# What is the force on a moving particle in a magnetic field?

1. Feb 24, 2015

### acdurbin953

1. The problem statement, all variables and given/known data
A charged particle of mass m = 6.9E-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 1.7T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.99 m, 0) and leaves the region at (x,y) = 0, 0.99 m a time t = 402 μs after it entered the region.

What is Fx, the x-component of the force on the particle at a time t1 = 134 μs after it entered the region containing the magnetic field.

2. Relevant equations
F=QvB
d=vt

3. The attempt at a solution
Solved for the velocity and got v=3855.7 m/s.
Solved for charge and got -1.58E-4
Then using d=vt, I solved for how far along the circular motion path the particle would be at t1, and got 0.516. Then I used θ=s/r=0.52/0.99 to get the angle in radians (0.52) and then converted to degrees and got 29.8° (I used an online converter just to be sure). From there I used F=QvBsin(θ). The answer was wrong, and so I plugged in that answer into the next question asking for Fy and it was wrong also.

Last edited: Feb 24, 2015
2. Feb 24, 2015

### BiGyElLoWhAt

how did you solve for velocity?
How did you solve for the charge?

3. Feb 24, 2015

### acdurbin953

For velocity: d=πr/2=1.55 then d/t=v. 1.55/402μS=3855.7 m/s. That part was marked correct.

For charge I used Q=vm/rB by relating F=QvB and F=ma. Then used right hand rule to determine that the charge was negative. That part was marked correct also.

4. Feb 24, 2015

### BiGyElLoWhAt

Hmmm... so you're looking for the force components?
One thing you need to realize is this: s/r = theta gives you the angle of the subtended arc. That is, the fraction of the circle that you traversed. The theta in your f=qvbsin(theta) is the angle between your velocity and the field. It comes from the cross product (not sure if that's in your scope or not).
This gives you the magnitude of the force, the direction of the force is perpendicular to the velocity and the field simultaneously, always.

5. Feb 24, 2015

### acdurbin953

Yes, looking for force components.
Oh, I didn't make that connection. So the angle would just be 90° then if it's always perpendicular?

6. Feb 24, 2015

### BiGyElLoWhAt

The velocity and field are not necessarily perpendicular, but the resultant force is always perpendicular to both. (with 2 lines, you can always lay a piece of paper down so that both lines are on the paper), the force is then either into or out of the paper.