electromagnetismmaxwell-equations
The magnetic monopoles does not exist which can be shown by $ \int {\vec{B} \cdot d\vec{A}} = 0 $.
But in Faraday's Law of electromagnetic induction, we clearly show the EMF induced is the time rate of change of the magnetic flux, which is $ E = -\frac{d\Phi_B}{dt} = -\frac{d\int{\vec{B}\cdot d\vec{A}}} {dt}$.
Now if $ \int {\vec{B} \cdot d\vec{A}} = 0 $ then shouldn't the induced emf be zero?
I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn.
Self inductance should be considered, but is left out for simplicity. So for example, if you have a planar circuit with inductance $L$, resistance $R$, area $A$, and there is a magnetic field of strength $B$ normal to the plane of the circuit, then the EMF is given by $\mathcal{E}=-L \dot{I} - A \dot{B}$.
This means, for example, that if $\dot{B}$ is constant, then, setting $IR=\mathcal{E}$, we find $\dot{I} = -\frac{R}{L} I - \frac{A}{L} \dot{B}$. If the current is $0$ at $t=0$, then for $t>0$ the current is given by $I(t)=-\frac{A}{R} \dot{B} \left(1-\exp(\frac{-t}{L/R}) \right)$. At very late times $t \gg \frac{L}{R}$, the current is $-\frac{A \dot{B}}{R}$, as you would find by ignoring the inductance. However, at early times, the inductance prevents a sudden jump of the current to this value, so there is a factor of $1-\exp(\frac{-t}{L/R})$, which causes a smooth increase in the current.
You are worried that EMF caused by the circuit's inductance will prevent any current from flowing. Consider the planar circuit as in part one, and suppose there is a external emf $V$ applied to the circuit (and no longer any external magnetic field). The easiest way to see that current will flow is by making an analogy with classical mechanics: the current $I$ is analogous to a velocty $v$; the resistance is analogous to a drag term, since it represents dissipation; the inductance is like mass, since the inductance opposes a change in the current the same way a mass opposes a change in velocity; and the EMF $V$ is analogous to a force. Now you have no problem believing that if you push on an object in a viscous fluid it will start moving, so you should have no problem believing that a current will start to flow.
To analyze the math, all we have to do is replace $-A \dot{B}$ by $V$ in our previous equations, we find the current is $I(t) = \frac{V}{R} \left(1-\exp(\frac{-t}{L/R}) \right)$, so as before the current increases smoothly from $0$ to its value $\frac{V}{R}$ at $t=\infty$.
This is where it's a good time to "converse with the math". Let's look at the equation:
$$\mathcal{E} = \oint_\gamma \mathbf{E} \cdot d\mathbf{l}$$
which in this case equals $\mathcal{E}_\mathrm{ind}$. This is based on the work formula:
$$W = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$
Thus, what your question is, is essentially, asking "how can you have negative work". Looking at the above equation and recalling how a dot product behaves, there's only one way: $\mathbf{F}$ and the element of displacement $d\mathbf{r}$ must be aimed at cross purposes with each other. In the case of the integral for $\mathcal{E}_\mathrm{ind}$, the same goes only with $\mathbf{E}$ and $d\mathbf{l}$.
Hence, the answer to your question is: when $\mathbf{E}$ points opposite to $d\mathbf{l}$.
But what does that mean? Well, the key here is that we have to think a little more closely about the work formula. I believe what you are imagining it means is "the work done by the force as the force pulls the particle along with it". It actually is more general - in a work integral, the particle can be moved in any direction, including against the force. Of course, to make that motion happen in real life, you need to supply a source of contrary force, but that doesn't change the maths. This is why, say, in a more elementary example, you can talk of "negative work" done by the gravitational force when you lift an object off the floor.
(Why is it defined that way? Well, for one, because we often can't solve for the "real" trajectory the particle follows! If we stipulated that as a precondition, it would make work an extremely non-trivial concept!)
Where your mistake lies in, then, is in missing that. The displacement around the wire $d\mathbf{l}$ that we use to describe emf is not the displacement that necessarily occurs in reality to a real positive charge (after all, in many applications we aren't "really" dealing with positive charges anyways!).
Rather, it is a hypothetical one where we imagine that we grab a charge and move it around all the way through the circuit in a specific, fixed direction, and ask what the work - whether positive or negative - done by the electric force for that movement is. If we ask about the work done in the actual movements of charges, we will get a different answer, and yes, this one will always be positive, at least provided we don't get into looking at the situation in too-fine detail.
From an intuitive point of view, if you want to figure when the force is doing "positive" work and when it's doing "negative" work, imagine that you can feel the electric force tugging on the positive charge in your hand as you move it through the circuit. When you feel it helping you, i.e. the tug is with the motion of your hand, at that moment (i.e. that small increment $d\mathbf{l}$) the electric force is doing positive work, and you are doing negative work (to retard the charge, if you were to try and not naturally speed up your hand as you'd likely tend to, of course). When you feel it is fighting you, i.e. the tug is against the motion of your hand, the electric force is doing negative work, and you are doing positive work (to help it against the contrary pull). Total negative work, and hence negative emf, will be if, in moving the charge, you had to fight more often than flow.
Best Answer
This can be resolved by being clear about what surface you're integrating over. In the first equation, $$ \oint {\overrightarrow{B} . \overrightarrow{dA}} = 0, $$ you're integrating over any closed surface, i.e. a surface without a hole in it, such as a sphere. The equation says that the magnetic flux coming in must equal the magnetic flux going out.
But in the second equation, $$ E = \frac{d}{dt} \int_\Sigma {\overrightarrow{B} . \overrightarrow{dA}}, $$ you're integrating over a surface with a hole in it, where the hole is a loop of wire, as shown in this diagram from Wikipedia:
This equation says that the rate of change of flux passing through the surface must equal the EMF in the wire loop. You can imagine it as coming in through the hole, and out through the surface. (Or the other way round or a bit of both.)
You can see a connection between the two equations if you imagine making the wire loop smaller and smaller until the hole closes completely. Then you're back at a closed surface again, where the first equation applies - and this infinitely small loop of wire can never experience an EMF.