Magnetic fields across lines and surfaces

[IniquiTrance]

I know that \int_{S}^{}\int_{}^{}\vec{B}\cdot d\vec{A} = 0 because \textbf {div} \vec{B}=0

IE, because \Phi_{B} leaving a closed surface must equal \Phi_{B} entering.

Yet how is it then that \int_{C}^{}\vec{B}\cdot d\vec{l} isn't also equal to zero?

Shouldn't it be true for any closed path that the amount of magnetic field lines leaving the perimeter of the path be equal to the amount entering, so that there be no net amount of field lines across it?

If it is true for a closed surface, shouldn't it be true for a closed path?

Thanks!

 

[AUMathTutor]

I think that the first thing you have there is for a closed surface, ie, a volume enclosed by a surface. The magnetic flux is zero for a surface enclosing a volume because magnetic field lines exit and enter in equal numbers, because there are no magnetic monopoles.

The second thing is a line integral around a curved path. A path can never enclose a volume. It's not true that the magnetic field lines passing through *any* surface must cancel out. In fact, I can give you an example of the second that isn't zero...

Let B = M(-sin theta, cos theta) / r^2 and let C : x=cos theta, y=sin theta, 0<t<2PI. Then C': x=-sin theta, y=cos theta, and the integral becomes the integral from 0 to 2 of M(sin^2 theta + cos^2 theta) = M which turns out to be (2PI)M.

I think...

 

[Phrak]

Shouldn't it be true for any closed path that the amount of magnetic field lines leaving the perimeter of the path be equal to the amount entering, so that there be no net amount of field lines across it?

What is the 'perimeter' of a path?

 

[Phrak]

Shouldn't it be true for any closed path that the amount of magnetic field lines leaving the perimeter of the path be equal to the amount entering, so that there be no net amount of field lines across it?

Think what you're asking. What is the perimeter of a path?