## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $v = 6.0\times 10^7~m/s$ (b) $r = 17.1~cm$
(a) We can write an expression for the force on the electron from the electric field: $F_E = E~q = \frac{\Delta V~q}{d}$ We can write an expression for the force on the electron from the magnetic field: $F_B = q~v~B$ If the electron is not deflected, then $F_E$ and $F_B$ must have the same magnitude while directed in opposite directions. We can find the electron's speed $v$: $F_B = F_E$ $q~v~B = \frac{\Delta V~q}{d}$ $v = \frac{\Delta V}{d~B}$ $v = \frac{600~V}{(5.0\times 10^{-3}~m)(2.0\times 10^{-3}~T)}$ $v = 6.0\times 10^7~m/s$ (b) We can find the electron's radius of curvature: $F_B = \frac{mv^2}{r}$ $q~v~B = \frac{mv^2}{r}$ $r = \frac{mv}{q~B}$ $r = \frac{(9.109\times 10^{-31}~kg)(6.0\times 10^7~m/s)}{(1.6\times 10^{-19}~C)~(2.0\times 10^{-3}~T)}$ $r = 0.171~m$ $r = 17.1~cm$