Problem Set 1

Problem 6

Part 1

The radius of curvature of a charged particle in a magnetic field is given by:

	p = q B r  -->  pc = q B r c
	
or   B = (pc/q) 1/rc = 1e12 V / (.75*500 m 3e8 m/sec)

B  =  10. ^12/( .75*500.  3. *10^8)

(Actually, the Fermilab ring is 1 km in radius and the magnets actually run at about 4 T.)

Part 2

The particles are going at the speed of light.  The time it takes to go around the ring once is:

period  =  10^3 * N[Pi]/(3. *10^8)

Or a frequency of:

frequency  =  1/period

If the machine had 24 bunches, then the frequency would be 24 times as great:

bunchfrequency  =  24. *frequency

Or 2.3 MHz.

Part 3

Number of times the protons transverse an RF cavity:

numb  =  10/period

Energy change:

deltaE  =  1. *10^12  -  120. *10^9

Energy per transit:

voltseen  = deltaE/numb

(There are actually many cavities rather than just one.  One MegaVolt on a cavity is very high!!)

Part 4

phase  = N[Sin[Pi/4]]

vmax  =  voltseen/phase

Part 5

An experimenter sees the same bunch every 10 micorseconds.   In that time, that bunch and another bunch came by.  Thus bunches come by and interactions can occur every 5 microseconds.