- #1
folgorant
- 29
- 0
Hi all!
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)
I know that:
1Tesla=10^4Gauss
1Coulomb=3x10^9Gauss
1m=10^2cm
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)
...who can find the error? please help!
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)
...who can find the error? please help!
