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| − | <math>\fbox{sadfs}</math>
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| | <math>1 MeV = 1.6\cdot 10^{-13} J = 1.6\cdot 10^{-13} \frac{m^2\cdot kg}{s^2}</math> | | <math>1 MeV = 1.6\cdot 10^{-13} J = 1.6\cdot 10^{-13} \frac{m^2\cdot kg}{s^2}</math> |
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| | <math>B = 0.01358 T = 135.8 G</math> | | <math>B = 0.01358 T = 135.8 G</math> |
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| | + | [http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back] |
Latest revision as of 06:17, 5 February 2009
[math]1 MeV = 1.6\cdot 10^{-13} J = 1.6\cdot 10^{-13} \frac{m^2\cdot kg}{s^2}[/math]
[math]c = 2.998 \cdot 10^8 \frac{m}{s}[/math]
[math]\frac{MeV}{C} = 0.534 \cdot 10^{-21} \frac{m\cdot kg}{s}[/math]
[math]p_e = 14 \frac{MeV}{c} = 7.47 \cdot 10^{-21} \frac{m\cdot kg}{s}[/math]
[math]B = \frac{p_e}{q_e \cdot R}[/math]
[math]1T=\frac{kg}{C\cdot s}[/math], [math]q_e = 1.6\cdot 10^{-19} C[/math], [math]1T=10^{-4}G[/math]
[math]B(T) = \frac{p_e (\frac{MeV}{c})\cdot 0.33\cdot 10^{-2}}{R(m)}[/math]
[math]B(T) = \frac{4.67\cdot 10^{-2}}{R(m)}[/math]
[math]180^0 = \kappa + 90^0 + \beta[/math]
[math]180^0 = \gamma + 90^0+ \beta[/math]
[math]\kappa = \gamma[/math]
[math]R = \frac{a}{cos(\beta)} = \frac{a}{cos(90^0 - \kappa)} = \frac{a}{sin(\kappa)}[/math]
[math]d = R \cdot (1 - cos(\kappa)) = \frac{a \cdot (1 - cos(\kappa))}{sin(\kappa)}[/math]
[math]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]B(T) = \frac{4.67\cdot 10^{-2}\cdot sin(\kappa)}{a(m)}[/math]
If [math]\kappa = 2^0[/math] then [math]sin(\kappa) = 0.0348995[/math] and our B-field becomes:
[math]B(T) = \frac{0.163\cdot 10^{-2}}{a(m)}[/math]
[math]a \simeq 0.12 m[/math] for the coils under consideration. Hence, B-field is:
[math]B = 0.01358 T = 135.8 G[/math]
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