An electron moves with a speed of 5.0 * 104 m/s perpendicular to a uniform magnetic field of magnitude 0.20 T. What is the magnitude of the magnetic force on the electron? (e = 1.60 * 10-19 C)

F = 1.6 * 10⁻¹⁵ N

Explanation:

Given: V = 5.0 * 10⁴ m/s, B = 0.20 T, q = 1.60 * 10⁻¹⁹ C and θ = 90°

To Find: F = ?

Solution:

we have a Formula for this question

F = q (V x B) ∴ x is used for cross product

F = q V B sin θ

F = 1.60 * 10⁻¹⁹ C * 5.0 * 10⁴ m/s * 0.20 T sin 90°

F = 1.6 * 10⁻¹⁵ N