A proton synchrotron accelerates protons to a kinetic energy of 500 GeV. At this energy, calculate (a) the Lorentz factor, (b) the speed parameter, and (c) the magnetic field for which the proton orbit has a radius of curvature of 750 m.

The energy of proton is $K=500GeV$

The radius of curvature of orbit is $750m$

The magnetic field for the electron is found by equating the necessary centripetal force by magnetic force on the electron.

(a)

The Lorentz factor is given as:

$K=\left(\gamma -1\right)m{c}^2$

Here, $q$is the charge of proton and its value is $1.6\times {10}^{-19}C$ , $m$ is the mass of proton and its value is $1.67\times {10}^{-19}C$ , $c$ is the speed of light and its value is $3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Substitute all the values in the above equation.

$\begin{array}{rcl}\left(500GeV\right)\left(\frac{1.6\times {10}^{9}J}{1GeV}\right)& =& \left(\gamma -1\right)\left(1.67\times {10}^{-27}kg\right){\left(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2\\ \gamma & =& 534\\ & & \end{array}$

Therefore, the Lorentz factor for proton is $534.$.

(b)

The speed parameter for proton is given as:

$\beta =\sqrt{1-\frac{1}{{\gamma }^2}}$

Substitute all the values in the above equation.

$\beta =\sqrt{1-\frac{1}{{\left(534\right)}^2}}$

$\beta =0.999999$

Therefore, the speed parameter for proton is 0.999999.

(c)

The magnetic field for proton is given as:

$B=\frac{mc\sqrt{{\gamma }^2-1}}{qr}$

Substitute all the values in the above equation.

$$\begin{gathered} B=\frac{\left(1.67\times {10}^{-27}kg\right)\left(3\times {10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right)\sqrt{{\left(534\right)}^2-1}}{\left(1.6\times {10}^{-19}C\right)\left(750m\right)} \\ B=2.23T \end{gathered}$$

Therefore, the magnetic field for proton is $2.23T$