Answer
$B = (0.025~T)~\hat{k}$
Work Step by Step
When $v = 50.0~m/s$, the net force is zero.
Then 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.
In part (a), we found that $E = 1.25~N/C$
We can find the magnitude of the magnetic field:
$F_B = F_E$
$qvB = qE$
$B = \frac{E}{v}$
$B = \frac{1.25~N/C}{50.0~m/s}$
$B = 0.025~T$
The force on the electron due to the electric field is in the -y direction. Therefore, the force on the electron due to the magnetic field must be in the +y direction.
The velocity is in the +x direction, and the charge on the electron is negative. Therefore, by the right hand rule, the magnetic field is out of the page, which is the +z direction.
We can express $B$ in unit-vector notation:
$B = (0.025~T)~\hat{k}$
