Question

At a certain point on Earth, the horizontal component of Earth’s magnetic field is 2.5 × 10-5 T due to north. An electron moves eastward with just the right speed, so the magnetic force on it balances its weight. Find the speed of the electron.

Answer 1

The balance of force requires

$$mg=qvB$$

$$v=\frac{mg}{qB}\\ =\frac{9.1*10^{-31}*9.8}{1.6*10^{-19}*2.5*10^{-5}}=2.2*10^{-6}\:\rm m/s$$