Kicker Magnets

Field needed to bend 44 MeV electrons 0.47 degrees

$1 MeV = 1.6\cdot 10^{-13} J = 1.6\cdot 10^{-13} \frac{m^2\cdot kg}{s^2}$

$c = 2.998 \cdot 10^8 \frac{m}{s}$

$\frac{MeV}{C} = 0.534 \cdot 10^{-21} \frac{m\cdot kg}{s}$

$p_e = 44 \frac{MeV}{c} = 23.5 \cdot 10^{-21} \frac{m\cdot kg}{s}$

$B = \frac{p_e}{q_e \cdot R}$

$1T=\frac{kg}{C\cdot s}$, $q_e = 1.6\cdot 10^{-19} C$, $1T=10^{-4}G$

$B(T) = \frac{p_e (\frac{MeV}{c})\cdot 0.33\cdot 10^{-2}}{R(m)}$

$B(T) = \frac{4.67\cdot 10^{-2}}{R(m)}$

$180^0 = \kappa + 90^0 + \beta$

$180^0 = \gamma + 90^0+ \beta$

$\kappa = \gamma$

$R = \frac{a}{cos(\beta)} = \frac{a}{cos(90^0 - \kappa)} = \frac{a}{sin(\kappa)}$

$d = R \cdot (1 - cos(\kappa)) = \frac{a \cdot (1 - cos(\kappa))}{sin(\kappa)}$

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

$B(T) = \frac{7.83\cdot 10^{-2}\cdot sin(\kappa)}{a(m)}$

If $\kappa = 0.47^0$ then $sin(\kappa) = 0.0082$ and our B-field becomes:

$B(T) = \frac{1.2\cdot 10^{-3}}{a(m)}$

$a \simeq 0.125 m$ for the coils under consideration. Hence, the B-field needed is:

$B = 0.00964 T = 96.4 G$

Magnetic Field Produced by Coils

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$z'=L+z$

$z=z'-L$

$B_z = \frac {μ_0 }{2} \frac {R^2 \cdot I}{(z_n^{'2} + R^{'2})^{\frac{3}{2}}$

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