What is the radius of the path of an electron moving with velocity of $3 × 10^7 m/s$ in a magnetic field of $6 × 10^{-4} T$ perpendicular to it?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 28 cm
The radius of the path of an electron is -
$r=\frac{mv}{qB}$
$=\frac{9.1×10^{-31}×3×10^7}{1.6×10^{-19}×6×10^{-4}}$
$≃28cm$