CUET Preparation Today
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? |
14 cm 28 cm 56 cm 112 cm |
28 cm |
The correct answer is Option (2) → 28 cm The radius of the path of an electron is - $r=\frac{mv}{qB}$ $=\frac{9.1×10^{-31}×3×10^7}{1.6×10^{-19}×6×10^{-4}}$ $≃28cm$ |
