1
$\begingroup$

I'm trying to make sense of the integrals which occur in Maxwell's equations and feel pretty comfortable with the notion of Gauß's law and Gauß's law for magnetism as (closed) surface integrals over flux densities, as well as with $\int_{x_1}^{x_2} \vec E \cdot \text{d}\vec s$ being the energy per charge which is needed/released to move a test charge $q$ from $x_1$ to $x_2$ in the presence of the elctric field.

But I'm struggling with the quantity $$\int_{x_1}^{x_2} \vec B \cdot \text{d}\vec s$$

Its unit is $\frac{N}{A}=\frac{Ns}{C}$ but this didn't lead to a picture for it. Neither did trying to connect it with work which helped me with understanding the corresponding quantity for the $\vec E$-field. I don't know how to invoke the Lorentz force $\vec F_L = q \vec v \times \vec B$ and even if I did, it doesn't even do any work.

So what kind of quantity is this integral? My goal is to really understand this quantity and not the other side of the corresponding Maxwell equation.

Improve this question
$\endgroup$
2
  • $\begingroup$ maybe multiplying by the speed of light might help. $\endgroup$
    – Phoenix87
    Commented Jan 3, 2017 at 15:56
  • $\begingroup$ Thanks for your suggestion. It didn't directly lead to something but it made me think about $\vec H$ and the unit of $\int \vec H \cdot d \vec s$ is simply $A$. So this is a current. Maybe it's easier to understand this first. Unfortunately, I don't really get the nature of $\vec H$ and $\vec D$ yet. My plan was to understand the microscopic Maxwell equations first. $\endgroup$
    – Marc
    Commented Jan 3, 2017 at 16:12

1 Answer 1

Reset to default
1
$\begingroup$

The integral $\int\vec{B}.\mathrm{d}\vec{s}$(or $\int\vec{B}.\mathrm{d}\vec{l}$ as I like to call it) has no significance what so ever as compared to $\int\vec{E}.\mathrm{d}\vec{l}$. This is because $\vec{E}$ is necessarily a gradient with a potential defined, provided $\vec{E}$ is electrostatic (If however $\vec{E}$ is created from a changing magnetic flux, it is not electrostatic and $\vec{E}\neq -\nabla V$ ). It is defined as $\vec{E}=-\nabla V$, where $V$ is the electrostatic potential. Now, the gradient for any function $f(\vec{r})$ is defined as $\nabla f(\vec{l}).\mathrm{d}\vec{l}=\mathrm{d}f(\vec{l})$, hence $\nabla V.\mathrm{d}\vec{l}=\mathrm{d}V$. This implies that $-\nabla V.\mathrm{d}\vec{l}=-\mathrm{d}V; \vec{E}.\mathrm{d}\vec{l}=-\mathrm{d}V; \int\mathrm{d}V=-\int \vec{E}.\mathrm{d}\vec{l}$ So, $-\int\vec{E}.\mathrm{d}\vec{l}$ is the amount of energy needed for a unit positive charge to travel through the said limits of the integration. This is how you define potential. The vector $\vec{B}$ on the other hand is not a gradient. What that means is that you cannot find a function $f(\vec{r})$ such that $\vec{B}=\nabla f$. But, the fact that $\vec{B}\neq \nabla f$ means $\oint\vec{B}.\mathrm{d}\vec{l}\neq 0$(unlike $\oint\vec{E}.\mathrm{d}\vec{l}=0$) and this quantity signifies a current in ampere's law. The law goes $$\oint_D\vec{B}.\mathrm{d}\vec{l}=\iint_S\mu \vec{J}.\mathrm{d}\vec{A}$$. It means that the loop, $D$ parallel to which you integrate your $\vec{B}$ The integration equals the amount of current flowing perpendicular to the surface area of that loop.

Improve this answer
$\endgroup$
9
  • 1
    $\begingroup$ Hmm. $\int \vec E \cdot \text{d} \vec s$ still has the property of being the energy a unit charge needs to travel along the path even if $\vec E$ cannot be derived of a potential because a changing magnetic flux makes the electric field non-conservative. So I'm not sure if your chain of reasoning is correct. $\endgroup$
    – Marc
    Commented Jan 3, 2017 at 16:53
  • $\begingroup$ Yes but you cannot define a potential for the electric field that arises due to changing magnetic fields $\endgroup$
    – Ghosal_C
    Commented Jan 3, 2017 at 16:55
  • $\begingroup$ That is my point. If I understood you correctly you say that we cannot hope for a physical interpretation of $\int \vec B \cdot d \vec s$ analogous to the verrsion involving $\vec E$ because $\vec E$ can be derived from a potential and $\vec B$ can't. But the line integral still has physical interpretation in terms of energy even in situations where $\vec E$ cannot be derived from a potential. $\endgroup$
    – Marc
    Commented Jan 3, 2017 at 17:02
  • $\begingroup$ And also in that case the closed circular integral of the field will not be zero. It is equal to the rate of change magnetic flux. The reason I was comparing to electrostatic fields is because you drew a comparison to Gauß's law and then went on to define potential which is the energy needed to transfer a unit charge from infinity to a certain point in presence of the charges field. In case of a field to changing magnetic flux, there ARE no charges. Hence no Gauß's law or potential the electric field then is similar to magnetic field with closed loops as lines force $\endgroup$
    – Ghosal_C
    Commented Jan 3, 2017 at 17:02
  • $\begingroup$ See, the line integral makes sense as a loop because it allows you to integrate the amount of current passing through it, but not as an energy since you do not have monopoles and the amount energy required to transfer a monopole in a non conservative field is path conscious and makes no concrete physical sense. So when you say from infinity to that point, you better make sure you specify which path you take because talking about the starting and ending points over a non conservative integral is absolutely pointless and half information $\endgroup$
    – Ghosal_C
    Commented Jan 3, 2017 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.