radius of path of electron in a magnetic field

Amith2006

# Electrons are moving in a uniform magnetic field of 50 Oersted having a velocity of 8.8 x 10^6 cm/sec. What is the radius of the circular path they follow?
I solved it in the following way:
In C.G.S system 1 Gauss = 1 Oersted (In vacuum)
So, B = 50 Gauss
V = 8.8 x 10^6 cm/sec
Let e be the charge of the electron, r be radius of its circular path and m be its mass.
e = 1.6 x 10^(-19) Coulomb = 4.8 x 10^(-10) Stat Coulomb
m = 9.1 x 10^(-28) grams
In a perpendicular magnetic field,
BeV = {m(V^2)}/r
r = (mV)/(Be)
Solving I get,
r = 3.3 x 10^(-13) cm
I think somewhere I have gone wrong as the radius is too small. Please help.

BoTemp

In cgs units the formula for the Lorentz force is (qv/c) x B. The final formula should read r = (mVc/Be). This gaves a much more reasonable answer.

Amith2006

I have learnt the expression for Lorentz force in C.G.S system.Thanks.