Radius of Curvature Calculation - Revision history

Oborn at 08:57, 17 February 2009

2009-02-17T08:57:08Z

Oborn at 08:56, 17 February 2009

2009-02-17T08:56:38Z

Oborn at 08:55, 17 February 2009

2009-02-17T08:55:38Z

Oborn at 08:54, 17 February 2009

2009-02-17T08:54:02Z

Oborn at 08:51, 17 February 2009

2009-02-17T08:51:31Z

Oborn at 08:50, 17 February 2009

2009-02-17T08:50:15Z

Oborn at 08:49, 17 February 2009

2009-02-17T08:49:29Z

Oborn at 08:48, 17 February 2009

2009-02-17T08:48:31Z

Oborn at 08:42, 17 February 2009

2009-02-17T08:42:14Z

Oborn at 08:42, 17 February 2009

2009-02-17T08:42:01Z

@@ Line 10: / Line 10: @@
 For this sample equation 1 MeV will be used to determine the radius of curvature per MeV of the incident beam.
 Energy = 1 MeV = <math> 1 * 10^6 \frac{J}{C} * 1.6022 * 10^{-19}C = 1*10^6\frac{kg * m^2}{s^2 * C} * 1.6022 * 10^{-19}C </math>
 Magnetic Field (B) = 0.35 Tesla = <math>0.35\frac{kg}{C*s}</math>
 Charge of an electron (q) = <math>1.6022 * 10^{19} Coulombs</math>
 To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac
 {m}{s})</math>
 Substituting the numbers above into the radius equation gives the following
 <math> \frac{(1*10^6 \frac{kg*m^2}{s^2*C}) * (1.6022 * 10^{-19}C)}{(2.99*10^8\frac{m}{s})*(0.35\frac{kg}{C*s})*(1.6022 * 10^{-19}C)} =
 .009556 meters = 0.9556 cm</math>
 So basically the radius of curvature for the electrons/positrons is ~0.9556cm per MeV
 So for a 7 MeV electron/positron pair the radius of curvature for either particle in a 0.35 T field would be 6.69cm
 [http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

@@ Line 1: / Line 1: @@
 Below are my calculations done for determining the radius of curvature of an electron/positron in the magnetic field for the pair spectrometer.
@@ Line 12: / Line 10: @@
 For this sample equation 1 MeV will be used to determine the radius of curvature per MeV of the incident beam.
 Energy = 1 MeV = <math> 1 * 10^6 \frac{J}{C} * 1.6022 * 10^{-19}C = 1*10^6\frac{kg * m^2}{s^2 * C} * 1.6022 * 10^{-19}C </math>
 Magnetic Field (B) = 0.35 Tesla = <math>0.35\frac{kg}{C*s}</math>
 Charge of an electron (q) = <math>1.6022 * 10^{19} Coulombs</math>
 To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac
-To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac{m}{s})</math>
+{m}{s})</math>
 Substituting the numbers above into the radius equation gives the following
 <math> \frac{(1*10^6 \frac{kg*m^2}{s^2*C}) * (1.6022 * 10^{-19}C)}{(2.99*10^8\frac{m}{s})*(0.35\frac{kg}{C*s})*(1.6022 * 10^{-19}C)} =
-<math> \frac{(1*10^6 \frac{kg*m^2}{s^2*C}) * (1.6022 * 10^{-19}C)}{(2.99*10^8\frac{m}{s})*(0.35\frac{kg}{C*s})*(1.6022 * 10^{-19}C)} = 0.009556 meters = 0.9556 cm</math>
+.009556 meters = 0.9556 cm</math>
 So basically the radius of curvature for the electrons/positrons is ~0.9556cm per MeV
-So for a 7 MeV electron/positron pair the radius of curvature for either particle in a 0.35 T field would be 6.69cm
+[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

← Older revision		Revision as of 08:55, 17 February 2009
Line 25:		Line 25:
	<math> \frac{(110^6 \frac{kgm^2}{s^2C}) (1.6022 * 10^{-19}C)}{(2.9910^8\frac{m}{s})(0.35\frac{kg}{Cs})(1.6022 * 10^{-19}C)} = 0.009556 meters = 0.9556 cm</math>		<math> \frac{(110^6 \frac{kgm^2}{s^2C}) (1.6022 * 10^{-19}C)}{(2.9910^8\frac{m}{s})(0.35\frac{kg}{Cs})(1.6022 * 10^{-19}C)} = 0.009556 meters = 0.9556 cm</math>

−	So basically the radius of curvature for the electrons/positrons is ~0.~~96cm~~ per MeV	+	So basically the radius of curvature for the electrons/positrons is ~0.9556cm per MeV
		+
		+	So for a 7 MeV electron/positron pair the radius of curvature for either particle in a 0.35 T field would be 6.69cm

← Older revision		Revision as of 08:54, 17 February 2009
Line 21:		Line 21:
	To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac{m}{s})</math>		To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac{m}{s})</math>

−	Substituting the numbers above into the radius equation gives the following<math> \frac{(110^6 \frac{kgm^2}{s^2C}) (1.6022 * 10^{-19}C)}{(2.9910^8\frac{m}{s})(0.35\frac{kg}{Cs})(1.6022 * 10^{-19}C)}</math>	+	Substituting the numbers above into the radius equation gives the following
		+
		+	<math> \frac{(110^6 \frac{kgm^2}{s^2C}) (1.6022 * 10^{-19}C)}{(2.9910^8\frac{m}{s})(0.35\frac{kg}{Cs})(1.6022 * 10^{-19}C)} = 0.009556 meters = 0.9556 cm</math>
		+
		+	So basically the radius of curvature for the electrons/positrons is ~0.96cm per MeV

← Older revision		Revision as of 08:51, 17 February 2009
Line 1:		Line 1:
−	[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]Below are my calculations done for determining the radius of curvature of an electron/positron in the magnetic field for the pair spectrometer.The Lorentz force is the centripetal force acting upon the electron/positron in the magnetic field which gives the following equation.<math>\frac{(mv^2)}{r} = q v * B </math>This equation can be rearranged to solve for r (and given that mv = momentum) to give the following equation<math>r = \frac{\rho}{qB}</math>For this sample equation 1 MeV will be used to determine the radius of curvature per MeV of the incident beam.Energy = 1 MeV = <math> 1 10^6 \frac{J}{C} * 1.6022 * 10^{-19}C = 110^6\frac{kg m^2}{s^2 * C} * 1.6022 * 10^{-19}C </math>Magnetic Field (B) = 0.35 Tesla = <math>0.35\frac{kg}{Cs}</math>Charge of an electron (q) = <math>1.6022 10^{19} Coulombs</math>To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac{m}{s})</math>Substituting the numbers above into the radius equation gives the following<math> \frac{(110^6 \frac{kgm^2}{s^2C}) (1.6022 * 10^{-19}C)}{(2.9910^8\frac{m}{s})(0.35\frac{kg}{Cs})(1.6022 * 10^{-19}C)}</math>	+	[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
		+
		+	Below are my calculations done for determining the radius of curvature of an electron/positron in the magnetic field for the pair spectrometer.
		+
		+	The Lorentz force is the centripetal force acting upon the electron/positron in the magnetic field which gives the following equation.
		+
		+	<math>\frac{(mv^2)}{r} = q v * B </math>
		+
		+	This equation can be rearranged to solve for r (and given that mv = momentum) to give the following equation
		+
		+	<math>r = \frac{\rho}{q*B}</math>
		+
		+	For this sample equation 1 MeV will be used to determine the radius of curvature per MeV of the incident beam.
		+
		+	Energy = 1 MeV = <math> 1 * 10^6 \frac{J}{C} * 1.6022 * 10^{-19}C = 110^6\frac{kg m^2}{s^2 * C} * 1.6022 * 10^{-19}C </math>
		+
		+	Magnetic Field (B) = 0.35 Tesla = <math>0.35\frac{kg}{C*s}</math>
		+
		+	Charge of an electron (q) = <math>1.6022 * 10^{19} Coulombs</math>
		+
		+	To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac{m}{s})</math>
		+
		+	Substituting the numbers above into the radius equation gives the following<math> \frac{(110^6 \frac{kgm^2}{s^2C}) (1.6022 * 10^{-19}C)}{(2.9910^8\frac{m}{s})(0.35\frac{kg}{Cs})(1.6022 * 10^{-19}C)}</math>

← Older revision		Revision as of 08:50, 17 February 2009
Line 1:		Line 1:
−	[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]	+	[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]Below are my calculations done for determining the radius of curvature of an electron/positron in the magnetic field for the pair spectrometer.The Lorentz force is the centripetal force acting upon the electron/positron in the magnetic field which gives the following equation.<math>\frac{(mv^2)}{r} = q v * B </math>This equation can be rearranged to solve for r (and given that mv = momentum) to give the following equation<math>r = \frac{\rho}{qB}</math>For this sample equation 1 MeV will be used to determine the radius of curvature per MeV of the incident beam.Energy = 1 MeV = <math> 1 10^6 \frac{J}{C} * 1.6022 * 10^{-19}C = 110^6\frac{kg m^2}{s^2 * C} * 1.6022 * 10^{-19}C </math>Magnetic Field (B) = 0.35 Tesla = <math>0.35\frac{kg}{Cs}</math>Charge of an electron (q) = <math>1.6022 10^{19} Coulombs</math>To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac{m}{s})</math>Substituting the numbers above into the radius equation gives the following<math> \frac{(110^6 \frac{kgm^2}{s^2C}) (1.6022 * 10^{-19}C)}{(2.9910^8\frac{m}{s})(0.35\frac{kg}{Cs})(1.6022 * 10^{-19}C)}</math>
−
−	Below are my calculations done for determining the radius of curvature of an electron/positron in the magnetic field for the pair spectrometer.
−
−	The Lorentz force is the centripetal force acting upon the electron/positron in the magnetic field which gives the following equation.
−
−	<math>\frac{(mv^2)}{r} = q v * B </math>
−
−	This equation can be rearranged to solve for r (and given that mv = momentum) to give the following equation
−
−	<math>r = \frac{\rho}{q*B}</math>
−
−	For this sample equation 1 MeV will be used to determine the radius of curvature per MeV of the incident beam.
−
−	Energy = 1 MeV = <math> 1 * 10^6 \frac{J}{C} * 1.6022 * 10^{-19}C = 110^6\frac{kg m^2}{s^2 * C} * 1.6022 * 10^{-19}C </math>
−
−	Magnetic Field (B) = 0.35 Tesla = <math>0.35\frac{kg}{C*s}</math>
−
−	Charge of an electron (q) = <math>1.6022 * 10^{19} Coulombs</math>
−
−	To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac{m}{s})</math>
−
−	Substituting the numbers above into the radius equation gives the following
−
−	<math> \frac{(110^6 \frac{kgm^2}{s^2C}) (1.6022 * 10^{-19}C)}{(2.9910^8\frac{m}{s})(0.35\frac{kg}{Cs})(
−	1.6022 * 10^{-19}C)}</math>

← Older revision		Revision as of 08:42, 17 February 2009
Line 19:		Line 19:
	Charge of an electron = <math>1.6022 * 10^{19} Coulombs</math>		Charge of an electron = <math>1.6022 * 10^{19} Coulombs</math>

−	To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac{m}{s}</math>	+	To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie <math> 2.99 * 10^8 \frac{m}{s})</math>

@@ Line 10: / Line 10: @@
 <math>r = \frac{\rho}{q*B}</math>
 For this sample equation 1 MeV will be used to determine the radius of curvature per MeV of the incident beam.
@@ Line 20: / Line 18: @@
 Charge of an electron = <math>1.6022 * 10^{19} Coulombs</math>