Faraday Cup Temperature

Calculating the temperature of a Faraday Cup Rod

Number of particles per second and corresponding beam power

Assume electron beam parameters at FC location are:

Frequency:     f=300 Hz
Peak current:  I=3 Amps
Pulse width:   t= 50 ps
Beam energy:   E=45 MeV

The number of electrons per second at FC location are:

[math] N = \frac {Q} {e} = \frac {f I ∆t} {e} = \frac {(300\ Hz)(3\ A) (50\ ps)} {1.6*10^{-19}\ C} = 0.28*10^{12} [/math]

The corresponding beam power for 45 MeV electron beam is:

[math] P = (0.28*10^{12}\ \frac{e}{sec}) (44\ MeV) (1.6*10^{-19} \frac{J}{eV}) = 1.9\ W[/math]

Temperature calculation

Now apply the Stefan-Boltzmann Law for one Faraday cup rod

[math] P = (0.924)\ (A)\ (\sigma)\ (T^4) [/math]

Here,

 A is the radiated area of the rode.radiatedanradiated
 [math] \sigma = 5.67*10^{-8} \frac {W}{m^2 K^4} [/math] is the Stefan-Boltzmann constant.  

Assume the all beam power goes to one FC rod and area of radiation is 2 cm^2 (two back sides) we can calculate the corresponding temperature:

[math] T^4 = \frac {P}{0.924\ A\ \sigma} = \frac {1.9 W}{(0.924)(2*10^{-4}\ m^2)(5.67*10^{-8} \frac {W}{m^2\ K^4})} = 0.18\ K^4[/math]

So

[math] T = 424\ K [/math]

Conclusion

An Aluminum converter that is 1/2 mil thick being struck by a 44 MeV electron beam with a 50 picosecond pulse width, 300 Hz rep rate, and 50 Amp peak current is found to be safe from melting.

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