# Magnetic Force on an a Particle

• mlsohani
In summary, a particle with a charge of 35.7 uC moving at a speed of 61.1 m/s in the positive x direction experiences a magnetic force of 2.13 e-3 N in the direction relative to the positive z-axis, determined by the direction of the velocity and magnetic field.

## Homework Statement

A particle with a charge of 35.7 uC moves with a speed of 61.1 m/s in the positive x direction. The magnetic field in this region of space has a component of 0.527 T in the positive y direction, and a component of 0.820 T in the positive z direction. What is the magnitude of the magnetic force on the particle?

magnetic F=[q]vB
electric F=qE

## The Attempt at a Solution

I really do not know what I am doing and need help
Here is an attempt:

F=qvB
(3.57 e-6 C)(61.1 m/s)(0.527 T)= 1.1495 e-4 N?

It is almost correct, however you took the wrong value for B. What is the magnitude of B?

ok! SO I figured it out! You use pathagoreum Theorem and you end up getting that
B= 0.9756 T
I pluged that in and got the answert to be: 2.13 e-3 N

NOW though I don't understand what they are asking for when they ask:
What is the direction of the magnetic force on the particle relative to the positive z-axis?
is it just the + or - infront of the value I received for the force? Or is there a completely different asnwer?

Do you know how to determine the direction of the magnetic force when given the direction of v and B? Just draw a picture so you can see how the vector is placed with respect to the z-axis.