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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>