Ampere's law in differential form

[DottZakapa]

According to this image, in the attached files there is the demonstration of the ampere's law in differential form. Bur i have some difficulties in understanding some passages. Probably I'm not understanding how to consider those two magnetic vectors oriented and why have different name.

in particular:

the following, i don't get why do i need such partial derivative

following, i don't understand why that minus in front of the partial derivative

Is there anyone so kindly patient to help?
thanks

 

[Delta2]

By the scheme, we can see that $B_y$ is the y-component of magnetic field at location $x+dx$ (along the segment QR), and $B'_y$ is the y-component of magnetic field at location $x$ (along the segement PS) (I consider the point $P$ has coordinates $(x,y)$). The definition of the partial derivative of $B_y$ with respect to $x$ is (for infinitesimal $dx$)
$$\frac{\partial B_y}{\partial x}=\frac{B_y(x+dx)-B_y(x)}{dx}=\frac{B_y-B'_y}{dx}$$
from which it follows that $$B_y-B'_y=\frac{\partial B_y}{\partial x}dx$$.

The minus sign is there because we do the convention to consider the positive direction of walking through the loop, the counter clockwise direction. So we take as positive the $B_x(y)dx$ and negative the $B_x(y+dy)dx$, so the contribution will be $(B_x(y)-B_x(y+dy))dx$ and by similar reasoning as that in the previous paragraph (with the roles of x and y swapped) we can see that $(B_x(y)-B_x(y+dy))=-\frac{\partial B_x}{\partial y} dy$

 

[DottZakapa]

Is this the decomposition of the two vectors B' and B that i have to consider?
If not, my main problem is that I'm not understanding how the two vectors ore oriented in the lines. I understand what is the positive circulation of the integral, the main thing is that i am not seeing where on each segment the vector component is pointing.

 

[Delta2]

Yes that is the decomposition of the two vectors. What's the problem with it?

We don't have to know if they point left or right or up or down. The reason that they become negative when we take the dot product $\vec{B}\cdot d\vec{l}$ is that the vector $d\vec{l}$ (which always points counterclockwise) becomes negative with respect to the positive unit vector $u_x$ or $u_y$. So for example at the segment PQ it is $d\vec{l}=dx\vec{u_x}$ while at the segment RS it is $d\vec{l}=-dx\vec{u_x}$

 

[DottZakapa]

Yes that is the decomposition of the two vectors. What's the problem with it?

We don't have to know if they point left or right or up or down. The reason that they become negative when we take the dot product $\vec{B}\cdot d\vec{l}$ is that the vector $d\vec{l}$ (which always points counterclockwise) becomes negative with respect to the positive unit vector $u_x$ or $u_y$. So for example at the segment PQ it is $d\vec{l}=dx\vec{u_x}$ while at the segment RS it is $d\vec{l}=-dx\vec{u_x}$

i agree with that so as you say it is
$\int_p^q B_x\,dx= Bdx$

$\int_r^s B_x'\,dx=\space -B'dx$

therefore
$(B_x-B_x')dx$

then

$B_x-B'_x=dB_x$

$dB_x=$
how the minus comes up in front of the partial?

 

[Delta2]

it is actually $B_x-B'_x=-dB_x$ because $B'_x=B_x(y+dy)$ and $B_x=B_x(y)$.