1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: [Magnetism] Tricky task on self-inductance of solenoid

  1. Mar 1, 2014 #1
    1. The problem statement, all variables and given/known data
    A Solenoid has N turns of Copper wire (Cross-section area of the wire is 10-6 m2) The Length of the solenoid is 25*10-2 m and its Resistance is 0.5 Ω. What's the Self-Inductance of the solenoid?

    2. Relevant equations
    Given answer: 5.5*10-5 H

    Please help me with the solution.
    3. The attempt at a solution
    How do you relate Resistance with Self-inductance to eliminate the need for (or find) the number of turns? All I could find by using R was the length of the whole wire(an unrolled solenoid). But this gives me nothing on the number of turns cause the diameter of the solenoid is not given.
  2. jcsd
  3. Mar 1, 2014 #2


    User Avatar

    Staff: Mentor

    Hi wetback! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    I reckon you should assume the turns in the winding are single-layer, tightly-wound and close-packed. This allows you to determine the diameter of the coil. :smile:

    What formula will you then use to calculate inductance?
    Last edited by a moderator: May 6, 2017
  4. Mar 1, 2014 #3
    Hey! As it often is, when you are desperate enough to look for help in a forum, you miraculously find the solution yourself! Eureka! And as always, it turns out to be pretty simple. Just some algebra.

    If anyone ever stumbles upon this post looking for help, I'll explain what I did:

    I used the inductance equation for a solenoid L=([itex]\mu[/itex]N2A)/l
    where [itex]\mu[/itex] - permeability constant(4[itex]\pi[/itex]*10^-7), N - number of turns, A - the cross-section area of the solenoid cylinder shape and l - the length of the solenoid
    I also used the resistance equation for a conductor(this case the copper wire): R=([itex]\rho[/itex]*l1)/A1 where [itex]\rho[/itex]-the resistivity of copper(1.7*10-8), l1 - the length of the wire, A1- the cross-section area of the wire.
    I expressed l1: l1=R*A1/[itex]\rho[/itex], then expressed N using l1: N=l1/(2[itex]\pi[/itex]*r) where r - the radius of the solenoid cylinder shape. Then r=l1/(2[itex]\pi[/itex]N) Now I expressed A: A=[itex]\pi[/itex]*r2. When you insert this expression in the L equation, N disappears and we are left with just the given numbers and L really is 5.5*10-5 H.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted