Difference between revisions of "B-field calculation"

From New IAC Wiki
Jump to navigation Jump to search
 
(21 intermediate revisions by the same user not shown)
Line 1: Line 1:
−
[[Image:B_field_trajectory.jpg |300px]]
+
[[Image:B_field_trajectory2.jpg |300px]]
  
  
Line 5: Line 5:
  
 
<math>c = 2.998 \cdot 10^8 \frac{m}{s}</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>p_e = 14 \frac{MeV}{c} = 7.47 \cdot 10^{-21} \frac{m\cdot kg}{s}</math>
Line 11: Line 13:
 
<math>B = \frac{p_e}{q_e \cdot R}</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=\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>B(T) = \frac{4.67\cdot 10^{-2}}{R(m)}</math>
Line 24: Line 29:
  
 
<math>d = R \cdot (1 - cos(\kappa)) = \frac{a \cdot (1 - cos(\kappa))}{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>
 
<math>B(T) = \frac{4.67\cdot 10^{-2}\cdot sin(\kappa)}{a(m)}</math>
Line 31: Line 38:
 
<math>B(T) = \frac{0.163\cdot 10^{-2}}{a(m)}</math>
 
<math>B(T) = \frac{0.163\cdot 10^{-2}}{a(m)}</math>
  
−
<math>a \simeq 12 cm</math> for the coils under consideration. Hence, B-field is:
+
<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>
 +
 
 +
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

Latest revision as of 06:17, 5 February 2009

B field trajectory2.jpg


[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]

Go Back

Retrieved from "https://wiki.iac.isu.edu/index.php?title=B-field_calculation&oldid=40707"