This answer explains that for an unloaded secondary, the natural phase relationship between primary voltage and secondary voltage is is zero degrees.
It therefore follows that if there is a secondary load current (due to a resistive load), the current in the primary due to that secondary resistive load must be 180 degrees out of phase with the secondary load current i.e. as current flows into the primary, current flows out from the secondary.
This of course is for an ideal transformer and a resistive load.
If you ignore the leakages and magnetization inductance of the transformer, and the load is reactive, then there will be a 90 degrees phase shift.
Bringing in leakage inductance and DC coil resistance will/can muddy the waters. Bringing in magnetization inductance muddies the water a bit more.
The low frequency transformer equivalent circuit is this: -
As you should be able to see, if you considered all the leakages, magnetization inductance and losses and then added a semi-reactive load, the phase angle is quite complex to calculate.
However I do-not know, in which-way Lenz's law acts in transformer;
because the law states the induced current will try to hinder the
cause.
Strictly speaking, it is voltage that is induced and any current that flows is subject to the that voltage, the load and the leakage inductance.
but when the secondary circuit of a step-up transformer turned On
(closed) , so-far I've know , the current in primary-coil goes-up
In normal usage, for a voltage transformer it is non-ideal to consider the secondary being short circuited. However, it makes no difference to the phase angle providing you obey the rules inherent to the model.
Best Answer
No, it doesn't do that; the changing magnetic field induces a voltage into the other coil. Any current in that coil is due to that induced voltage driving current through any load connected to that winding. Current cannot be induced, only voltage.
1 turn of wire wound on a bobbin might have 1 uH inductance. At 50 Hz, the impedance of that 1 turn is 0.000314 ohms (an inductor's impedance is \$2\pi f L\$). Clearly that is far too low to connect to any 50 Hz AC voltage source. So you wind more turns and what you find is that inductance increases as the turns are squared so, ten turns yields 100 uH inductance and 1000 turns yields 1 henry of inductance.
1 henry has an impedance of 314 ohms at 50 Hz and, with a little bit of generalistic hand waving and over-simplification, a transformer primary aims to be about 10 henry (3142 ohms).
If you applied 230 V 50 Hz to this primary winding you would find that the current is about 73 mA and won't cause a fire.
So, it's got nothing to do with the coil's resistance. It's all to do with getting enough inductance.