Crossed Fields: The Hall Effect — Question 6

Question 6

Only the field component normal to the Hall strip contributes to its transverse voltage

A Hall strip carries current along x. The field of magnitude B lies in the yz plane and makes angle theta with +z. Its measured voltage convention is upper minus lower along y. Assume one carrier type and no transverse conduction.

1. Derive VH(theta) and compare it with the perpendicular-field reading at theta=0.

2. At theta=60.0 degrees, find the density error if the experimenter mistakenly uses the full B in the perpendicular-field formula.

Question 6 - Solution

Only the field component normal to the Hall strip contributes to its transverse voltage

(a) Method and calculation

$$\vec v_d\times\vec B=v_dB_y\hat z-v_dB_z\hat y,\quad E_y=v_dB_z.$$$$V_H(\theta)=-\frac{IB\cos\theta}{nqt}=V_H(0)\cos\theta.$$

(b) Interpretation and further calculation

$$n_{\mathrm{inferred}}=\frac{IB}{|q|t|V_H(\theta)|}=\frac{n}{|\cos\theta|}.$$$$\theta=60.0^\circ:\quad n_{\mathrm{inferred}}=2n.$$

Physical check

The in-plane field can produce a thickness-direction Hall field, but it does not replace Bz in the specified transverse voltage.