An electron has a circular path of radius 0.01 m in a perpendicular magnetic induction $10^{-3} T$. The speed of the electron is nearly
Answer & explanation
Correct answer: option 2
$R = \frac{mv}{qB}$
$\Rightarrow v = \frac{qBR}{m} = \frac{ 1.6\times 10^{-19} \times 10^{-3} \times 0.01}{9.11\times 10^{-31}} = 1.76\times 10^6 m/s$