Current flow through inductor (large inductance) in AC circuit
The transient occurs because, when you turn on the signal generator, the inductor's current started at zero when the sine voltage starts at zero.  That's not a normal, pure-AC situation where the inductor's voltage is always 90deg shifted.

Remember:  inductor current in a pure AC circuit should always be at 90deg phase relative to inductor voltage.  For pure AC situations, inductor current and voltage are never zero at the same time.  But when you turned on your signal generator, V &amp; I were both zero. That means that you've started out with some sort of DC-based situation, and it has to decay away, until finally the pure-AC signals are left.

Or said another way:  For the pure-AC wave, when the inductor current is zero, what's the inductor voltage at that instant?  To start up in pure-AC mode, a voltage must already be applied at t=0.   To get rid of the transient pulse, you'd have to start up your sine-voltage generator not at phase=0deg, but with a phase shift.  What phase?  It's the phase which gives the proper sig.gen. voltage at the instant that I=0 for the circuit!  :)  Starting up with any other phase ...that would be the same as applying a DC pulse.

In other words, the inductor is seeing the slow-rise of the first half of your initial sine wave as a DC pulse, a big voltage with the wrong current for producing 90deg phase.  This pulse must fade away before the magnitudes and phases will be correct for pure-AC.

Simple version:  remove the 10ohm resistor, then apply a cosine voltage signal (or, a sine signal with 90deg shift either pos or neg phase.)  That hits the inductor with a large initial voltage while its initial current still remains zero.   The transient will be gone.  Why?