91Ó°ÊÓ

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:

$K=\left(\mathrm{γ}-1\right)m{c}^2$

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

Substitute all the values in the above equation.

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

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

(b)

The speed parameter for proton is given as:

$\mathrm{β}=\sqrt{1-\frac{1}{{\mathrm{γ}}^2}}$

Substitute all the values in the above equation.

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

$\mathrm{β}=0.999999$

Therefore, the speed parameter for proton is 0.999999.

(c)

The magnetic field for proton is given as:

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

Substitute all the values in the above equation.

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

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