Question

A cyclotron with dee radius 45.0 cm is operated at an oscillator frequency of 11.0 MHz to accelerate protons. (a) What magnitude B of magnetic field is required to achieve resonance? (b) At that field magnitude, what is the kinetic energy of a proton emerging from the cyclotron? MeV Suppose, Instead, that B = 1.58 T. (c) What oscillator frequency is required to achieve resonance now? MHz (d) At that frequency, what is the kinetic energy of an emerging proton? MeV

Answer 1

let the proton of charge 49 ware the ion source hednien que deo arderber the negedire dee at the insturry with verouty o normal to the magnetic canes of force ? so, for nequetic field I, the torce Facting on the proper to given by it the magnetic fields perpendicular to the direction of fore, Fraub. The rath of proper innert oto a circeo of radio (radion of dee) so b giren by globe me 97. it f is the trequering of oscillatot, to ar ( for resonne) no 2205f. Bra mirant a B= 24fm aro for profer, ma 1167x10 + kg, tall.o mitz = 110X0 Hz ar 11b81513 0.72T -- BC 215x 11.0X10% H2 X 667 X10-27 - 16810-19 magnitude B of magnetic field is required to achine resonance is 0.72 T.

mw, the Propa, kinetic energy of the Kf = meer 1.6X1019 are roa radio of dee a 45cm., 0.45m. Elle a 5*1.17X10127 per anar 10-45 m) <[200+ MB) . - 8.677x10! now leur lo6810 mera 106 eN = 116210 13 g. k 8.077 *:00137 = 5.05 men. 1.6.x26s (1/mon) so, Kinedic energy of the proper emerging from the cyclotron o 5.05 mer. T2Tm teoo it. BC 1.58 T, Ba 2 fm og tager som Li libriolc*1158.1 24 x10*42 - 24 MHZ. 2 * 1.67 xloretag I now the to achire resonance, the usullarr frequerey will he 24 mer 24 muz. So, 27 at t= 24 MHZ, ke atm (21900f). C .67X10 Ping x 4 x 10.44 m) x (24210 ) - 3.845810-12 g. - (4,18 m gmper 24.1 mer. kinetic energy of emergery proton is 24.1 men 2.so
We know for resonance, the maximum radius of the circular path of the proton in a cyclotron is given by

$r_0=\frac{mv}{qB}$

Again, $f=\frac{v_0}{2\pi r_0}$

Using these we can find the magnetic field.