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!

Expression for the time average power dissipated in a resistance R

  1. Jul 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A square loop with side-length a is positioned at the centre of a long thin solenoid, which has
    radius r (with r > a), length l and N turns. The plane of the loop is perpendicular to the
    5axis of the solenoid. A current I = I0 sin ωt flows through the loop. Derive an expression for
    the time-averaged power dissipated in a resistance R connected between the terminals of the
    solenoid. You may assume that R is much greater than the resistance of the solenoid and
    that the self-inductance of the solenoid is negligible.


    2. Relevant equations


    P = J E d (this is meant to be the integral of the total volume of the shape in question)
    P = I^2 * R

    3. The attempt at a solution

    I understand how the power dissipated is calculated over a macroscopic object. But how does this change when it is the referring to a solenoid and the terminals?
     
  2. jcsd
  3. Jul 20, 2011 #2

    Philip Wood

    User Avatar
    Gold Member

    The power is dissipated in R. Neglect any dissipated in the solenoid, because its resistance is negligible compared to that of R. [Not that there would be any problem with calculating the power in the solenoid if we had to - it's only a piece of wire, but wound into a coil !]

    But first you need to find the emf induced in the solenoid. Look up 'mutual inductance', or work from Faraday's Law.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Expression for the time average power dissipated in a resistance R
Loading...