The discussion revolves around the integral form of the magnetic flux equation, specifically focusing on the role of the dot product in the integrand and the implications of varying angles across a surface.
Some participants have provided guidance on the correct expression for magnetic flux and discussed the nuances of the dot product in relation to varying angles. There seems to be an ongoing exploration of the implications of these concepts without a clear consensus.
Participants note that the cosine term is not constant across the surface, which complicates the integration process. There is also mention of the compactness of the notation used in the equation, suggesting a deeper complexity beneath the surface.
Yes, it is ##\phi =\iint_{S}B\cdot dS##TSny said:Can you give us the explicit expression for the integral form? Does the integrand have a dot product of vectors?
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 said: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.
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.Einstein44 said: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. 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 said:So you are saying that it is implied due to the dot product, but just not shown in the equation.
Got it, thanks!TSny said: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.