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Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
Motion of particles in magnetic fields 08
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Motion of particles in magnetic fields 08

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SACE Physics Section 2 Topic 4

SACE Physics Section 2 Topic 4

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  • 1. Motion of Particles in Magnetic Fields Section 2 Topic 4
  • 2. Forces on Moving Charges <ul><li>Charges that are stationary ; </li></ul><ul><ul><li>have no magnetic force applied to it. </li></ul></ul><ul><li>A wire that has no P.D. applied to its ends ; </li></ul><ul><ul><li>has no magnetic force associated. </li></ul></ul><ul><li>We have investigated current carrying conductors ; </li></ul><ul><ul><li>and the magnetic force associated with it. </li></ul></ul>
  • 3. Forces on Moving Charges <ul><li>Another way to produce an electric current is ; </li></ul><ul><ul><li>to have a moving beam of charged particles. </li></ul></ul><ul><li>If the beam were to move perpendicularly into a magnetic field ; </li></ul><ul><ul><li>then every charged particle would experience a force. </li></ul></ul>
  • 4. Forces on Moving Charges <ul><li>It must be perpendicular as from ; </li></ul><ul><ul><li>F = BIl sin  , sin  = 0 . </li></ul></ul><ul><li>Force would be zero ; </li></ul><ul><ul><li>if the motion was parallel to the field . </li></ul></ul><ul><li>If the beam was visible ; </li></ul><ul><ul><li>seen to be deflected by , </li></ul></ul><ul><ul><li>magnetic interaction. </li></ul></ul>
  • 5. Forces on Moving Charges <ul><li>This deflection occurs because ; </li></ul><ul><ul><li>the charged particles are no longer constrained by , </li></ul></ul><ul><ul><li>the lattice of metal ions in the wire. </li></ul></ul>
  • 6. Forces on Moving Charges <ul><li>The deflection of the beam is determined ; </li></ul><ul><ul><li>by the right hand palm rule. </li></ul></ul><ul><li>Be careful as the thumb must point in the direction of conventional current ; </li></ul><ul><ul><li>i.e. +ive to -ive. </li></ul></ul>
  • 7. Forces on Moving Charges <ul><li>For +ively charged particles ; </li></ul><ul><ul><li>this is the direction of the beam. </li></ul></ul><ul><li>For -ively charged particles ; </li></ul><ul><ul><li>the conventional current is , </li></ul></ul><ul><ul><li>opposite to the direction of the beam. </li></ul></ul>
  • 8. Forces on Moving Charges <ul><li>If the beam moves in a plane that is perpendicular to the magnetic field ; </li></ul><ul><ul><li>the path will describe an arc of a circle. </li></ul></ul><ul><li>The force always acts on the particles ; </li></ul><ul><ul><li>always at 90 o to the direction of motion. </li></ul></ul>
  • 9. Forces on Moving Charges <ul><li>There is no component in the direction of motion ; </li></ul><ul><ul><li>so the force will not change the speed of the particles, </li></ul></ul><ul><ul><li>only the direction. </li></ul></ul>
  • 10. Forces on Moving Charges <ul><li>As the beam is always at 90 o to the field lines ; </li></ul><ul><ul><li>force acts as a centripetal force , </li></ul></ul><ul><ul><li>produces uniform circular motion . </li></ul></ul>
  • 11. Forces on Moving Charges <ul><li>The magnitude of the force acting on the beam is determined by: </li></ul><ul><li>F = B I l sin  </li></ul>
  • 12. Forces on Moving Charges <ul><li>Il needs to be determined for a beam of particles ; </li></ul><ul><ul><li>each of charge q , </li></ul></ul><ul><ul><li>moving at a constant speed v . </li></ul></ul>
  • 13. Forces on Moving Charges <ul><li>Consider a small length of the beam l . </li></ul><ul><li>Assume that the number of particle past a point in 1 second is n </li></ul><ul><li>The current I = q/  t = nq </li></ul><ul><li>If the charges travel a distance l in one second, their velocity is l /1=v </li></ul><ul><li>The force on the beam is F=I l B=nqvB </li></ul><ul><li>Thus the force on each particle = qvB </li></ul>
  • 14. Forces on Moving Charges <ul><li>Alternatively, imagine a single charge q moving a distance l in a time  t. </li></ul><ul><li>We know F=IlB. </li></ul><ul><li>But, I = q/  t and so </li></ul><ul><li>Which gives F=qvB </li></ul>
  • 15. Forces on Moving Charges <ul><li>Substituting into F = B I l sin  ; </li></ul><ul><li>F = q vB sin  </li></ul><ul><li>Where  is the angle between v and B . </li></ul><ul><li>This equation gives the magnitude ; </li></ul><ul><ul><li>direction is determined by the right hand palm rule. </li></ul></ul>
  • 16. Forces on Moving Charges <ul><li>A beam of charged particles in a magnetic field ; </li></ul><ul><ul><li>can follow a semi-circular path , </li></ul></ul><ul><ul><li>with uniform circular motion. </li></ul></ul><ul><li>The radius and other features can easily be determined. </li></ul><ul><li>The magnetic force, </li></ul><ul><ul><li>supplies a centripetal force, therefore: </li></ul></ul>
  • 17. Forces on Moving Charges <ul><li> F B  =  F c  </li></ul><ul><li>and rearranging gives the equation : </li></ul>
  • 18. Forces on Moving Charges <ul><li>The period of the motion and the frequency of revolution can be deduced from: </li></ul>
  • 19. Forces on Moving Charges Try Example 1
  • 20. Solution <ul><li>B = 6.0 T up the page F = 4.8 x 10 -14 N west </li></ul><ul><li>q = 1.6 x 10 -19 C m = 9.1 x 10 -31 kg </li></ul>
  • 21. Solution <ul><li>In B field, protons move in circular path. </li></ul><ul><li>Centripetal force supplied by; </li></ul><ul><ul><li>Magnetic force. </li></ul></ul><ul><li>| F c | = | F B | </li></ul><ul><li>F B = B q v </li></ul>
  • 22. Solution <ul><li>v = 5.0 x 10 4 m s -1 </li></ul><ul><li>For direction, </li></ul><ul><ul><li>use right hand palm rule. </li></ul></ul>
  • 23. Solution
  • 24. Solution <ul><li>Current is south. </li></ul><ul><li>As charge is +ive; </li></ul><ul><ul><li>Motion is in same direction </li></ul></ul><ul><li>v = 5.0 x 10 4 m s -1 </li></ul>
  • 25. Forces on Moving Charges Try Example 2
  • 26. Solution <ul><li>B = 2.5 x 10 -3 T north q = 1.6 x 10 -19 C </li></ul><ul><li>m = 9.1 x 10 -31 kg v = 1.0 x 10 6 m s -1 east </li></ul>
  • 27. Solution – Part (a) <ul><li>Use right hand palm rule. </li></ul><ul><li>Field north; </li></ul><ul><ul><li>Fingers. </li></ul></ul><ul><li>Current west; </li></ul><ul><ul><li>Current in opposite direction to path of electrons. </li></ul></ul><ul><li>Force into page; </li></ul><ul><ul><li>Palm faces into page. </li></ul></ul>
  • 28. Solution – Part (b) <ul><li>F c supplied by magnetic force. </li></ul><ul><li>| F c | = | F B | </li></ul>
  • 29. Solution – Part (b) <ul><li>r = 2.28 x 10 -3 m </li></ul><ul><li>r = 2.3 x 10 -3 m (2 sig. figs.) </li></ul>
  • 30. Solution – Part (c) <ul><li>Assume electrons travel; </li></ul><ul><ul><li>In circular path, </li></ul></ul><ul><ul><li>Uniform circular motion. </li></ul></ul>
  • 31. Solution – Part (c) <ul><li>T = 1.4 x 10 -8 s </li></ul>
  • 32. Charged Particles Moving at an Angle to the Field <ul><li>If a beam of charged particles ; </li></ul><ul><ul><li>travelling perpendicular , </li></ul></ul><ul><ul><li>to the magnetic field, </li></ul></ul><ul><li>A semi-circular path may be achieved ; </li></ul><ul><ul><li>if beam is in a vacuum. </li></ul></ul>
  • 33. Charged Particles Moving at an Angle to the Field <ul><li>If air were present ; </li></ul><ul><ul><li>the collisions would cause , </li></ul></ul><ul><ul><li>particles to lose energy , </li></ul></ul><ul><ul><li>and slow down. </li></ul></ul><ul><li>Now consider a beam at some other angle (such as 45 o ). </li></ul>
  • 34. Charged Particles Moving at an Angle to the Field
  • 35. Charged Particles Moving at an Angle to the Field <ul><li>The velocity is split into two components v par and v perp </li></ul><ul><li>From diagram, </li></ul><ul><li>v perp = v sin  </li></ul><ul><li>v par = v cos  . </li></ul>
  • 36. Charged Particles Moving at an Angle to the Field <ul><li>Parallel component will remain constant . </li></ul><ul><li>Perpendicular component will cause the beam ; </li></ul><ul><ul><li>to move in uniform circular motion. </li></ul></ul>
  • 37. Charged Particles Moving at an Angle to the Field <ul><li>The radius of the circular motion is given by: </li></ul>
  • 38. Charged Particles Moving at an Angle to the Field <ul><li>When the two components are added together ; </li></ul><ul><ul><li>path of the charged particles describe , </li></ul></ul><ul><ul><li>a spiral. </li></ul></ul><ul><li>If the magnetic field is constant ; </li></ul><ul><ul><li>path is helical </li></ul></ul><ul><ul><ul><li>Helix is a spiral with uniform radius and uniform pitch. </li></ul></ul></ul>
  • 39. Charged Particles Moving at an Angle to the Field
  • 40. Charged Particles Moving at an Angle to the Field Try Example 3
  • 41. Solution
  • 42. Solution <ul><li>B = 4.0 x 10 -3 T </li></ul><ul><li>E = 10 eV = 1.6 x 10 -18 J </li></ul><ul><li>m = 9.1 x 10 -31 kg </li></ul><ul><li>q = 1.6 x 10 -19 C </li></ul><ul><li> = 60 o </li></ul>
  • 43. Solution – Part (a) <ul><li>v = 1.88 x 10 6 m s -1 </li></ul><ul><li>v = 1.9 x 10 6 m s -1 (2 sig. Figs.) </li></ul>
  • 44. Solution – Part (a) <ul><li>v = v para + v perp </li></ul>
  • 45. Solution – Part (a) <ul><li>Parallel Component: </li></ul><ul><li>v para = v cos 60 o </li></ul><ul><li>v para = 1.88 x 10 6 x 0.5 </li></ul><ul><li>v para = 9.4 x 10 5 m s -1 (2 sig.figs.) </li></ul>
  • 46. Solution – Part (a) <ul><li>Perpendicular component: </li></ul><ul><li>v perp = v sin 60 o </li></ul><ul><li>v perp = 1.88 x 10 6 x 0.866 </li></ul><ul><li>v perp = 1.6 x 10 6 m s –1 (2 sig.figs.) </li></ul>
  • 47. Solution – Part (b) <ul><li>F c for motion of electrons supplied by; </li></ul><ul><ul><li>Magnetic force. </li></ul></ul><ul><li>| F c | = | F B | </li></ul>
  • 48. Solution – Part (b) <ul><li>r = 2.32 x 10 -3 m </li></ul><ul><li>r = 2 .3 x 10 -3 m (2 sig. figs.) </li></ul>
  • 49. Solution – Part (c) <ul><li>Assuming circular motion: </li></ul><ul><li>8.9 x 10 -9 s (2 sig. figs.) </li></ul>
  • 50. Solution – Part (d) <ul><li>In direction parallel to axis of helical motion; </li></ul><ul><ul><li>Velocity is uniform. </li></ul></ul>
  • 51. Solution – Part (d) <ul><li>Pitch = v para T </li></ul><ul><li>Pitch = 9.4 x 10 5 x 8.94 x 10 -9 </li></ul><ul><li>Pitch = 8.4 x 10 -3 m </li></ul>
  • 52. Application- Magnetic Fields in Cyclotrons <ul><li>Consider the motion of a positive ion ; </li></ul><ul><ul><li>inside the dees of a cyclotron. </li></ul></ul>
  • 53. Application- Magnetic Fields in Cyclotrons <ul><li>The dees are placed between ; </li></ul><ul><ul><li>poles of an electromagnet. </li></ul></ul><ul><li>The ions are not shielded from ; </li></ul><ul><ul><li>magnetic field , </li></ul></ul><ul><ul><li>unlike the electric field. </li></ul></ul><ul><li>This means the ions are affected ; </li></ul><ul><ul><li>inside the dees , </li></ul></ul><ul><ul><li>in the gap between them. </li></ul></ul>
  • 54. Application- Magnetic Fields in Cyclotrons <ul><li>To make the ions move in circular path ; </li></ul><ul><ul><li>uniform magnetic field is needed , </li></ul></ul><ul><ul><li>perpendicular to the plane of the dees. </li></ul></ul><ul><li>Polarity of the field is important ; </li></ul><ul><ul><li>t o make the ions move in the right direction. </li></ul></ul><ul><li>The force is such that ; </li></ul><ul><ul><li>always acting towards the centre of the circle , </li></ul></ul><ul><ul><li>causing centripetal acceleration. </li></ul></ul>
  • 55. Application- Magnetic Fields in Cyclotrons <ul><li>In the diagram below ; </li></ul><ul><ul><li>at the particular instant shown, </li></ul></ul><ul><ul><li>its velocity is down. </li></ul></ul><ul><li>As the force must always act towards the centre of the circle ; </li></ul><ul><ul><li>determine the direction of the magnetic field , </li></ul></ul><ul><ul><li>using the right hand rule. </li></ul></ul>
  • 56. Application- Magnetic Fields in Cyclotrons <ul><li>The field must be in a direction which is ; </li></ul><ul><ul><li>OUT of the page. </li></ul></ul>
  • 57. Period of Circular Motion <ul><li>Previously we stated: </li></ul><ul><li>The radius of the circular arc described by an ion is proportional to its speed. </li></ul><ul><li>We can apply Newton’s second law ; </li></ul><ul><ul><li>F = ma , </li></ul></ul><ul><ul><ul><li>as it applies to circular motion </li></ul></ul></ul>
  • 58. Period of Circular Motion <ul><li>And F = qvB </li></ul><ul><li>therefore, upon rearrangement : </li></ul>
  • 59. Period of Circular Motion <ul><li>As the mass m ; </li></ul><ul><ul><li>charge q , </li></ul></ul><ul><ul><li>magnetic field B , </li></ul></ul><ul><ul><li>are all constant , </li></ul></ul><ul><li>r  v </li></ul>
  • 60. Period of Circular Motion <ul><li>We also stated: </li></ul><ul><li>The time for the ion to complete one semicircle is the same irrespective of the speed of the ion. </li></ul><ul><li>From before, if the speed doubles ; </li></ul><ul><ul><li>radius doubles. </li></ul></ul>
  • 61. Period of Circular Motion <ul><li>This also doubles the ; </li></ul><ul><ul><li>circumference (2  r). </li></ul></ul><ul><li>Mathematically, this can also be shown to be true. </li></ul><ul><li>The velocity of an object undergoing circular motion is given by: </li></ul>
  • 62. Period of Circular Motion <ul><li>Rearranging for T </li></ul><ul><li>From  we can substitute for r . </li></ul>
  • 63. Period of Circular Motion <ul><li>This shows that the period is ; </li></ul><ul><ul><li>independent of speed , </li></ul></ul><ul><ul><li>or radius . </li></ul></ul>
  • 64. Period of Circular Motion <ul><li>Constant for ions of ; </li></ul><ul><ul><li>a constant charge , </li></ul></ul><ul><ul><li>Mass , </li></ul></ul><ul><ul><li>in a given magnetic field. </li></ul></ul><ul><li>Be careful. </li></ul>
  • 65. Period of Circular Motion <ul><li>The period refers to the time ; </li></ul><ul><ul><li>for one complete revolution, </li></ul></ul><ul><ul><li>not the time in a dee. </li></ul></ul><ul><ul><ul><li>That is half a period. </li></ul></ul></ul>
  • 66. Kinetic Energy of Ions <ul><li>Any nuclear reaction caused by a collision ; </li></ul><ul><ul><li>requires a certain amount of energy. </li></ul></ul><ul><li>It is critical to know the kinetic energy ; </li></ul><ul><ul><li>with which the ions leave the cyclotron. </li></ul></ul>
  • 67. Kinetic Energy of Ions <ul><li>We previously stated that if we know the P.D ; </li></ul><ul><ul><li>and the number of revolutions, </li></ul></ul><ul><ul><li>we can calculate the kinetic energy. </li></ul></ul><ul><li>While this is correct ; </li></ul><ul><ul><li>the number of revolutions is not normally known. </li></ul></ul>
  • 68. Kinetic Energy of Ions <ul><li>An alternative method is required. </li></ul><ul><li>K = ½mv 2 </li></ul><ul><li>If we rearrange equation  </li></ul>
  • 69. Kinetic Energy of Ions <ul><li>Substituting into formula for K : </li></ul>
  • 70. Kinetic Energy of Ions <ul><li>This indicates that the kinetic energy of an ion ; </li></ul><ul><ul><li>of given charge , </li></ul></ul><ul><ul><li>and mass . </li></ul></ul><ul><li>Only depends on the radius ; </li></ul><ul><ul><li>of the final circle , </li></ul></ul><ul><ul><li>magnitude of the magnetic field. </li></ul></ul><ul><li>This can be understood due to two points. </li></ul>
  • 71. Kinetic Energy of Ions <ul><li>Point 1 </li></ul><ul><li>If the magnetic field increases ; </li></ul><ul><ul><li>the radii decreases , </li></ul></ul><ul><ul><li>ions make more revolutions , </li></ul></ul><ul><ul><li>more crossings of the gap between the dees. </li></ul></ul>
  • 72. Kinetic Energy of Ions <ul><li>At each crossing ; </li></ul><ul><ul><li>they are accelerated , </li></ul></ul><ul><ul><li>to higher kinetic energy. </li></ul></ul><ul><li>Increasing the magnetic field results in ; </li></ul><ul><ul><li>increase of the kinetic energy , </li></ul></ul><ul><ul><li>of the emerging ions , </li></ul></ul><ul><ul><li>at a given radius. </li></ul></ul>
  • 73. Kinetic Energy of Ions <ul><li>Point 2 </li></ul><ul><li>If the P.D. is increased ; </li></ul><ul><ul><li>the ions gain more speed with each crossing of the gap , </li></ul></ul><ul><ul><li>and so make circles with larger radii , </li></ul></ul><ul><ul><li>and make fewer revolutions. </li></ul></ul>
  • 74. Kinetic Energy of Ions <ul><li>This means that a larger P.D. does not result in ; </li></ul><ul><ul><li>a larger kinetic energy , </li></ul></ul><ul><ul><li>of the emerging ions at a given radius. </li></ul></ul>
  • 75. Cyclotrons Try Example 4
  • 76. Solution <ul><li>T = 4.4 x 10 -8 s </li></ul>
  • 77. Solution <ul><li>f = 1 /T </li></ul><ul><li>f = 1/4.4 x 10 -8 </li></ul><ul><li>f = 2.3 x 10 7 Hz </li></ul><ul><ul><li>or 23MHz </li></ul></ul>
  • 78. Cyclotrons Try Example 5
  • 79. Solution <ul><li>K = 1. 125 x 10 -12 J </li></ul>
  • 80. Solution <ul><li>K = 7.03 x 10 6 e V </li></ul><ul><ul><li>or 7.03 Me V </li></ul></ul>

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