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

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. 

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

rva.raghav · Answer

C

ujjwal · Answer

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