# What kind of quantity is $\int_{x_1}^{x_2} \vec B \cdot \text{d}\vec s$?

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.

• maybe multiplying by the speed of light might help. – Phoenix87 Jan 3 '17 at 15:56
• 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. – Marc Jan 3 '17 at 16:12

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.
• 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. – Marc Jan 3 '17 at 16:53
• 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. – Marc Jan 3 '17 at 17:02