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Question Re: the application of your own field.help me. 

Keywords: field
Forum: Fields: Magnetic and Otherwise
Re: Angry the application of your own field.help me.
Re: Question Re: the application of your own field.help me. (Gumplinger Peter)
Date: 05 Mar, 2005
From: <yelongchun2005>

thanks for your answer,good man,

the E-field is composed of a cathode cylinder and a anode which the axes of the cathode. E=V/(r*ln(b/a)) b:the radius of cylinder,a:the radius of axes.( a<r<b)

//include
#ifndef radialEfield_H
#define radialEfield_H
#include "G4Types.hh"
#include "G4ThreeVector.hh"
#include "G4ElectricField.hh"
class radialEfield : public G4ElectricField
{
  public: 
   radialEfieldfield( G4double Voltage, G4double b, G4double a);

   virtual void GetFieldValue(const    G4double pos[4], 
                              G4double *field) const;
  private: 
   G4double fv;
   G4double fba;
}
#endif 

//src
#include "radialEfield.hh"
radialEfield::radialEfield(G4double Voltage, G4double b, 
                      G4double a)
{  fv=Voltage
   fba=std::log(b/a);
 }
radialEfield::GetFieldValue(const    G4double para[4], 
                              G4double *field) const
{
  G4ThreeVector  pos(para[0], para[1], para[2]); 

  G4double radius =sqrt(pow(para[1],2)+pow(para[2],2);

  G4double fieldMag= fv / ( radius * fba )  
  //E -direction
  G4ThreeVector  unitDir(0,para[1]/radius,para[2]/radius);//
  G4ThreeVector  Efield= fieldMag * unitDir; 
  // Return the field value
  field[0]= field[1]= field[2]= 0.0; 
  field[3]= Efield.x(); 
  field[4]= Efield.y();
  field[5]= Efield.z();
}

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1 Warning: Re: the application of your own field.help me.   (John Apostolakis - 07 Mar, 2005)
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