Question about the Magnetic Flux equation in Integral form

[Einstein44]

Homework Statement

I was a bit confused with all the formulas for Flux, so here's my question...
I know that for a uniform B field flux is defined as $\phi =BAcos\Theta$, however when looking at the integral form I never saw it written with $cos\Theta$... Is there a reason? Could it still be written with a cos at the end??

Relevant Equations

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[TSny]

Can you give us the explicit expression for the integral form? Does the integrand have a dot product of vectors?

 

[Delta2]

Yes, @TSny implies that the $\cos\theta$ term is hidden inside the dot product of the vectors. But is not a constant term like it is in the case of a uniform magnetic field and a plane surface, it varies as we move from point to point on the surface, which is what makes the integration difficult.

 

[Einstein44]

Can you give us the explicit expression for the integral form? Does the integrand have a dot product of vectors?

Yes, it is $\phi =\iint_{S}B\cdot dS$

 

[Einstein44]

Yes, @TSny implies that the $\cos\theta$ term is hidden inside the dot product of the vectors. But is not a constant term like it is in the case of a uniform magnetic field and a plane surface, it varies as we move from point to point on the surface, which is what makes the integration difficult.

So you are saying that it is implied due to the dot product, but just not shown in the equation. I am not sure I fully understand what is going on, but I get what you are saying.

 

[Delta2]

So you are saying that it is implied due to the dot product, but just not shown in the equation. I am not sure I fully understand what is going on, but I get what you are saying.

Yes well you wrote correctly the formula at post #4, but the whole point there is that $B\cdot dS$ is a very cute and compact formalism that hides a mini can of worms underneath: If you want to exactly write its equal expression using cosine of angle its a kind of trouble.

 

[TSny]

So you are saying that it is implied due to the dot product, but just not shown in the equation.

Yes. Note that for a uniform B field, you can write the flux in terms of a dot product as $\Phi = B S \cos \theta = \vec B \cdot \vec S$ where in the last expression the $\cos \theta$ is "hidden" in the dot product. You just want to make sure that you understand the meaning of an area vector.

 

[Einstein44]

Yes. Note that for a uniform B field, you can write the flux in terms of a dot product as $\Phi = B S \cos \theta = \vec B \cdot \vec S$ where in the last expression the $\cos \theta$ is "hidden" in the dot product. You just want to make sure that you understand the meaning of an area vector.

Got it, thanks!