How do the effects of semiconductor doping affect the Hall effect? For instance, consider number 4 and 5 in the following sample:

Using the right hand rule, B points downwards, conventional current points to the right (because of the 5V battery), and therefore, the force on electrons points into the page. Electrons are going into the page from the red wire to the black wire and conventional current is going from the black wire to the red wire. But when conventional current goes from ground (black wire) to the higher voltage (red wire), then the voltage must be negative. Therefore, the voltmeter would read a negative reading.
However, I am unsure what kind of effects doping the semiconductor would have on the voltmeter.
 A: *

*Only charges moving inside the main conductor (germanium here) represent the Hall effect, so you should be careful not to lead with statements about charges moving within wires of a voltmeter. 

*Electrons move from Q to A due to the electric field, and due to the Lorentz force are driven INTO the page along the germanium, thereby producing negative charge at the negative terminal of the voltmeter. This actually causes a positive reading. 

*In metals the charge carrier is electrons, so to refer to electrons in the analysis of th Hall effect as above, is appropriate. But In doped semiconductors either the positive (P) or negative (N) charge carrier is dominant by several orders of magnitude. Depending on doping only one reference is valid.
So it is clear that Q4 requires you to analyse movement of "holes" (P doped) going in the general direction A to Q due to the electric field, and collecting closer to the negative terminal of the voltmeter due to Lorentz force leading to negative voltmeter reading
Q5 deals with n doping, so the analysis is the same as in a metal, as per (2)
