Saved Bookmarks
| 1. |
What is the radius of the path of an electron (mass 9xx10^(-31)kg and charge 1.6xx10^(-19)C) moving at a speed of 3xx10^(7)m//s in a magnetic field of 6xx10^(-4)T perpendicular to it? What is its frequency ? Calculate its energy in keV. (1eV=1.6xx10^(-19)J) |
|
Answer» Solution :1. When a PARTICLE of mass m and charge q moves in a uniform magnetic field B with speed v radius of its circular path is GIVEN by, `r=(mvsintheta)/(Bq)` (where `theta` = angle between `vecvandvecB`) `thereforer=0.2812m=28.12cm` 2. We have, `v=romega` `thereforev=r(2pif)` `thereforef=v/(2pir)` = `(3xx10^(7))/((2)(3.14)(0.2812))` `thereforef=1.699xx10^(7)Hz~~17MHz` 3. Now, kinetic energy of electron is, `K=1/2mv^(2)` = `1/2xx9xx10^(-31)xx(3xx10^(7))^(2)` `thereforeK=4.05xx10^(-16)J` `thereforeK=(4.05xx10^(-16))/(1.6xx10^(-19))EV` `thereforeK=2.531xx10^(3)eV` |
|