Heat flux is a vector beceause it has a magnitude and a direction. Furthermore it has these properties in every point in space, which makes it a vector field. You can think of an analogy with the mass flux in a medium with inhomogenius density; diffusion will tend to equalize the denisty everywhere, so there will be specific motion of mass at every point determined by its imediate surroundings.

The direction of the heat flux specifies for each point the direction of the fastest drop in temperature.

Finally, heat flux is normal to a isothermal surface, because if it wasn't it would have a tangential component along the isothermal surface at that point. That would in turn mean that there would be a non-zero temperature gradient (difference) along the surface, which would mean that it isn't a isothermal surface.

Further resources:

http://www.et.byu.edu/~vps/ME340/ME340.htm

http://www.amazon.com/books/dp/0470501960

http://freevideolectures.com/Course/3005/Heat-Transfer/1

So intuitively $\nabla \cdot B = 0$ is best understood by appeal to the continuity equation $\dot \rho = -\nabla \cdot J,$ in other words if you imagine $\vec B$ as a flow field $\vec v$ for some fluid of constant density, this describes a flow of stuff which does not accumulate at any particular points: it is not "flowing into" or "flowing out of" any of those places.

The expression $\nabla\times B = 0$ is a little harder to understand when you imagine $\vec B$ as a flow field $\vec v$, but if you imagine that you insert a little pinwheel into the flow field at a point, the curl says something about how much torque there will be on that pinwheel due to the fluid trying to drag its tines, with the direction of the curl being the direction that makes the pinwheel spin the most. So $\nabla\times B=0$ means literally "this flow field does not make a pinwheel spin." That helps to understand the strange flow field $\vec v = \alpha ~ \hat \theta / r,$ which seems to "curl" around the origin but has $\nabla \times \vec v = 0$ at all points that are not the origin. If you think about a tiny little area $dr~r~d\theta,$ it has a slightly longer "outside edge" than its "inside edge" and so the flow along that outside edge drags the pinwheel slightly more; but if the flow drops like $1/r$ this slight lengthening is exactly compensated by the slight attenuation in the flow field, meaning that the pinwheel is not rotated at all.

Helmholtz's theorem says that any "nice" vector field $X$ can be
"integrated" into a sum of two fields, a curl and a gradient: $$X = \nabla \times Y - \nabla \zeta.$$Since $\nabla \cdot X = -\nabla^2 \zeta$ and $\nabla \times X = \nabla \times (\nabla \times Y)$ due to standard vector identities, whenever we know that $\nabla \cdot X = 0$ we can solve this trivially with $\zeta = 0$ and we know that $X = \nabla \times Y$ for this "vector potential" $Y.$ Similarly when we know that $\nabla\times X = 0$ we know that $Y=0$ works and that $X = -\nabla \zeta$ for this "scalar potential" $\zeta.$

# How this plays out in classical E&M

The full Maxwell Equations are (SI units) $$\begin{array}{ll}\nabla \cdot E = \rho/\epsilon_0&~~\nabla \times E = -\dot B\\
\nabla \cdot B = 0&~~\nabla \times B = \mu_0 J + \mu_0\epsilon_0 \dot E.
\end{array}$$
Clearly we always have $B = \nabla \times A$ for some vector potential $\vec A$ by the third equation. More subtly, the second equation then works out to $\nabla \times (E + \dot A) = 0$ and therefore $E = -\dot A - \nabla \phi$ for some scalar potential $\phi.$ The choices that you make for these are not unique as you can add any $\nabla \psi$ to $A$ and still preserve $B,$ because the curl of a gradient is zero. If you work out the effect of this on $E$ you'll see that you have to also subtract $\dot \psi$ from $\phi$ to preserve $E$ as well. This is called the "gauge freedom" but it really reflects this freedom to choose your "constants of integration" in the Helmholtz decomposition.

The remaining equations work out to $$\begin{array}{rl}-\nabla\cdot \dot A -\nabla^2 \phi ~=& \rho/\epsilon_0\\
\nabla(\nabla \cdot A) - \nabla^2 A ~=& \mu_0 J - \mu_0\epsilon_0\ddot A - \mu_0\epsilon_0\nabla \dot \phi,\end{array}
$$ once you use this identity that $\nabla \times (\nabla \times X) = \nabla(\nabla\cdot X) - \nabla^2 X.$ These can be put into a very elegant form by recognizing the wave operator $\square X = \mu_0\epsilon_0 \ddot X - \nabla^2 X$ and then incorporating the other spare terms in the second equation into a scalar field $\lambda = \mu_0\epsilon_0\dot \phi + \nabla\cdot A.$

Note that your gauge freedom maps $\lambda \mapsto \lambda - \square \psi$ so you can basically choose any functional form you want for $\lambda$ by solving a wave equation, in theory. Note also that $\lambda$ enters into the first equation via replacing $\nabla\cdot \dot A$ with $\dot \lambda - \mu_0 \epsilon_0 \ddot \phi,$ so you get $$\begin{array}{rl}\square \phi ~=& \rho/\epsilon_0 + \dot \lambda\\
\square A ~=& \mu_0 J - \nabla\lambda.\end{array}$$The "Lorentz gauge" sets $\lambda = 0$ because $(c\rho, J)$ is a 4-vector in relativity where $c = 1/\sqrt{\mu_0\epsilon_0}$ and hence this gives you an equation $\Box(\phi/c, \vec A) = \mu_0 (c\rho, \vec J),$ and since $\Box$ is relativistically covariant you get that $(\phi/c, \vec A)$ is also a 4-vector in relativity. If that doesn't make much sense to you just accept "hey, it makes these equations look exactly alike" and that's good enough.

The "Coulomb gauge" sets $\lambda = \mu_0\epsilon_0 \dot\phi$ so that the first equation reduces to just $-\nabla^2\phi = \rho/\epsilon_0,$ and we can basically assume that the scalar potential $\phi$ updates instantaneously to all of the charges, which makes it super-simple to calculate. If you look up at the definition of $\lambda$ you find out that it also sets $\nabla\cdot A = 0$ so $A$ can be integrated in terms of a vector potential potential, $A = \nabla \times W.$

## Best Answer

The relation between the scalar $j$ and the vector $\vec j$ is simply :

$$j = \vec j \cdot \vec n$$

where $\vec n$ is a unit vector normal to the surface $\mathrm{d}A$.

So, you could write:

$$j ~ \mathrm{d}A = \vec j \cdot \vec n ~ \mathrm{d}A = \vec j \cdot \vec {\mathrm{d}A}$$

with the notation $\vec {\mathrm{d}A} = \vec n ~ \mathrm{d}A$

So, choose the notation that you prefer, but it must be clear for you.

Suppose, for instance, that you are working with a conserved quantity $Q$. We may define a quantity density $\rho$, and a quantity flux $\vec j$, so that the local law of the conservation of the quantity is written:

$$\frac{\partial \rho}{\partial t} + \operatorname{div} \vec j = 0$$

Considering now a volume $V$, with boundary surface $A$. The variation of the quantity $Q$, in the volume V, between 2 times $t_1$ and $t_2$ is:

$$\Delta Q = \int_{t_1}^{t_2} \int_V \frac{\partial \rho(\vec x,t)}{\partial t} \mathrm{d}^3x = - \int_{t_1}^{t_2}\int_V \operatorname{div} \vec j \mathrm{d}^3x = - \int_{t_1}^{t_2} \int_A \vec j.\vec {\mathrm{d}A}$$

The last equality comes from Stokes Theorem.