Question

An electron experiences a magnetic force of magnitude 4.60 X 10^-15 N when moving at an angle of 60.0 with respect to a magnetic field of magnitude 3.50 X 10^-3 T. Find the speed of electron

Answer 1

The Lorentz force is

$F = q [ v \times B]$

The modulus of this force is

$F = qvB \sin \alpha$ where $\alpha$ is the angle between velocity and magnetic field.

The speed of the electron is

$\displaystyle v = \frac{F}{qB\sin \alpha} = \frac{4.6 \cdot 10^{-15}}{1.6 \cdot 10^{-19} \cdot3.5 \cdot 10^{-3} \cdot 0.866}= 0.9485 \cdot 10^7 = 9.485 \cdot 10^6 \; m/s.$

Answer: $9.485 \cdot 10^6$ m/s.