To see how this is done we must first discuss which units to use and study the relativistic equation connecting total energy and momentum. The charged particles experience a force given by equation (1):

(1)

because B and v lie in perpendicular directions this force will cause the particles to travel in circular trajectories. Hence, for circular motion, equation (1) gives

mass x centripetal acceleration = centripetal force

	(2)
	(3)
( units: Kg m s-1 )	(4)

The unit of Q is Coulombs (C), B is Tesla (T) and r is meters (m). If we multiply both sides of the equation by the speed of light, c = 3x10⁸ms^-1, then the units are now in Joules because: Momentum x Speed = Energy

( units: Joules (J) )

(5)

One electron volt, 1 eV = 1.6x10^-19 J or, expressed another way, 1 J = (1/1.6x10^-19)eV.

Therefore the units of the equation, above, can be converted to eV as follows:

( units: Electronvolts (eV) )

(6)

but Q is equal to the charge on the particle moving in the magnetic field. For this exercise Q is equal to the charge on one electron or proton = 1.6x10^-19 C. Therefore the equation above reduces to:    

( units: eV )

(7)

by substituting in the value for c, on the right hand side, we get

( units: eV )

(8)

or, because 1GeV = 1x10⁹ eV   

( units: GeV )

(9)

Finally, by expressing the units in terms of c we obtain:

( units: GeV/c )

(10)

Note: Even though we have derived this formula using classical mechanics it is also valid for the HEP case, provided we use the relativistic momentum, p=γm₀v

Hence the units of momentum are GeV/c, B still has units of Tesla, r of meters. 

As a summary:

These are very convenient units in which to measure energy, mass and momentum when working with particles of very light mass accelerated to very high energies.  For example, it is much easier to remember that the rest mass of the proton is roughly 1 GeV/c², than it is to remember that its value is 1.67x10^-27 Kg!

  
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