Question

Calculate the period of a proton moving at 2.5 x 10⁷ m/s perpendicularly to a 1.2T field of a cyclotron.

mass of a proton = 1.67 x 10^-27 kg

charge on a proton = 1.602 x 10^-19 C

Answer 1

The time period of a particle moving perpendicular to a field of a cyclotron, taking in relativistic considerations into account, is given by:

$T=\frac{2\pi\gamma m}{Bq}$\

where,it's given, m = 1.67 x 10^-27 kg ; B = 1.2 T ; q = 1.602 x 10^-19 C and (gamma) is Lorentz Factor.

Lorentz Factor is given by:

$\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}$

where,it's given, v = 2.5 x 10⁷ m/s and c = speed of light = 3 x 10⁸ m/s [general fact]

The value of Lorentz Factor is :

$\gamma=\frac{1}{\sqrt{1-(\frac{2.5\times10^7}{3\times10^8})^2}}=1.00349527178103$

Therefore, the time period of the particle is:

T = 2 x 3.14159 x 1.00349527178103 x (1.6 x 10 1.2 x (1.602 x 10-19)

On simplification, we get:

T = 0.5248 ns

SUMMARY:

The time period of the proton particle is T=0.5248 ns