What is the wire's linear charge density from a proton?

[Bigworldjust]

Homework Statement

A proton orbits a long charged wire, making 1.30*10^6 revolutions per second. The radius of the orbit is 1.20cm.

What is the wire's linear charge density?

Homework Equations

- q E = m w^2 r
- 9*10^9 [2 λ /r] q = m w^2 r

The Attempt at a Solution

λ = linear charge dens
w = 2pi/period = 2pi/T
w = d theta/dt = 1.4* 2pi /1

I'm confused on how to solve for λ here though. Can anyone help me out?

 

[frogjg2003]

You said you know:
F=m v^2/r=m r ω^2=mr(2π/t)^2
F=q E= q 2 k λ/r

You correctly said that you have to set them equal to each other
2qkλ/r=4π^2mr/t^2
You were given r and t. q, k, and m are known physical constants. The only unknown variable is λ. Just solve for it and plug in numbers.

 

[Bigworldjust]

You said you know:
F=m v^2/r=m r ω^2=mr(2π/t)^2
F=q E= q 2 k λ/r

You correctly said that you have to set them equal to each other
2qkλ/r=4π^2mr/t^2
You were given r and t. q, k, and m are known physical constants. The only unknown variable is λ. Just solve for it and plug in numbers.

Ah, okay, so I have it set up like this:

(2*1.6*10^-19*9*10^9*λ)/(.012)=(4π^2*1.67*10^-27*.012)/(1.30*10^6)^2

I got 1.95056*10^-33 nC/m but that seems to be wrong, anywhere I went wrong?

 

[frogjg2003]

Don't plug in numbers until the final answer. Solve for λ, then plug in numbers. Also, just because it's small doesn't mean it's wrong.

 

[Bigworldjust]

Don't plug in numbers until the final answer. Solve for λ, then plug in numbers. Also, just because it's small doesn't mean it's wrong.

Alright so I got λ = 2π^2mr^2/kqt^2. I plugged in the numbers and got the same answer as before though, lol.

 

[frogjg2003]

Ok, you got the same answer. Why did you think it was wrong?

 

[Bigworldjust]

Ok, you got the same answer. Why did you think it was wrong?

I submitted it through Mastering Physics and it said it was wrong, lol. I'm not sure if its the units because the answer is suppose to be in nC/m, but I am pretty sure it cancels out accordingly. Hmm..

 

[frogjg2003]

I see the problem. You weren't given the period, you were given the rotation speed. "1.30*10^6 revolutions per second"

 

[Bigworldjust]

I see the problem. You weren't given the period, you were given the rotation speed. "1.30*10^6 revolutions per second"

Oh, so I have to convert the rotation speed to its period? So it would be 1/1300000?

 

[frogjg2003]

Or you could use the frequency 1.6*10^6 instead of 1/period. Either way, you're multiplying by (1.3*10^6)^2 instead of dividing.

 

[Bigworldjust]

Or you could use the frequency 1.6*10^6 instead of 1/period. Either way, you're multiplying by (1.3*10^6)^2 instead of dividing.

Hmm I got 5.57*10^-33 and that still seems to be wrong, lol.

 

[frogjg2003]

Let's go back to the equation that you found:
\lambda = \frac{2\pi^2mr^2}{kqt^2}
Replace t by 1/f to get
\lambda = \frac{2\pi^2mr^2f^2}{kq}.
Now, what you have been plugging in as t (1.6*10^6s) isn't t, it's the frequency f (1.6*10^6 Hz). Also, make sure that when you're plugging in units, you're using only base units. Only m should be in kg, r should be in m, f should be in Hz, k should be in N*m^2/C^2. Your answer should be in C/m.

 

[Bigworldjust]

Let's go back to the equation that you found:
\lambda = \frac{2\pi^2mr^2}{kqt^2}
Replace t by 1/f to get
\lambda = \frac{2\pi^2mr^2f^2}{kq}.
Now, what you have been plugging in as t (1.6*10^6s) isn't t, it's the frequency f (1.6*10^6 Hz). Also, make sure that when you're plugging in units, you're using only base units. Only m should be in kg, r should be in m, f should be in Hz, k should be in N*m^2/C^2. Your answer should be in C/m.

Alright I got 8.439 nC/m and that is still wrong, lol. Damn, I don't know where I'm going wrong. I'm plugging in everything:

I got:
m = 1.67*10^-27 kg
r = .012m
f = 1.6*10^6 Hz
k = 9*10^9 N*m^2/C^2
q = 1.6*10^-19 C

 

[frogjg2003]

That's because I had a typo and I wrote 1.6 MHz instead of 1.3 MHz.

 

[Bigworldjust]

That's because I had a typo and I wrote 1.6 MHz instead of 1.3 MHz.

Ah, I finally got it. Thanks a lot for all of your help, I really appreciate it.

 

[frogjg2003]

You're welcome.