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.