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.
F = 0 newtons
F = -6.0 × 10-15 newtons
F = -9.6 × 10-15 newtons
F = -3.0 × 10-16 newtons
F = -3.2 × 10-21 newtons

Question

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.
F = 0 newtons
F = -6.0 × 10-15 newtons
F = -9.6 × 10-15 newtons
F = -3.0 × 10-16 newtons
F = -3.2 × 10-21 newtons

anonymous · Answer

$$F=q ( v \times B) $$

v and B are vectors and the x is cross product

Since v and B are perpendicular 

F=qvB 

Plug in the values to get the answer (including the negative sign of charge)