Why Voltage Increases in a step up Transformer rather than the Current? In a step up transformer, we all know voltage increases and current decreases, the answer to more voltage is because it has more windings, but why does more windings induce more voltage, can't the secondary windings generate more current but less voltage? 
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I've seen answers that there is less displacement current and more conduction current, some said the electrons are more congested due to more windings, and these answers were apparently wrong (said the experts from this forum).
 A: Transformers are made so that the same magnetic flux that goes through one coil goes through the other. Using the relation
$$\epsilon=N\frac{\text d\Phi}{\text d t}$$
for the magnitude of the induced EMF, we see that more coils induces a larger voltage. This is because each loop receives the same flux, so each loop gets an induced EMF of $\frac{\text d\Phi}{\text d t}$, and you can think of an entire coil as a system of loops in series.
Since the power in each coil must be equal, we have $P_1=P_2$, or $I_1V_1=I_2V_2$, so if one coil, say $1$, has a higher voltage, then the other coil ($2$) must have a higher current amplitude.
A: 
can't the secondary windings generate more current but less voltage?

Yes, certainly. But this would be called a step down transformer. That is just a step up transformer run in reverse. It is not a matter of physics but just a matter of which sides you label as in and out. So a step up transformer by definition is one that increases the voltage and decreases the current. Take the same piece of hardware and turn it around and it becomes a step down transformer which decreases voltage and increases current. 
A: …but why does more windings induce more voltage, can't the secondary windings generate more current but less voltage?
More windings (more turns, $N$) results in more voltage because of Faraday’s Law of induction given in @Aaron Stevens answer.  
The question of current vs. voltage in the primary winding vs. the secondary winding of an ideal (lossless) transformer is a matter of conservation of energy, which requires that:
$$V_{primary}I_{primary}=V_{secondary}I_{secondary}$$
That is, watts into the transformer equals watts out of the transformer or, for a given time $t$, Joules into the transformer equals Joules out of the transformer.
Hope this helps.
A: Aaron Stevens answer gives you the basic answer about more windings having a higher voltage. Faraday's Law is part of EM theory, summarized in the more mathematically sophisticated Maxwell's equations. Basically, a changing current on the input side (primary) of the transformer produces a changing magnetic field (Ampere's circuit law). The changing magnetic field passes through the windings in the output (secondary) side and by Faraday's law induces an electromotive force in the secondary side which results in a voltage around the closed loop of the secondary. 
The number of windings on the primary side determines the amplitude of the magnetic field and hence the size of the rate of change.  Each coil/winding on the secondary side responds to the changing magnetic field. The mathematical result is that the voltage induced on the secondary side is proportional to the ratio of the secondary coils to the primary coils. As mentioned in Bob D's answer, the power must be conserved (you can't get something for nothing!) so the current must be have inversely. In short, the secondary voltage is influenced directly by the windings ratio.
