Question

Two particles, an electron and a proton, move in a circular path in a uniform magnetic field of intensity B=1.23 T. Find the ratio between the time period of the proton Tp to the electron Te (i.e., find 'p)

Answer 1

The ratio of the time period of the proton to the electron is 1835.16.

Step-by-step explanation:

Given that,

Two particles, an electron and a proton, move in a circular path in a uniform magnetic field of intensity B=1.23 T

We need to find the ratio between the time period of the proton Tp to the electron Te.

The time period in magnetic field is given by :

`T=(2\pi m)/(qB)`

For proton, time period is :

`T_P=(2\pi m_P)/(q_pB)\ ....(1)`

For an electron, the time period is :

`T_e=(2\pi m_e)/(q_eB)\ ....(2)`

From equation (1) and (2) :

`(T_p)/(T_e)=((2\pi m_p)/(q_pB))/((2\pi m_e)/(q_eB))\\\\As\ q_e=q_p\\\\(T_p)/(T_e)=(m_p)/(m_e)\\\\=(1.67* 10^(-27))/(9.1* 10^(-31))\\\\=1835.16`

So, the ratio of the time period of the proton to the electron is 1835.16.