{"id":64,"date":"2015-09-27T22:39:56","date_gmt":"2015-09-27T22:39:56","guid":{"rendered":"http:\/\/heidib.me\/cyclotron\/wordpress\/?page_id=64"},"modified":"2016-03-10T09:51:52","modified_gmt":"2016-03-10T09:51:52","slug":"64-2","status":"publish","type":"page","link":"http:\/\/koethcyclotron.org\/?page_id=64","title":{"rendered":"Theory of Operation"},"content":{"rendered":"<div id=\"attachment_65\" style=\"width: 710px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/i1.wp.com\/heidib.me\/cyclotron\/wp-content\/uploads\/2015\/09\/theory_of_oper.jpg\"><img aria-describedby=\"caption-attachment-65\" loading=\"lazy\" class=\"size-full wp-image-65\" src=\"https:\/\/i1.wp.com\/heidib.me\/cyclotron\/wp-content\/uploads\/2015\/09\/theory_of_oper.jpg?resize=700%2C539\" alt=\"Fig.1 Conceptual diagrm showing operation of a cyclotron\" width=\"700\" height=\"539\" srcset=\"https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/uploads\/2015\/09\/theory_of_oper.jpg?w=700 700w, https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/uploads\/2015\/09\/theory_of_oper.jpg?resize=300%2C231 300w\" sizes=\"(max-width: 700px) 100vw, 700px\" data-recalc-dims=\"1\" \/><\/a><p id=\"caption-attachment-65\" class=\"wp-caption-text\">Fig.1 Conceptual diagrm showing operation of a cyclotron<\/p><\/div>\n<p>&nbsp;<\/p>\n<p>The operation of a cyclotron is based on the fact that the period of the motion of a charged particle in a uniform magnetic field is independent of the velocity of the particle, as can be seen in the following derivation:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_a6c7acfc672d8535b8d9d849465f458e.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"F = ma\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>F = ma<\/script><\/p><\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_f617e62f859b8927ae2ec64971a5bc68.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"qvB = mv^2\/r\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>qvB = mv^2\/r<\/script><\/p><\/p>\n<p>Solve for <span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_4b43b0aee35624cd95b910189b3dc231.gif?w=1170' style='vertical-align: middle; border: none; padding-bottom:2px;' class='tex' alt=\"r\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>r<\/script>:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_a61d1ac9a7d162b8f21140561a8ca59b.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"r=mv\/qB\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>r=mv\/qB<\/script><\/p><\/p>\n<p>Now find the period, <span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_b9ece18c950afbfa6b0fdbfa4ff731d3.gif?w=1170' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt=\"T\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>T<\/script>:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_83de07f1fa21dfa9431ce96457b7c56b.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"T=1\/f=2\\pi\/\\omega=2\\pi r\/v=2\\pi mv\/(qBv)\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>T=1\/f=2\\pi\/\\omega=2\\pi r\/v=2\\pi mv\/(qBv)<\/script><\/p><\/p>\n<p>The <span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_9e3669d19b675bd57058fd4664205d2a.gif?w=1170' style='vertical-align: middle; border: none; padding-bottom:2px;' class='tex' alt=\"v\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>v<\/script>'s cancel:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_9401da3dc476495a846646b12f1a1951.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"T=2\\pi m\/qB\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>T=2\\pi m\/qB<\/script><\/p><strong>\u00a0<\/strong><\/p>\n<p>The 'Cyclotron Frequency' <span class='MathJax_Preview'><img src='https:\/\/i2.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_8fa14cdd754f91cc6554c9e71929cce7.gif?w=1170' style='vertical-align: middle; border: none; ' class='tex' alt=\"f\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>f<\/script>\u00a0immediately follows:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i2.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_7bdd4c7d6429cf9a3cd1321f69edc7c6.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"f = qB\/2\\pi m\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>f = qB\/2\\pi m<\/script><\/p><\/p>\n<p>&nbsp;<\/p>\n<p>Fig. 1 is a schematic drawing of a cyclotron. The particles move in two semicircular metal containers called DEEs (because of their shape). The dees are housed in a vacuum chamber that is in a uniform magnetic field provided by an electromagnet. (The region in which the particles move must be evacuated so that the particles will not lose energy and be scattered in collisions with air molecules.) Between the dees there is maintained a potential difference <span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_def0fceff3b277a685bdb2936e614835.gif?w=1170' style='vertical-align: middle; border: none; ' class='tex' alt=\"\\Delta V\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>\\Delta V<\/script>\u00a0that alternates in time with a period <span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_b9ece18c950afbfa6b0fdbfa4ff731d3.gif?w=1170' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt=\"T\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>T<\/script>, which is chosen to be equal to the cyclotron period that was found in the above derivation. This potential difference creates an electric field across the gap between the dees. At the same time, there is no electric field within each dee because of the shielding of the metal dees.<\/p>\n<p>&nbsp;<\/p>\n<p>The charged particles are initially injected into dee 1 with a small velocity from an ion source near the center of the dees. They move in a semicircle in dee 1 and arrive a the gap between dee 1 and dee 2 after time <span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_b9ece18c950afbfa6b0fdbfa4ff731d3.gif?w=1170' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt=\"T\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>T<\/script>, where <span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_b9ece18c950afbfa6b0fdbfa4ff731d3.gif?w=1170' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt=\"T\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>T<\/script> is the cyclotron period, and is also the period with which the potential across the dees\u00a0is alternated. The alternation of the potential is adjusted so that dee 1 is at a higher potential than dee 2 when the particles arrive at the gap between them. Each particle is therefore accelerated across the gap by the electric field across the gap and gains energy equal to q. Because it has more kinetic energy, the particle moves in a semicircle of larger radius in dee 2, and again arrives at the gap after a time \u00a0<span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_b9ece18c950afbfa6b0fdbfa4ff731d3.gif?w=1170' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt=\"T\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>T<\/script>. By this time the potential between the dees has been reversed so that dee 2 is now at the higher potential. Once more the particle is accelerated across the gap and gains additional kinetic energy equal to <span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_12890c6bff3146c81f7d5db71b2853a1.gif?w=1170' style='vertical-align: middle; border: none; ' class='tex' alt=\"q\\Delta V\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>q\\Delta V<\/script>. Each time the particle arrives at the gap, it is accelerated and gains kinetic energy <span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_12890c6bff3146c81f7d5db71b2853a1.gif?w=1170' style='vertical-align: middle; border: none; ' class='tex' alt=\"q\\Delta V\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>q\\Delta V<\/script>. Thus, it moves in larger and larger semicircular orbits until it eventually leaves the magnetic field. In the typical cyclotron, each particle may make up to 50 to 100 revolutions before reaching its final energy.<\/p>\n<p>&nbsp;<\/p>\n<p>The kinetic energy of a particle leaving a cyclotron can be calculated by the following derivation:<\/p>\n<p>&nbsp;<\/p>\n<p>1<sup>st<\/sup> set the following equal to the <span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_5e325655534500f29aa7d840a65162be.gif?w=1170' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt=\"r_{max}\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>r_{max}<\/script>\u00a0of the dees and solve for the particle's velocity:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_c4a72481c7bd4659a0c85b085073e773.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"r_{max} = mv\/qB\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>r_{max} = mv\/qB<\/script><\/p><\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_1984da082fdb24c43a78969e24a92e4d.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"v = qBr\/m\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>v = qBr\/m<\/script><\/p><strong>\u00a0<\/strong><\/p>\n<p>Next solve for the Kinetic Energy:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i2.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_9137303c5abe5e8f8d076dd49571992a.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"K = \\frac12 mv^2 = \\frac12 m (q^2B^2r^2\/m^2)\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>K = \\frac12 mv^2 = \\frac12 m (q^2B^2r^2\/m^2)<\/script><\/p><\/p>\n<p>Cancel the <span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_6f8f57715090da2632453988d9a1501b.gif?w=1170' style='vertical-align: middle; border: none; padding-bottom:2px;' class='tex' alt=\"m\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex'>m<\/script>'s and finally get:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_8f0d491a41c1274380236f1bdb87cca8.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"K = q^2B^2r^2\/2m\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>K = q^2B^2r^2\/2m<\/script><\/p><strong>\u00a0<\/strong><\/p>\n<p>Now for an example with the Rutgers 12-Inch Cyclotron running protons, determine the cyclotron frequency and maximum Kinetic Energy:<\/p>\n<p>&nbsp;<\/p>\n<p><strong><u>12-INCH CYCLOTRON PARAMETERS:<\/u><\/strong><\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_1d3e570a29a00deef0df132e3d706049.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"r_{max} = 5'' = 0.127 \\mathrm{m}\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>r_{max} = 5'' = 0.127 \\mathrm{m}<\/script><\/p><\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_5d8239399f5652e11a4f40a92b988592.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"B_{max} = 1.2 \\mathrm{T}\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>B_{max} = 1.2 \\mathrm{T}<\/script><\/p><\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i2.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_a20d16329cb3baf66b403001ea538bfc.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"q = -1.6\\times 10^{-19} \\mathrm{C}\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>q = -1.6\\times 10^{-19} \\mathrm{C}<\/script><\/p><\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i0.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_238458f1e6bf0c37fb5f6e1e92a66394.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"m =1.67 \\times 10^{-27} \\mathrm{kg}\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>m =1.67 \\times 10^{-27} \\mathrm{kg}<\/script><\/p><strong>\u00a0<\/strong><\/p>\n<p>1<sup>st<\/sup> Determine the Cyclotron Frequency:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i2.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_d9bc36dd0321b957451ddda919bb964f.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"f = qB\/2\\pi m = (1.6\\times 10^{-19} \\mathrm{C}) (1.2 \\mathrm{T}) \/ (2\\pi)(1.67\\times 10^{-27} \\mathrm{kg}) = 18.3 \\mathrm{MHz}\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>f = qB\/2\\pi m = (1.6\\times 10^{-19} \\mathrm{C}) (1.2 \\mathrm{T}) \/ (2\\pi)(1.67\\times 10^{-27} \\mathrm{kg}) = 18.3 \\mathrm{MHz}<\/script><\/p><\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p>2<sup>nd<\/sup> Find the maximum Kinetic Energy of the protons:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='https:\/\/i1.wp.com\/koethcyclotron.org\/wp-content\/plugins\/latex\/cache\/tex_e19682a45b157e11d0338b17fe6841e5.gif?w=1170' style='vertical-align: middle; border: none;' class='tex' alt=\"K = (1.6\\times 10^{-19} \\mathrm{C})^2(1.2 \\mathrm{T})^2 (0.127 \\mathrm{m})^2\/2(1.67\\times 10^{-27} \\mathrm{kg}) = 1.78\\times 10^{-13} \\mathrm{J} = 1.1 \\mathrm{MeV}\" data-recalc-dims=\"1\" \/><\/span><script type='math\/tex;  mode=display'>K = (1.6\\times 10^{-19} \\mathrm{C})^2(1.2 \\mathrm{T})^2 (0.127 \\mathrm{m})^2\/2(1.67\\times 10^{-27} \\mathrm{kg}) = 1.78\\times 10^{-13} \\mathrm{J} = 1.1 \\mathrm{MeV}<\/script><\/p><strong>\u00a0<\/strong><strong><sub>\u00a0<\/sub><\/strong><\/p>\n<p><em>- adapted from Tipler's Physics for Scientists and Engineers, Vol. 2.<\/em><\/p>\n","protected":false},"excerpt":{"rendered":"<p>&nbsp; The operation of a cyclotron is based on the fact that the period of the motion of a charged particle in a uniform magnetic field is independent of the velocity of the particle, as can be seen in the following derivation: Solve for : Now find the period, : The 's cancel: \u00a0 The [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"spay_email":""},"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/P6SHXY-12","_links":{"self":[{"href":"http:\/\/koethcyclotron.org\/index.php?rest_route=\/wp\/v2\/pages\/64"}],"collection":[{"href":"http:\/\/koethcyclotron.org\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"http:\/\/koethcyclotron.org\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"http:\/\/koethcyclotron.org\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/koethcyclotron.org\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=64"}],"version-history":[{"count":12,"href":"http:\/\/koethcyclotron.org\/index.php?rest_route=\/wp\/v2\/pages\/64\/revisions"}],"predecessor-version":[{"id":726,"href":"http:\/\/koethcyclotron.org\/index.php?rest_route=\/wp\/v2\/pages\/64\/revisions\/726"}],"wp:attachment":[{"href":"http:\/\/koethcyclotron.org\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=64"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}