as the rod will move, the positive chrg will get "accumulated" on one side and negative on other(due to force by magnetic field)..find the direction using q(vXB)..positive chrg will go to B and -ve to A..chk again!!
hence,
B is at higher potential
as the rod will move, the positive chrg will get "accumulated" on one side and negative on other(due to force by magnetic field)..find the direction using q(vXB)..positive chrg will go to B and -ve to A..chk again!!
hence,
B is at higher potential
ok thanx.....and what will be direction of emf.....i have no idea if we take dir. from + to - or - to +........plz tell!