An electron enters perpendicularly into a uniform magnetic field which has a flux density of 0.12 T This results in a magnetic force of 9.6 x 10\(^{-2}\) N on the electron. Calculate the speed of the electron as it enters the magnetic field. (e =1.6 x \(10^{19}\) C)
Explanation
F = evB sin \(\theta\)
V = \(\frac{F}{evB sin \theta} = \frac{9.6 \times 10^{12}}{1.6 \times 10^{-19} \times 0.12 sin 90^o}\)
= \(\frac{9.6 \times 10^{-12}}{1.6 x 10^{-19} \times 0.12 \times 1}\)
= 5.0 x 10\(^8 ms{-1}\)