# Acceleration of a Charged Particle

1. Jul 19, 2007

### rum2563

1. The problem statement, all variables and given/known data
A positively charged particle of mass 3.0 X 10 ^-12 kg and charge +6.4 X 10^-5 C is perpendicular to a 1.5 T magnetic field. If the particle has a speed of 4.8 X 10^3 m/s, the acceleration of the particle...?

2. Relevant equations
e = kg / r^2
qB = mv / r

3. The attempt at a solution

Given information:
m = 3.0 X 10 ^-12 kg
q = 6.4 X 10^-5 C
B = 1.5 T
v = 4.8 X 10^3 m/s

To find the acceleration, I started reading my book, and I found the equation:
q = mg / e
So if I rewrite this to find "g", it should be like this:
g = qe / m

BUT, I don't know e, so to find e (epsilon), I must first know the radius "r"
r = mv / qB
= (3.0 X 10 ^-12)(4.8 X 10^3) / (6.4 X 10^-5)(1.5)
= 1.5 X 10 ^-4 m

Now, I can find e:
e = kq / r^2
= (9 X 10^9)(6.4 X 10^-5) / (1.5 X 10^-4)^2
= 2.56 X 10^13 N/C

Now, I can find "g"

g = qe / m
= (6.4 X 10^-5)(2.56 X 10^13) / (3.0 X 10 ^-12)
= 5.46 X 10^20 m/s^2

But this is not the correct answer.

The correct answer should be one of the following:
3.4 X 10^-4
2.6 X 10^-20
1.5 X 10^11
1.5 X 10^-4
1.7 X 10^19

2. Jul 19, 2007

### G01

The relationships you are using are not correct in this case. The first one is the definition of an electric field (e stands for the electric field). The second is incorrect. "qB" is not a useful quantity. I think the relationship you were thinking of was qE=mv^2/r, which describes circular motion with the electric force as the centripetal force. Either way, it doesn't apply to this problem.

I think the best way to solve this would be to go back to fundamentals. Remember Newton's Second Law:

$$\Sigma F = ma$$

You want the find a, you know the mass, m, and the force here is the magnetic force. HINT: Do you know a relationship to find the magnetic force on a particle moving through a constant magnetic field?

I hope this helped. If you need more help, I'll answer as soon as I can.
Good Luck!

Last edited: Jul 19, 2007