Chapter 20, Problem 15P

To determine

The ratio of the radii of the circular paths of a proton and an electron on entering a region of constant magnetic field.

Answer to Problem 15P

Solution:

The ratio of the radii of the circular paths of a proton and an electron on entering a region of constant magnetic field is 42.8.

Explanation of Solution

The radius(r)of the trajectory of a charge particle moving with a velocity(v)perpendicular to magnetic field(B)is given by, $r=\frac{mv}{Bq}$

If K is the kinetic energy of the particle then, $mv=\sqrt{2mK}$

$\therefore r=\frac{mv}{Bq}=\frac{\sqrt{2mK}}{Bq}$

$\therefore \frac{r_P}{r_e}=\frac{\frac{\sqrt{2m_{P}K}}{Bq}}{\frac{\sqrt{2m_{e}K}}{Bq}}=\sqrt{\frac{m_P}{m_e}}$

Given:

A proton and an electron have the same kinetic energy upon entering a region of constant magnetic field.

Formula used:

$\frac{r_P}{r_e}=\sqrt{\frac{m_P}{m_e}}$

Calculation:

$\frac{r_P}{r_e}=\sqrt{\frac{m_P}{m_e}}=\sqrt{\frac{1.67\times {10}^{-27}}{9.1\times {10}^{-31}}}=42.8$