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: Time Dependent Current in a Wire

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data

    Problem is attached

    2. Relevant equations

    A formula sheet is also attached

    3. The attempt at a solution

    flux=[itex]\int[/itex]B dA from .31m to .82
    B=u I(enclosed)/2(pi)(d)
    dA=dx L
    so ∫ (u)(I)(L)dx / 2(pi)(x) from .31m to .82m remember x=d in the pic. My answer is 3.9687e-7 and its wrong

    integral came out to be (u)(I)(L)/2pi (ln(.82) - ln(.31))

    Attached Files:

  2. jcsd
  3. Mar 23, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    That "dA" indicates an area - so why is this not a double integral?

    I dont think you've used the formula correctly or you got confused between two different uses.
    $$B=\frac{\mu_0 I}{2\pi d}$$... would be the magnetic field strength a distance d from a long straight wire.

    The magnetic flux through area dA at position (x,y) would be ##d\Phi = B(x,y,t)\;dA##
    You'd have to integrate over the whole LxW area to find the total flux.
  4. Mar 23, 2014 #3
    I just got the right answer by multiplying by the width, not length. so dA= dx w, since dx is the length thats changing times the width which gives area. What do you mean by double integral? Doesn't that give volume?
  5. Mar 23, 2014 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Well done.

    Triple integrals give volume. dV=dx.dy.dz,

    dA=W.dx is only true when the thing you are integrating does not vary with y
    - which is what you have.
    Last edited: Mar 23, 2014
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted