# Gauss units relation

Hi all!

## Homework Statement

The problem is:

For a particle with electric charge equal to that of one electron, the trajectory radius R is related to the absolute value of the perpendicular component $$P_{\bot}$$of the relativistic moment perpendicular to $$\textbf{B}$$ from the relation:

$$P_{}^{} (MeV/c) = 3.00 \times10^{-4} BR (Gauss \times cm)$$

## Homework Equations

I know that:

1Tesla=10^4Gauss
1Coulomb=3x10^9Gauss
1m=10^2cm

## The Attempt at a Solution

So:

$$\textbf{F}=q\textbf{v}\times\textbf{B}$$

$$\textbf{F}_{\bot}=\frac{mv^2}{R}=qvB$$

$$\textbf{P}_{\bot}= qBR$$

where on the left side I transform P in MeV/c units obtaining a new value of P,
and on the right side I have qBR initially in Coulomb x Tesla x m.
So

$$qBR= (1.6 \times 10^{-19} \times 3\times10^9) (B \times 10^4 )(R \times 10^2)= 4.8 \times 10^{-4} (Gauss^2 \times cm)$$

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pam
"1Coulomb=3x10^9Gauss" is wrong.

precision

okay, it's 1 Coulomb = 3 x 10^9 esu (or Cstat)

but the problem is not solved!!!