• Support PF! Buy your school textbooks, materials and every day products Here!

Simple Self-Inductance Problem

  • Thread starter ttiger2k7
  • Start date
  • #1
58
0
[SOLVED] Simple Self-Inductance Problem

1. Homework Statement

Find the inductance L of a long solenoid of length 39.0 cm, and radius 2.30 cm, which has 3.10E2 turns. (Express you answer in units of H for henrys).

2. Homework Equations

[tex]L=\frac{N\Phi_{B}}{i}[/tex]
[tex]\Phi_{B}=BA=\frac{\mu_{0}NiA}{2\pi*r}[/tex]

3. The Attempt at a Solution

Okay, so the way I am thinking to solve this problem is to first get the magnetic flux. But I have no idea how to find the current [tex]i[/tex] for a solenoid. Since I don't know how to find [tex]i[/tex], I simply plugged in the formula for flux into the formula for inductance to get:

[tex]L=\frac{N\Phi_{B}}{i}=\frac{\mu_{0}N^{2}A}{2\pi*r}[/tex]

That's where I get stuck. *If* I am in fact going about this the right way, do I just find area by using pi*r^2?
 

Answers and Replies

  • #2
360
21
Try

[tex]
L=\frac{\mu_{0}N^{2}A}{l}
[/tex]

Where A is cross sectional area and l is the length of the solenoid.

I think...
 
  • #3
58
0
Thanks for the quick response. I got it.
 

Related Threads for: Simple Self-Inductance Problem

  • Last Post
Replies
0
Views
889
  • Last Post
Replies
8
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
827
  • Last Post
Replies
1
Views
12K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
2K
Top