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TOWARDS MULTI-WAVELENGTH OBSERVATIONS OF ...


4.3 Poynting Flux, GEM and LEM Variation with s in Self-Similar Model . . . . . . . . 47. 4.4 Streamplot of Magnetic Field between 10M and 100M in Self-Similar ...


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Motion in Combined Electric and Magnetic Field: What is Magnetic ...


Magnetic fields are used to guide the motion of charged particles in ... The Van Allen radiation belts in space around the earth consist of these energetic charges trapped in ... to a uniform magnetic field B (no E) moves in a circular path with the radius ... force, q v × B, acts as a centripetal force and produces a circular motion  ...


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Veda
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a)

Given that

A monoenergetic electron beam speed v = 5.20 * 10^{6} m/s

Magnetic field B = 1.30 * 10^{-4} T

Specific charge of electron \frac{e}{m} = 1.76*10^{11} C/kg  

Magnetic field is perpendicular to the beam velocity

                       So \theta = 90 \degree

  

Step 1: Set up an equation for the radius of the circular path of charged particle

              Formulas required

                  Force acting on electron in magnetic field F = evB \sin \theta

                  Where, e - charge, B - magnetic field, v - velocity, \theta - angle between the velocity and the magnetic field. 

  

                               F = evB   .........................(1)                        \because    \theta = 90 \degree                                         

                

              A charged particle moving in a circular path, so it acquires centripetal force

                      Centripetal force   F = \frac{mv^2}{r} ...............(2)

                        Where,  r - radius of the circular path

  

                From equation(1) and (2) we can write as follows

                             evB = \frac{mv^2}{r}

                                  r = \frac{mv^2}{evB}

                                  r = \frac{mv}{eB}


              Hence,  radius of the circular path of charged particle r = \frac{mv}{qB} ................(3)


Step 2: Plug in the given values in the equation (3) for radius of circular path

                       r = \frac{v}{\frac{e}{m} B}

                        r = \frac{5.20 * 10^{6}}{1.76*10^{11} * 1.30 * 10^{-4}}   

                        r = 0.227 m

                        r = 22.7 cm

  

                Therefore, radius of the circular path of charged particle   r = 22.7 cm


b)

Given that

Energy of the electron beam E = 20 MeV = 20* 10^{6} eV   

                                               E = 20 * 10^{6} * 1.6 * 10^{-19} J

              

Step 1: Set up an equation for the velocity of the electron

                Kinematic energy K.E = \frac{1}{2} mv^2.   

                             v^2 = \frac{2 K.E}{m}

                             v = \sqrt{\frac{2 K.E}{m}}


Step 2: Substitute the given values in the velocity equation

                  Mass of an electron m = 9.1 * 10^{-31} kg  

                     v = \frac{2 * 20 * 10^{6} * 1.6 * 10^{-19}}{9.1 * 10^{-31}}

                     v = 2.65 * 10^{9} m/s

              

Step 3: Observation

                Speed of light c = 38 10^{8} m/s

                Speed of electron is greater than the speed of the light which is practically not possible.

                                  K.E = \frac{1}{2} mv^2.   is valid only for v << c


              So modify the formula of radius by replacing the relativistic mass

                             Modified mass m=m_0*(1-\frac{v^2}{c^2})^{-\frac{1}{2}}

                                       m_o - mass of particle at rest position


                 Thus, valid formula for calculating the radius of circular path is as follows                  

                                Radius  r=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}\cdot\frac{v}{eB}