# Validity of equation for $E \times B$ drift in SI and Gaussian unit systems

I derive $$E \times B$$ drift $$\vec v_{E \times B} = \dfrac{\vec E \times \vec B}{B^2}$$ and in SI I use V/m and T for the fields. I do not use Gaussian (cgs) units often, but I thought this equation holds for cgs. However, the conversion factors for statvolt/cm and gauss are $$1/3 \times 10^2$$ and $$10^4$$, so one gets additional 1/3 factor. Is it really there in cgs equation for the drift or do I miss something?

• Quick conversion: |E| = |VxB| ~ $10^{-3} \ \lvert V \rvert \ \lvert V \rvert$ mV/m (if V is in km/s and B in nT). So a 400 km/s flow speed and 10 nT magnetic field give E ~ 4 mV/m. Then we can recover 400 km/s by seeing that if E is orthogonal to B, we have u ~ E/B = $4 \times 10^{-3}/\left( 10 \times 10^{-9} \right)$ m/s = 400 km/s. Jan 16 at 22:46

The equation for $$\mathbf{E} \times \mathbf{B}$$ drift in Gaussian units is $$\mathbf{u} = c \frac{\mathbf{E} \times \mathbf{B}}{B^2};$$ see, for example, Jackson's Classical Electrodynamics, 3rd ed. Also, your conversion for the electric field is incorrect; it should be $$1 \text{ V/m} = \frac{1}{\text{"3"}} \times 10^{-4} \text{ statvolt/cm}$$ (again, see Jackson, and note that "3" = 2.99792458.)