up.

Edit : This integral can't be done since we no more have an infinitesimal to integrate with respect to it. My problem with this is that when I'm thinking that we're kind of distributing the force over that $$|\mathbf{dS}|$$ we'll be loosing "strength", $$|\mathbf{B}|*0.00000000.......1$$, I hope you're getting what I mean.

Edit 2 : I got it, I was thinking wrong from the beginning by ignoring the units, a fractional number of surface would still actually represent something because of the meaning of a square meter which is a finite quantity (a collection of points dare I say) and thus fractions of it are still finite quantities ($$n\to\infty\in \mathbb{N}$$ points forming an area) have escaped my thought, we are actually adding the strength "$$n$$ times", I was blind to the unit.

So you could try to do this, if $$\text d\mathbf S$$ is an infinitesimal area element whose magnitude is its area and direction normal to the surface. However this is not the standard definition of flux, and as you have seen this presents some problems. The standard form is to not divide by the magnitude of the area element: $$\Phi=\iint\mathbf B\cdot\text d\mathbf S$$ To more easily compare this to what you have written, we can express the integral as $$\Phi=\iint\mathbf B\cdot\left(\hat n|\text d\mathbf S|\right)$$ i.e. we have explicitly written out the magnitude $$|\text d\mathbf S|$$ and direction $$\hat n$$ of the area element $$d\mathbf S$$. Therefore, your definition becomes $$\Phi_{you}=\iint\frac{\mathbf B\cdot\left(\hat n|\text d\mathbf S|\right)}{|\text d\mathbf S|}=\iint\mathbf B\cdot\hat n$$

And this doesn't really make any sense. You will have an infinite sum of terms that do not go to $$0$$, and so your proposed flux approaches infinity.

This relates to your worry of "losing strength", but this is what we do in integrals we first see in introductory calculus: $$I=\int_a^b f(x)\ \text dx$$ this is the limit of an infinite sum, so we need the $$\text dx$$ in order to "remove the strength" so we have a converging sum. More explicitly we want $$I=\lim\limits_{N\to\infty}\sum_{i=0}^Nf(x_i)\Delta x$$ where $$x_i=a+i\Delta x$$ and $$\Delta x=\frac{b-a}{N}$$ so that $$\Delta x\to0$$ as $$N\to\infty$$. If we divide by our $$\Delta x$$ we get $$I=\lim\limits_{N\to\infty}\sum_{i=0}^Nf(x_i)=\lim\limits_{N\to\infty}\sum_{i=0}^Nf\left(a+\frac{b-a}{N}i\right)$$ which does not converge.

• I see how it can't be right mathematically, but I still can't make complete sens of multiplying $|\mathbf{B}|$ by $|\mathbf{dS}|$. Just now I was thinking about this and here's what went through my mind, the area is extremely small, we could say $|\mathbf{dS}|=\frac{1}{10^{999..}}m^2$, and thus $|\mathbf{B}|*|\mathbf{dS}|$ would be like dividing the strength many many times. What I am looking for exactly is how we see that multiplication, what does it represent, I can't make a sens out of it when the surface is a fractional number. – Luyw Apr 3 '19 at 15:14
• I got it, I was thinking wrong from the beginning by ignoring the units, a fractional number of surface would still actually represent something, the meaning of meters which is a finite distance and thus fractions of it are still finite distances have escaped my thought, I was blind to the unit. – Luyw Apr 3 '19 at 15:18
• @Luyw Sounds good :) Please consider upvoting any answers that have been useful as well as selecting an answer as the accepted answer to aid future readers. – BioPhysicist Apr 3 '19 at 15:21

Flux is analogous to number of things going perpendicular a surface and it is extensive in area (i.e. it doubles when the area doubles). As an example, normal force on a surface is the flux of pressure on it.

By definition, flux of $$\textbf{B}$$ through unit area vector parallel to it will be $$B$$, its magnitude, and for an area element $$\textbf{dS}$$, it will be the dot product. If you divide by the magnitude of the area as you suggested, then flux will not depend on the total amount of area that is considered.

In general we define flux as some field line say magnetic field lines passing through some surface. More the field lines, more is the flux. Its just a measure of flow.

• The OP knows this already – BioPhysicist Apr 3 '19 at 13:51
• I thought that he is confused with the definition, as he mentioned " some thing is going through volume/line ". It's only related to surface though we can relate it in terms of volume integral using Stokes theorem . – rabin sarkar Apr 3 '19 at 14:05
• Then you should be more specific. Right now your answer reads similar to the quoted Kahn academy definition. – BioPhysicist Apr 3 '19 at 14:06