Magnetic Force on a Moving Charge

[GodxIsInTheTV]

Homework Statement

A magnet produces a 0.5 T field between its poles, directed to the east. A dust particle with charge q = -8 times 10-18 C is moving straight down at 0.6 cm/s in this field. What is the magnitude and direction of the magnetic force on the dust particle?

Homework Equations

F= qv X b

The Attempt at a Solution

Tried converting .6 cm/s into m/s and solving since the sin (90) would still be 1, but it keeps telling me I am wrong.

 

[Defennder]

Hi, you need to be more explicit as to what you have tried which you said did not correspond to the correct answer. What is your answer? It's a vector isn't it?

 

[baba89]

I would approach this problem by first understanding the principles of magnetic force on a moving charge. The magnetic force on a moving charge is given by the equation F=qv x B, where q is the charge of the particle, v is its velocity, and B is the magnetic field. In this case, the charge is given as -8 x 10^-18 C and the magnetic field is 0.5 T directed to the east.

To find the magnitude of the force, we can substitute the values into the equation F=qv x B. Since the particle is moving straight down, its velocity is only in the y-direction and can be written as v=0.006 m/s. The magnetic field, being directed to the east, can be written as B=0.5 T in the x-direction. Thus, the magnitude of the force can be calculated as F= (-8 x 10^-18 C)(0.006 m/s)(0.5 T) = -2.4 x 10^-22 N.

To determine the direction of the force, we can use the right-hand rule. If we point our right thumb in the direction of the particle's velocity (downward), and our fingers in the direction of the magnetic field (east), our palm will point in the direction of the force (south). Therefore, the direction of the magnetic force on the dust particle is south.

In conclusion, the magnitude of the magnetic force on the dust particle is -2.4 x 10^-22 N and it is directed towards the south.