Faraday Cup Temperature
Calculating the temperature of a Faraday Cup Rod
Number of particles per second hitting one rod
Assume electron beam parameters at Faraday Cup 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 that hit the FC per second is:
[math]\displaystyle{ 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}\frac {e^-}{sec} }[/math]
So, we have around [math]\displaystyle{ 4.6875*10^{12} }[/math] electrons per second or [math]\displaystyle{ 15.625*10^9 }[/math] electrons per pulse.
Calculating the stopping power due to collision of one 44 MeV electron in Aluminum
From NIST ([1] see link here) the stopping power for one electron with energy of 44 MeV in Aluminum is [math]\displaystyle{ 1.78 MeV cm^2/g }[/math].
[math]\displaystyle{ 1 mil = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 0.00254 cm }[/math]
The effective length of 1/2 mil Al:
[math]\displaystyle{ (2.70 \frac {g}{cm^3})(0.00127 cm) = 0.003429 \frac {g}{cm^2} }[/math]
The total stopping power due to collisions on Al per incident electron:
[math]\displaystyle{ (1.78 MeV \frac {cm^2}{g})(0.003429 \frac {g}{cm^2}) = 0.0061 \frac {MeV}{electron} }[/math]
The energy deposited per pulse:
[math]\displaystyle{ (0.0061 \frac {MeV}{electron})(15.625*10^9 \frac {electrons}{pulse}) = 95.3125*10^6 \frac {MeV}{pulse} }[/math]
The energy deposited per second:
[math]\displaystyle{ (95.3125*10^6 \frac {MeV}{pulse})(300 \frac {pulses}{second}) = 28.6*10^9 \frac {MeV}{second} }[/math]
Calculating the temperature increase
The power deposited in 1/2 mil Al is:
[math]\displaystyle{ P = (28.6*10^{15} \frac {eV}{second})(1.6*10^{-19} \frac {C}{eV}) = 4.575*10^{-3} W }[/math]
Stefan-Boltzmann Law (Wien Approximation) says [math]\displaystyle{ P = (0.924)(Area)(\sigma)(T^4) }[/math]
Solving for Temperature and taking into account the two sides of the converter we get:
[math]\displaystyle{ T^4 = \frac {P}{(0.924)(2A)(\sigma)} }[/math]
where [math]\displaystyle{ \sigma }[/math] is the Stefan-Boltzmann constant, [math]\displaystyle{ \sigma = 5.67*10^{-8} \frac {W}{m^2 K^4} }[/math]. Assume a beam spot diameter on the converter surface of 5mm, or an area of [math]\displaystyle{ A = 19.62 mm^2 = 19.62*10^{-6} m^2 }[/math].
Plugging in the numbers we see that the temperature will increase [math]\displaystyle{ 217.2 K }[/math]. Now, adding in the temperature of the converter at room temperature we get :
[math]\displaystyle{ T = 300 + 217.2 = 517.2 K }[/math]
The melting temperature of Aluminum is [math]\displaystyle{ 933.5 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.