NPN circuit (breadboard)
It is simply a test circuit for you to test and observe.  It is not a &quot;complete&quot; circuit in the sense you expect - it is not a &quot;switch.&quot;
The point of the exercise is for you to carry out the experiment and develop an understanding of how to assemble the circuit and test setup, and then to gain an understanding of how the base current of a transistor behaves.  It does you no good for someone to describe it to you - you need to carry out the experiment.


If you have difficulty carrying out the experiment, then you should explain the problem you are having and ask what is going wrong.
Once you carry out the experiment, you could post your completed setup and the results and ask if you have done it correctly and if the results are reasonable.

Since there seems to be some dispute about what \$V_{bias}\$ plotted against \$I_B\$ would look like, I whipped up a DC sweep generator and generated the plot.
I used an Arduino Nano I had laying around handy to make the \$V_{bias}\$ generator.  The Nano puts out a 10kHz PWM signal that is filtered down to DC.  By varying the duty cycle of the PWM signal, I can change the output voltage.  The PWM generator has 1023 steps.
This is the circuit:

The red circled area generates the \$V_{bias}\$ from the PWM signal.
R2 takes the place of \$R_b\$ in the drawing in the question.  I used a 2N3904 transistor because I had one close to hand.
The Nano can only measure voltage, but \$I_b = I_{R_2} = (V_1-V_2)/R_2\$.
I wrote a simple Python program to drive the Nano, collect the data, and crunch the numbers.  It dumps the finished numbers into  CSV file, and I used LibreOffice to make the plot itself.
This is the plot of \$I_B\$ against \$V_{bias}\$:

As you can see, the statement &quot;\$I_b\$ vs \$V_{bias}\$ does not make much sense because practically, it is the IV curve of the resistor \$R_b\$&quot; is incorrect.
The plot of \$I_B\$ against \$V_{bias}\$ clearly shows the typical response of a diode junction.
For comparison, here's the plot of \$V_{B}\$ against \$I_B\$:

That shows the diode response of the junction as well.