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

  • #1

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?
 

Answers and Replies

  • #2
265
0
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.
 
  • #3
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?
 
Last edited:
  • #4
265
0
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.
 
  • #5
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.
 
  • #6
265
0
Ok, you got the same answer. Why did you think it was wrong?
 
  • #7
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..
 
  • #8
265
0
I see the problem. You weren't given the period, you were given the rotation speed. "1.30*10^6 revolutions per second"
 
  • #9
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?
 
  • #10
265
0
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.
 
  • #11
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.
 
  • #12
265
0
Let's go back to the equation that you found:
[tex]\lambda = \frac{2\pi^2mr^2}{kqt^2}[/tex]
Replace t by 1/f to get
[tex]\lambda = \frac{2\pi^2mr^2f^2}{kq}.[/tex]
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.
 
  • #13
Let's go back to the equation that you found:
[tex]\lambda = \frac{2\pi^2mr^2}{kqt^2}[/tex]
Replace t by 1/f to get
[tex]\lambda = \frac{2\pi^2mr^2f^2}{kq}.[/tex]
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
 
  • #14
265
0
That's because I had a typo and I wrote 1.6 MHz instead of 1.3 MHz.
 
  • #15
That's because I had a typo and I wrote 1.6 MHz instead of 1.3 MHz.
Ah, I finally got it. Thanks alot for all of your help, I really appreciate it.
 
  • #16
265
0
You're welcome.
 

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

Replies
0
Views
1K
Replies
10
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
1
Views
8K
Replies
3
Views
2K
Replies
5
Views
9K
Replies
4
Views
5K
Top