Question

A beam of electrons moving with a uniform speed of $4\times {10}^7{ms}^{-1}$ is projected normal to the uniform magnetic field where $B={10}^{-3}Wb/m^2$. What is path of the beam in magnetic field?

Answer 1

Given,
$V=4\times {10}^7{ms}^{-1}$
$B=1\times {10}^{-3}wb/m^2$
Let the path of the electron be $r$
Since, the electrons are released normally to the magnetic field, the electrons, travel in a circular path.
$\therefore BeV\frac{m{v}^2}{r}$ (or) $r=\frac{mv}{Be}$
$r=\frac{9.1\times {10}^{-31}\times 4\times {10}^7}{1\times {10}^{-3}\times 1.6\times {10}^{-19}}$
$r=0.2275m$