The Hall Effect and the Sign of the Carriers
Push a current down a conducting bar and put it in a magnetic field across the flow, and the moving charges feel the Lorentz force $q\mathbf{v}\times\mathbf{B}$, which shoves them sideways. They pile up against one edge until the electric field of the accumulated charge grows strong enough to cancel the magnetic push; from then on the carriers drift straight again, but the bar is left holding a transverse voltage, the Hall voltage $V_H = IB/(nqt)$. Its size measures the carrier density $n$, and crucially its sign reveals the sign of the carriers. That is the subtle part: positive carriers drifting one way and negative carriers drifting the other are deflected to the very same edge (flipping the charge also flips the velocity), so you cannot tell them apart by where they go. What differs is the charge they deposit, so the polarity of the Hall voltage flips. Toggle between holes and electrons and watch the edge charges and the voltmeter reverse while the deflection stays put; the panel below plots $V_H$ against $B$, the slope $1/nq$ carrying the carrier sign.
WHAT TO TRY
- Switch the carrier between holes and electrons: both deflect to the same edge, but the edge charges and the voltmeter sign reverse. That sign is how the Hall effect tells the carriers apart.
- Sweep the field $B$ through zero: the deflection, the edge charge, and $V_H$ all flip with it, and $V_H$ rides up and down the straight line in the panel below.
- Raise the current $I$ to deflect harder and grow $V_H$; raise the carrier density $n$ and $V_H$ shrinks (more carriers share the load).
- Watch the buildup after a change: the carriers swing toward the edge, then straighten out once the Hall field has grown to cancel the magnetic force.