Answer
$B = (-2.67\times 10^{-4}~T)~\hat{k}$
Work Step by Step
We can use the kinetic energy after accelerating through the potential difference to find the electron's speed:
$K = \vert q \vert~V_1$
$\frac{1}{2}mv^2 = \vert q \vert~V_1$
$v^2 = \frac{2~\vert q \vert~V_1}{m}$
$v = \sqrt{\frac{2~\vert q \vert~V_1}{m}}$
$v = \sqrt{\frac{(2)(1.6\times 10^{-19})(1000~V)}{9.109\times 10^{-31}~kg}}$
$v = 1.87\times 10^7~m/s$
The force on the electron due to the magnetic field must be equal in magnitude to the force on the electron due to the electric field between the plates.
We can find the magnitude of the magnetic field:
$qvB = qE$
$B = \frac{E}{v}$
$B = \frac{E}{v}$
$B = \frac{V_2}{d~v}$
$B = \frac{100~V}{(0.020~m)(1.87\times 10^7~m/s)}$
$B = 2.67\times 10^{-4}~T$
Since the lower plate is at a lower potential, the force on the electron due to the electric field is directed upward. Therefore, the force on the electron due to the magnetic field must be directed downward.
Then, by the right hand rule, the magnetic field must be directed into the page, which is the -z direction.
We can express the magnetic field in unit-vector notation:
$B = (-2.67\times 10^{-4}~T)~\hat{k}$
