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Next: Application to the SLS Up: Eddy current effects in Previous: The SLS Booster

   
Theoretical background

The ramping cycle of the booster is determined by the time-dependent dipolar field of the bending magnets. These fields follows a biased-sine shape given by:

 \begin{displaymath}
B_{F,D}(t)=B_{0\ F,D}*(\alpha_{E}-\cos(\omega*t))
\end{displaymath} (1)

where $\alpha_E$ and $B_{0\ F,D}$ determine the minimum and maximum energy possible in the booster. In the SLS case, the beam is injected at the minimum energy of the ramping cycle, and extracted at the maximum. The energy ramp is:

\begin{displaymath}E(t) = E_{0}*(\alpha_{E}-\cos(\omega*t))
\end{displaymath} (2)

And the parameters $\alpha_E$, $B_{0\ F,D}$ and E0 are given by::
$\displaystyle \alpha_{E}$ = $\displaystyle \frac{E_{extraction}+E_{injection}}{E_{extraction}-E_{injection}}$ (3)
$\displaystyle B_{0\ F,D}$ = $\displaystyle \frac{B_{extraction\ F,D}}{\alpha_{E}+1}$ (4)
E0 = $\displaystyle \frac{E_{extraction}}{\alpha_{E}+1}$ (5)

The time dependent fields produce eddy currents in the vacuum chamber of the bending magnet. These currents create a distortion of the magnetic field inside the vacuum chamber. In the center of the pipe, the distortion of the dipolar magnetic field is a decrease of the field, given by:

 \begin{displaymath}
\Delta B=\dot{B} \:
\frac{\mu_0\kappa D a^2}{h}\:\mathcal{J}
\end{displaymath} (6)

where $\dot{B}$ is the temporal derivative of the corresponding BF,D(t) field, $\kappa$ the conductivity of the vacuum chamber, D its thickness and 2h the magnet gap. The parameters a and bare the horizontal and vertical half apertures of the vacuum chamber. $\mathcal{J}$ is a form factor given by:

 \begin{displaymath}
\mathcal{J} = \int_0^{\pi/2} \left({\sin\varphi\sqrt{\cos^2...
...\left(\frac{b}{a}\right)^2\sin^2\varphi}
} \right)\: d\varphi
\end{displaymath} (7)

For a round vacuum pipe (a=b), $\mathcal{J}$ is 1. in the SLS case, b/a=2/3, and $\mathcal{J}\approx 0.8$.

The analysis of this field distortion shows[1] that it can be expressed as a sextupolar time-dependent component given by:

 
m $\textstyle \equiv$ $\displaystyle \frac{1}{2}\frac{1}{B\rho}\frac{d^2 B}{dx^2}$  
  = $\displaystyle \mathcal{J}\mu_0 \kappa \frac{D}{h}\frac{\dot{B}}{B\rho}$ (8)

If we replace in equation 8 the values of the time-dependent magnetic field( equation 1) and its derivative, we obtain the following expression for the the sextupolar component:

 \begin{displaymath}
m=\frac{\mu_0\kappa D}{h}\frac{\omega}{\rho}
\; \mathcal{J}\cdot F(t)
\end{displaymath} (9)

If we rearrange the terms in this last equation, we can write it as:

 \begin{displaymath}
m=\tilde{m}\cdot F(t)
\end{displaymath} (10)

where $\tilde{m}$ is a constant, given by:

 \begin{displaymath}
\tilde{m}=\frac{\mu_0\kappa D}{h}\frac{\omega}{\rho} \mathcal{J}
\end{displaymath} (11)

and F(t) contains the dependence in time:

 \begin{displaymath}
F(t)=\frac{\sin\omega t}{\alpha_{E}-\cos\omega t}
\end{displaymath} (12)

The function F(t) is plotted in figure 2. As show in the plot, the sextupolar contribution will be more important at the start of the ramping cycle (low energies).


  
Figure 2: Plot of F(t) for the SLS booster parameters (3 Hz repetition rate, Emin= 0.1 GeV and Emax= 2.4 GeV.)
\begin{figure}
\begin{center}
\mbox{\epsfig{figure=ft.eps, width=.9\textwidth} }
\end{center}\end{figure}


next up previous
Next: Application to the SLS Up: Eddy current effects in Previous: The SLS Booster
Marc Munoz
1998-11-18