Answer
$1.51\times10^{-5}m$
Work Step by Step
This is discussed in example 20-6. The magnetic force causes circular motion. The electron will move in a clockwise circular path. The radius of the motion is calculated in the book’s example.
$$r=\frac{mv}{qB}$$
$$=\frac{(9.11\times10^{-31}kg)( 1.70\times10^6m/s)}{ (1.60\times10^{-19}C) (0.640T)}$$
$$=1.51\times10^{-5}m$$
If the magnetic field exists only above the horizontal plane, the electron makes a half-circle and exits the field going downward.
