Crossed Fields: The Hall Effect — Question 1

Question 1

The Hall-voltage sign follows the carrier sign only after the contact convention is fixed

A strip carries conventional current I along +x in a field $B\hat z$ with B positive. Its width is w along y and thickness t along z. There is one carrier type of signed charge q and density n. Define $V_H=V(+w/2)-V(-w/2)$.

1. Derive the transverse electric field and signed Hall voltage in steady state.

2. For negative and positive carriers, determine which edge is at higher potential while keeping I and B fixed.

Question 1 - Solution

The Hall-voltage sign follows the carrier sign only after the contact convention is fixed

(a) Method and calculation

$$J_x=nqv_d,\quad E_y-v_dB=0\Rightarrow E_y=\frac{J_xB}{nq}.$$$$V_H=-E_yw=-\frac{IB}{nqt},\quad J_x=I/(wt).$$

(b) Interpretation and further calculation

$$q<0\Rightarrow V_H>0\Rightarrow V_U>V_L.$$$$q>0\Rightarrow V_H<0\Rightarrow V_U<V_L.$$

Physical check

Reversing the voltage-contact order reverses the reported Hall sign without changing the material.