Where the current's emitter BJT Transistor comes from?
You need to know what the opamp is doing first. It looks like the opamp is performing two tasks:


Translating the voltage V1 from 5v to 10v. (by comparating inputs and saturating)
It's also buffering V1.


All this assuming that the VCC of the opamp is 10v. And it seems like the opamp is saturating to VCC. This means the output of the opamp is at 10v volts (assuming full swing opamp).

Back to your question, about the transistor. It's not switching as you think, it's not completely saturated nor in cutoff, it's on the "Active Region", the current is coming from the transistor (that's the point of the circuit).



Source of image:
https://www.electronics-tutorials.ws/transistor/tran_4.html

To fully understand your circuit:
Ok, why would I make this circuit? What's the purpose?

Three reasons:


Voltage
Current
Control


The voltage of your apparent signal is 5V. But the circuit is translating it to 10v.

Opamps usually can't supply too much current (usually 20mA max). But your load consumes 50mA approx. That's why you need the BJT.

You may ask, why not connect the load directly to the 10v rail?
The answer is "Control": With this circuit you can control your load with the 5v signal, it may be turned on or off. Yes, this is a driver. And the good thing about it is that you can control 10v with 5v.