Amit & Deepali Sinha Foundation

The Hall Angle Measures the Mobility

Irodov 3.268DifficultyUniqueness

Problem Irodov 3.268

In a Hall effect measurement on a copper conductor in a magnetic field of induction B=100B = 100 mT, the transverse electric field is found to be η=3.1⋅103\eta = 3.1\cdot 10^3 times weaker than the longitudinal electric field that drives the current. Find the mobility of the conduction electrons in copper.

The copper sample carrying the current density j across the field B (its sense is not given; drawn into the page).

The copper sample carrying the current density j\vect j across the field B\vect B (its sense is not given; drawn into the page).

Given
  • B=100B = 100 mT
  • E∥/E⊥=η=3.1⋅103E_{\parallel}/E_{\perp} = \eta = 3.1\cdot 10^3
  • carriers electrons, charge −e-e
  • j⊥B\vect j \perp \vect B (assumed)
Find
the electron mobility u0u_0

Solution

Step 1Drift in the longitudinal field
The longitudinal field E_∥ drives the electrons against j with the drift velocity u. The magnetic force F_m drives them to the upper face, which charges negative, and the transverse field E_⊥ points from the positive face to the negative one.

The longitudinal field E∥\vect E_{\parallel} drives the electrons against j\vect j with the drift velocity u\vect u. The magnetic force Fm\vect F_m drives them to the upper face, which charges negative, and the transverse field E⊥\vect E_{\perp} points from the positive face to the negative one.

The mobility u0u_0 is the drift speed per unit driving field. The longitudinal field E∥E_{\parallel} along the conductor sets the electrons drifting against the current with the speed

u=u0E∥(1) u = u_0 E_{\parallel} \tag{1}
Step 2The transverse field

The carriers are electrons, of charge −e-e, moving at right angles to B\vect B (assumed). The magnetic force −e u×B-e\,\vect u\times\vect B pushes them toward one face, which charges negative, until the transverse field E⊥E_{\perp} of the separated charges stops the sideways drift:

eE⊥=euBeE_{\perp} = euB

and with (1),

E⊥=u0E∥BE_{\perp} = u_0 E_{\parallel} B
(2)
Step 3The mobility

The ratio of the two fields is given. Dividing (2) by E∥E_{\parallel}:

E⊥E∥=u0B=1η\frac{E_{\perp}}{E_{\parallel}} = u_0 B = \frac{1}{\eta}

Numerically, with B=0.100B = 0.100 T,

u0=13.1×103×0.100 m2/(V⋅s)u_0 = \frac{1}{3.1\times 10^3\times 0.100}\ \text{m}^2/(\text{V}\cdot\text{s})
u0=1ηB≈3.2×10−3 m2/(V⋅s)u_0 = \frac{1}{\eta B} \approx 3.2\times 10^{-3}\ \text{m}^2/(\text{V}\cdot\text{s})
(3)
Check

Units: a tesla is V⋅\cdots/m2^2, so 1/ηB1/\eta B is in m2^2/(V⋅\cdots), the unit of a mobility (a speed per field). An independent estimate: with one free electron per atom, copper (density 8.9×1038.9\times 10^3 kg/m3^3, M=63.5M = 63.5 g/mol) has n≈8.4×1028n \approx 8.4\times 10^{28} m−3^{-3}, and its resistivity ρ=1.7×10−8\rho = 1.7\times 10^{-8} Ω⋅\Omega\cdotm gives

u0=1ρne=11.7×10−8×8.4×1028×1.60×10−19≈4×10−3 m2/(V⋅s),u_0 = \frac{1}{\rho n e} = \frac{1}{1.7\times 10^{-8}\times 8.4\times 10^{28}\times 1.60\times 10^{-19}} \approx 4\times 10^{-3}\ \text{m}^2/(\text{V}\cdot\text{s}),

the same order as (3).

Remark

The ratio E⊥/E∥=u0BE_{\perp}/E_{\parallel} = u_0B is the tangent of the Hall angle, the angle by which the total field in the conductor is turned away from the current. Since u0=eτ/mu_0 = e\tau/m with τ\tau the mean time between collisions, it is also ωτ\omega\tau, with ω=eB/m\omega = eB/m the cyclotron frequency: here about 3×10−43\times 10^{-4}, so an electron is scattered thousands of times before it could turn appreciably in the field, which is why the Hall field in a metal is so small.