#### Answer

(a) $v = 6.0\times 10^7~m/s$
(b) $r = 17.1~cm$

#### Work Step by Step

(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:
$F_B = \frac{mv^2}{r}$
$q~v~B = \frac{mv^2}{r}$
$r = \frac{mv}{q~B}$
$r = \frac{(9.109\times 10^{-31}~kg)(6.0\times 10^7~m/s)}{(1.6\times 10^{-19}~C)~(2.0\times 10^{-3}~T)}$
$r = 0.171~m$
$r = 17.1~cm$