Question

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.

Short answer

(a) The Lorentz factor for proton is 534 .

(b) The speed parameter for proton is 0.99999 .

(c) The magnetic field for the electron is 0.178T.

Step-by-step solution

Identification of given data

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.

Determination of Lorentz factor

(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 $534.$.

Determination of speed parameter

(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.

Determination of magnetic field for proton

(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 $2.23T$