Charge accumulation at a bi-metallic junction When we join two straight cylindrical wires of two different metals say 'iron' and 'copper' together such that their circular faces are in contact. If we make a constant current say I through the wires then is any charge is accumulated at the junction . To me it seems that there would be some charge as we have $ I = neAv $ and the $ nAv $ term for both can't be different as I is same through both ? I am very confused any help would be appreciated. Thanks in advance
 A: 
If we make a constant current say I through the wires then is any charge is accumulated at the junction 

Yes, at the junction there will be a surface charge provided that the resistivity of the two materials is different. This is described as a Type 1 surface charge in this paper:
https://www.tu-braunschweig.de/Medien-DB/ifdn-physik/ajp000782.pdf 
Basically, the current density must be the same in both wires, and since they have different resistivities that means that the E field must be different. So the E field is discontinuous at the junction which implies a surface charge at the junction. 
A: Can't we treat the iron and copper like 2 resisters separated by a perfectly conducting wire? In which case, you would consider their joint resistance as the sum of their resistances (in series), which of course vary due to difference in electron density. Then it becomes a case of calculating the current, by considering one single 'average' metal bar of the two. 
Edit: I was wrong! Definitely still learning here
