Kicker Magnets

Field needed to bend 44 MeV electrons 0.47 degrees

[math]\displaystyle{ 1 MeV = 1.6\cdot 10^{-13} J = 1.6\cdot 10^{-13} \frac{m^2\cdot kg}{s^2} }[/math]

[math]\displaystyle{ c = 2.998 \cdot 10^8 \frac{m}{s} }[/math]

[math]\displaystyle{ \frac{MeV}{C} = 0.534 \cdot 10^{-21} \frac{m\cdot kg}{s} }[/math]

[math]\displaystyle{ p_e = 44 \frac{MeV}{c} = 23.5 \cdot 10^{-21} \frac{m\cdot kg}{s} }[/math]

[math]\displaystyle{ B = \frac{p_e}{q_e \cdot R} }[/math]

[math]\displaystyle{ 1T=\frac{kg}{C\cdot s} }[/math], [math]\displaystyle{ q_e = 1.6\cdot 10^{-19} C }[/math], [math]\displaystyle{ 1T=10^{-4}G }[/math]

[math]\displaystyle{ B(T) = \frac{p_e (\frac{MeV}{c})\cdot 0.33\cdot 10^{-2}}{R(m)} }[/math]

[math]\displaystyle{ B(T) = \frac{4.67\cdot 10^{-2}}{R(m)} }[/math]

[math]\displaystyle{ 180^0 = \kappa + 90^0 + \beta }[/math]

[math]\displaystyle{ 180^0 = \gamma + 90^0+ \beta }[/math]

[math]\displaystyle{ \kappa = \gamma }[/math]

[math]\displaystyle{ R = \frac{a}{cos(\beta)} = \frac{a}{cos(90^0 - \kappa)} = \frac{a}{sin(\kappa)} }[/math]

[math]\displaystyle{ d = R \cdot (1 - cos(\kappa)) = \frac{a \cdot (1 - cos(\kappa))}{sin(\kappa)} }[/math]

[math]\displaystyle{ B(T) = \frac{p_e (\frac{MeV}{c})\cdot 0.33\cdot 10^{-2}\cdot sin(\kappa)}{a(m)} }[/math] - general expression for B-field.

[math]\displaystyle{ B(T) = \frac{7.83\cdot 10^{-2}\cdot sin(\kappa)}{a(m)} }[/math]

If [math]\displaystyle{ \kappa = 0.47^0 }[/math] then [math]\displaystyle{ sin(\kappa) = 0.0082 }[/math] and our B-field becomes:

[math]\displaystyle{ B(T) = \frac{1.2\cdot 10^{-3}}{a(m)} }[/math]

[math]\displaystyle{ a \simeq 0.125 m }[/math] for the coils under consideration. Hence, the B-field needed is:

[math]\displaystyle{ B = 0.00964 T = 96.4 G }[/math]

Magnetic Field Produced by Coils

File:B field coil.jpg

[math]\displaystyle{ z'=L+z }[/math]

[math]\displaystyle{ z=z'-L }[/math]

[math]\displaystyle{ B_z^n=\frac{\mu_0}{2}\frac{R^2 \cdot I}{(z_n^{'2}+R^{'2})^{\frac{3}{2}} }[/math]

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