Run a current across a magnetic field and the moving charges inside the conductor feel a sideways Lorentz force. They drift toward one edge, and charge piles up there until it pushes back. The small voltage that appears across the conductor is the Hall voltage. It does something remarkable: it tells you the sign of the moving charges, and how many there are. It connects directly to the drift velocity you already know.
Key idea
For a strip of thickness $t$ (measured along $\vec{B}$) carrying current $I$:
$$V_H = \frac{IB}{nqt}.$$
Positive and negative carriers are pushed to the same edge, so the sign of $V_H$ reveals the sign of the carriers.
Why it works
The sideways magnetic force on each carrier is $qv_dB$. Charge builds up on the edge until its electric field $E_H$ pushes back just as hard: $qE_H = qv_dB$, so $E_H = v_dB$. Across a strip of width $w$, $V_H = E_Hw = v_dBw$. With $v_d = I/(nqwt)$ the width cancels: $V_H = IB/(nqt)$.
Why the same edge? Electrons drifting left feel $(-e)(-v\hat{\imath})\times\vec{B}$, which is the same as a positive charge drifting right. What differs is the sign of the charge that piles up, and so the sign of the voltage.
Why the same edge? Electrons drifting left feel $(-e)(-v\hat{\imath})\times\vec{B}$, which is the same as a positive charge drifting right. What differs is the sign of the charge that piles up, and so the sign of the voltage.
Worked examples
Worked example: In copper
A copper strip $0.10\ \text{mm}$ thick carries $10\ \text{A}$ in a $1.0\ \text{T}$ field. Copper has $8.5 \times 10^{28}$ free electrons per m³. Find $V_H$.
- $$V_H = \frac{IB}{nqt} = \frac{(10)(1.0)}{(8.5 \times 10^{28})(1.60 \times 10^{-19})(1.0 \times 10^{-4})} = 7.35 \times 10^{-6}\ \text{V}$$
- About $7.4\ \mu\text{V}$: tiny, because copper has so many carriers that they drift very slowly.
Answer: $V_H \approx 7.4\ \mu\text{V}$
Worked example: Counting carriers in a semiconductor
A semiconductor strip $0.20\ \text{mm}$ thick carries $5.0\ \text{mA}$ in a $0.50\ \text{T}$ field and shows $V_H = 1.2\ \text{mV}$. Find the carrier density.
- $n = \dfrac{IB}{qtV_H} = \dfrac{(5.0 \times 10^{-3})(0.50)}{(1.60 \times 10^{-19})(2.0 \times 10^{-4})(1.2 \times 10^{-3})} = 6.51 \times 10^{22}\ \text{m⁻³}$.
- Millions of times fewer carriers than copper, so a far bigger Hall voltage for a small current. That is why Hall sensors, used to sense fields in phones and in car speed sensors, are made from semiconductors.
Answer: $n \approx 6.51 \times 10^{22}\ \text{m⁻³}$
Worked example: Reading the sign
Current flows along $+x$ in a strip, $\vec{B}$ points along $+z$, and the $-y$ edge is at the higher potential. What are the carriers?
- Either kind of carrier is pushed toward $-y$ ($q\,\vec{v}\times\vec{B}$ points along $-y$ for both).
- The $-y$ edge is higher, so the charge piled up there is positive: the carriers are positive. (In a p-type semiconductor the carriers act as positive "holes".)
Answer: Positive
Common mistake
"The moving charges are always positive." Conventional current is drawn as if positive charges move, which works for circuits but hides who is really moving. In copper the carriers are electrons; in some semiconductors they behave as positive holes. Only the Hall voltage can tell the difference, because both kinds pile up on the same edge with opposite charges. Check: work out the force on both possibilities before deciding which edge is higher.