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

Answer 1

For plato users: Option C.

Step-by-step explanation:

Answer 2

`9.6\cdot 10^(-15) N`

Step-by-step explanation:

The magnetic force acting on a charged particle moving perpendicularly to the field is given by:

`F=qvB`

where

q is the charge

v is the speed of the particle

B is the magnetic field strength

In this problem, we have:

`q=-1.6\cdot 10^(-19) C` is the charge

`v=3.0\cdot 10^6 m/s` is the speed

`B=0.020 T`

so, the magnetic force is

`F=(-1.6\cdot 10^(-19) C)(3.0\cdot 10^6 m/s)(0.020 T)=9.6\cdot 10^(-15) N`