# Homework Help: Magnetic force

1. Oct 22, 2012

### wuffle

Once again I can't seem to do this simple problem, not sure why.

hate mechanics :(

1. The problem statement, all variables and given/known data

Charged particle enters a region containing constant magnetic field and leaves it after 708 micro seconds, what is Fx, the x component of the force after being 236 micro seconds long in the constant magnetic field
Also it travels a quarter circle(not sure if the right word, basically a half of a semicircle)

we're given:

B=1.2 T
R=0.95 m
m=5.7*10^-8 kg

2. Relevant equations

ma=qvb

3. The attempt at a solution

I calculated the constant velocity when the charge is in the magnetic field, the only problem i have is calculating Vx and Vy after 236μsec, particularly i cant find the angle, one way i tried solving it is realizing that 236/708 is 1/3, so distance that particle has traveled is

(2piR/4)/3 which i found to be 0.49742...then idk, i thought maybe the angle would be 60 deg or 30 deg since it has travelled 1/3 of the way so i tried finding Vx=V*cos(30) or Vx=V*cos(60) but it didn't work out, i think i know that on halfway the angle is 45(Vx=Vy), but that's it, basically the equation i got is

Fx=m*(V*cos(theta))^2/R and I can't find theta :(

2. Oct 22, 2012

### Simon Bridge

When the particle enters the B field, it has circular motion.
So you can use the kinematics of circular motion to find the angle that rotates through by the time it leaves the field.

3. Oct 23, 2012

### Basic_Physics

The force stays perpendicular to the motion so the speed stays constant. A quarter circle is 90 degrees so a third of it is 30 degrees.

4. Oct 25, 2012

### Simon Bridge

Though you are encouraged to think of angles in radians.

The size of an angle is the distance around the circumference of a unit circle that is inside the angle.

The total distance around the whole circle is $2\pi$.
A half circle is therefore $\pi$ and a quarter circle is $\pi/2$.
Use these and a lot of physics equations will make more sense - like the arc-length inside angle $\theta$ at distance $r$ is $s=r\theta$. If you used degrees you have to say that $s=\pi r/180$ .