Magnetic Force — Easy JEE physics MCQ
An electron moving with velocity 2 × 10⁶ m/s enters a uniform magnetic field of 0.5 T perpendicular to its direction of motion. The magnitude of the force experienced by the electron is: (e = 1.6 × 10⁻¹⁹ C)
- A. 1.6 × 10⁻¹³ N
- B. 3.2 × 10⁻¹³ N
- C. 1.6 × 10⁻¹² N
- D. 8 × 10⁻¹⁴ N
Solution
Using F = qvB sin θ, where θ = 90°: F = (1.6 × 10⁻¹⁹)(2 × 10⁶)(0.5) = 1.6 × 10⁻¹³ N.