Question

Chapter 21, Problem 26 amo e provo es background pertinent to this problem An electron has a kinetic energy o 2.0 x 10·7·r mo es en , orcuda path that is pepe d dr to , unterm mor te field d magnade oncettua 9.0 x 103 T. Determine the radius of the path. Number the tolerance is +/-2% Click if you would like to Show Work for this questions Units m Open Show Work By accessing this Question Assistance, you will lean whle you eam points based on the Pount Potentsal Pokcy set by your

Answer 1

The velocity of the electron is

$v=\sqrt{2 E/m}=6.6263\times 10^6$

where E is the kinetic energy and m is the mass of electron

$m=9.11\times 10^{-31}kg$

The Lorentz force on the electron balances the centripetal force on the electron moving in circular path

$\frac{mv^2}{r}=q vB$

where q is the charge of the electron and B is the given magnetic field

$q=1.602\times 10^{-19}C$

Thus the radius of the path is

$r=\frac{mv}{qB}=0.4187\,\,m$