## anonymous 3 years ago What is the magnetic force acting on an electron if its speed is 3.0 × 106 meters/second and the direction is perpendicular to a magnetic field of 0.020 teslas? The value of q = -1.6 × 10-19 coulombs. A. F = 0 newtons B. F = -6.0 × 10-15 newtons C. F = -9.6 × 10-15 newtons D. F = -3.0 × 10-16 newtons E. F = -3.2 × 10-21 newtons

1. anonymous

C

2. ujjwal

The magnitude of force is given by $F=q(v\times B)$